Amines - Class 12 Chemistry - Chapter 9 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Amines | NCERT | Chemistry | Class 12
$\mathrm{Ph}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3} \xrightarrow{\mathrm{Hg}^{2+} / \mathrm{H}^{+}} \mathrm{A}. \mathrm{A}$ is a) b) c) d) $\mathrm{Ph}$
A a
B $\mathrm{b}$
C $\mathrm{c}$
D $\mathrm{d}$
The reaction given involves the alkyne $\mathrm{Ph}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}$ treated with $\mathrm{Hg}^{2+}$ and $\mathrm{H}^{+}$. This setup typically indicates a reaction known as Mercury(II)-catalyzed hydration of alkynes, where the alkyne is converted into a ketone.
Following this reaction:
$$ \mathrm{Ph}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3} \xrightarrow{\mathrm{Hg}^{2+} / \mathrm{H}^{+}} \mathrm{Ph}-\mathrm{C}(=\mathrm{O})-\mathrm{CH}_3 $$
This product is a ketone with the ketone group ($\mathrm{C}=\mathrm{O}$) directly attached to the benzene ring and the methyl group on the other side of the carbonyl carbon.
Given the options:
A) a
The correct answer must be option A), as it correctly identifies the transformation and the product of the reaction as a ketone.
If $\mathrm{pK}{\mathrm{a}}\left(\mathrm{NH}{4}^{+}\right)=9.2$, then the amount of $\left(\mathrm{NH}{4}\right){2} \mathrm{SO}{4}$ to be added in 500 mL of 0.1 M $\mathrm{NH}{4} \mathrm{OH}$ for the preparation of a buffer of pH=8.5 would be: (Given: $\log 5=0.7$)
A) 0.175 mol B) 0.125 mol C) 0.150 mol D) 0.20 mol
To prepare a buffer with a pH of 8.5 using $\mathrm{NH}{4}\mathrm{OH}$ and $\left(\mathrm{NH}{4}\right){2}\mathrm{SO}{4}$, we use the Henderson-Hasselbalch equation for bases, given that $\mathrm{pK}{\mathrm{a}}(\mathrm{NH}{4}^+) = 9.2$.
First, we convert $\mathrm{pH}$ to $\mathrm{pOH}$, given that $\mathrm{pH} + \mathrm{pOH} = 14$: $$ \mathrm{pOH} = 14 - 8.5 = 5.5 $$ Then considering $\mathrm{pK}{\mathrm{b}} = 14 - \mathrm{pK}{\mathrm{a}} = 14 - 9.2 = 4.8$, the equation becomes: $$ 5.5 = 4.8 + \log \frac{n_{\text{salt}}}{0.05} $$ Where $0.05$ mol is the initial concentration of $\mathrm{NH}{4}\mathrm{OH}$ times the volume in liters (0.1 M $\times$ 0.5 L). Simplifying the equation: $$ \log \frac{n{\text{salt}}}{0.05} = 5.5 - 4.8 = 0.7 $$ With given $\log 5 = 0.7$, we find: $$ \frac{n_{\text{salt}}}{0.05} = 5 \quad \Rightarrow \quad n_{\text{salt}} = 5 \times 0.05 = 0.25 \text{ mol of } \mathrm{NH}{4}^+ $$ Each molecule of $\left(\mathrm{NH}{4}\right){2}\mathrm{SO}{4}$ provides two $\mathrm{NH}{4}^+$ ions, thus the amount needed of $\left(\mathrm{NH}{4}\right){2}\mathrm{SO}{4}$ is: $$ \frac{0.25}{2} = 0.125 \text{ mol} $$ Therefore, the correct amount of $\left(\mathrm{NH}{4}\right){2}\mathrm{SO}_{4}$ to be added is 0.125 mol, corresponding to option B.
Baby talcum powder contains
A. Benzoin, glyceryl diacetate
B. Zinc acetate, glyceryl diacetate
C. Zinc stearate, boric acid
D. Zinc stearate, cinnamic ester
The correct answer is C. Zinc stearate, boric acid.
Zinc stearate acts as a lubricant, giving a smooth texture to the powder, and boric acid is incorporated as an antiseptic to prevent minor skin infections.
Q74. Amit is older than Babloo but younger than Chintoo. Dinesh is younger than Esha but older than Amit. If Chintoo is younger than Dinesh, then who is the oldest of all?
A) Amit
B) Chintoo
C) Dinesh
D) Esha
Solution:
The problem describes a series of age comparisons among five individuals: Amit, Babloo, Chintoo, Dinesh, and Esha. Let's analyze the given information to determine who is the oldest:
-
Amit is older than Babloo but younger than Chintoo:
- Babloo < Amit < Chintoo
-
Dinesh is younger than Esha but older than Amit:
- Amit < Dinesh < Esha
-
Chintoo is younger than Dinesh:
- Chintoo < Dinesh
Combining the comparisons:
- We know Babloo < Amit and Amit < Dinesh, so Babloo < Amit < Dinesh.
- From Amit < Dinesh and Chintoo < Dinesh, and since Amit is younger than Chintoo, we can say Babloo < Amit < Chintoo < Dinesh.
- Finally, given that Dinesh < Esha, we conclude Babloo < Amit < Chintoo < Dinesh < Esha.
Esha is therefore the oldest of all. The correct answer is:
D) Esha
When excess of $\mathrm{Cl}_{2}$ reacts with ammonia, the products formed are:
A) $\mathrm{NH}{3}$ and $\mathrm{N}{2}$
B) $\mathrm{NCl}_{3}$ and $\mathrm{HCl}$
C) $\mathrm{NCl}{3}$ and $\mathrm{N}{2}$
D) $\mathrm{NH}{4}\mathrm{Cl}$ and $\mathrm{N}{2}$
Solution
The correct answer is Option B: $\mathrm{NCl}_3$ and $\mathrm{HCl}$.
When ammonia ($\mathrm{NH}_3$) reacts with an excess amount of chlorine ($\mathrm{Cl}_2$), the products formed are nitrogen trichloride ($\mathrm{NCl}_3$) and hydrogen chloride ($\mathrm{HCl}$). The chemical equation representing this reaction is:
$$ \mathrm{NH}_3 + 3\mathrm{Cl}_2 \rightarrow \mathrm{NCl}_3 + 3\mathrm{HCl} $$
This shows that each molecule of ammonia reacts with three molecules of chlorine to produce one molecule of nitrogen trichloride and three molecules of hydrogen chloride.
When $\mathrm{Na}$ is heated in dry ammonia, the compound formed is
A) Sodium amide
B) Sodium azide
C) Sodium nitride
D) Sodium hydride
When sodium ($\mathrm{Na}$) is heated in dry ammonia ($\mathrm{NH}_3$), the reaction results in the formation of sodium amide ($\mathrm{NaNH}_2$). This can be represented by the chemical equation: $$ \mathrm{Na} + \mathrm{NH}_3 \rightarrow \mathrm{NaNH}_2 $$ Therefore, the correct answer is A) Sodium amide.
The oxidation number of $\mathrm{N}$ in $\mathrm{N}{2} \mathrm{H}{5}^{+}$
A) -3 B) -2 C) -1 D) 2
Solution
Let's find the oxidation number of $\mathrm{N}$ in the ion $\mathrm{N}_2\mathrm{H}_5^+$.
We start by setting up an equation for the sum of the oxidation numbers in the ion, which must equal the charge on the ion:
Assume the oxidation number of $\mathrm{N}$ is $x$. Since there are two nitrogen atoms, the total oxidation contribution is $2x$.
For hydrogen, the oxidation number is typically +1. Therefore, with five hydrogen atoms, the total oxidation contribution from hydrogen is $5 \times (+1) = +5$.
Considering the overall charge of the ion is +1, we set up the equation: $$ 2x + 5 = +1 $$ To find $x$, isolate it: $$ 2x = +1 - 5 = -4 $$ $$ x = \frac{-4}{2} = -2 $$
Thus, the oxidation number of $\mathrm{N}$ in $\mathrm{N}_2\mathrm{H}_5^+$ is -2.
The correct answer is $\mathbf{B}-2$.
"The IUPAC name of:
A) o-bromoaniline
B) 4-bromoaniline
C) Aniline
D) 3-bromoaniline"
The IUPAC name for the given compound is 4-bromoaniline, so the correct option is B.
The lids of three containers containing ammonia gas, rose water, and chalk powder respectively were opened. The smell of will be sensed first.
A. ammonia gas
B. rose water
C. chalk powder
The correct answer is A. ammonia gas.
Ammonia gas molecules spread and diffuse the fastest among the options provided. Therefore, the order in which the smells will likely be sensed is: Ammonia gas > Rose water > Chalk powder.
A mole of $\mathrm{N}_{2}\mathrm{H}_{4}$ loses $10~\text{mol}$ of electrons to form a new compound $Y$. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in $Y$? (There is no change in the oxidation number of hydrogen.)
(A) -1
(B) -3
(C) +3
(D) +5
The correct answer is: Option (C) +3
First, let's determine the initial oxidation state of nitrogen in $\mathrm{N}_{2}\mathrm{H}_{4}$. Each $\mathrm{H}$ atom has an oxidation state of +1. Since there are 4 hydrogen atoms contributing a total of +4, the overall molecule is neutral.
The sum of the oxidation states in the molecule must be zero: $$ 2x + 4(+1) = 0 $$ where $x$ is the oxidation state of nitrogen. Solving for $x$, we get: $$ 2x + 4 = 0 \ 2x = -4 \ x = -2 $$ Therefore, each nitrogen atom initially has an oxidation state of -2.
Next, the problem states that a mole of $\mathrm{N}_{2}\mathrm{H}_{4}$ loses $10$ moles of electrons. The loss of electrons indicates an increase in oxidation state. The change in number of moles of electrons lost per nitrogen atom will be $\frac{10}{2} = 5$ moles per nitrogen atom. The new oxidation state of nitrogen, $y$, can be expressed and calculated as follows: $$ y = -2 + 5 = +3 $$
Hence, the oxidation state of nitrogen in the new compound $Y$ is +3.
Manju had a chain made of pearls which was gifted by her uncle. However, she came to know that it wasn't made of original pearls. From which material do you think the chain was made from?
A Melamine
B Polystyrene
C Bakelite
D PET
Solution
The correct option is C Bakelite
Bakelite is a type of thermosetting plastic. It's important to note that Bakelite can emulate the appearance and texture of natural materials like pearl and jade after it has been molded. This similarity, combined with its relatively low cost, makes Bakelite a popular alternative for expensive gemstones in various applications, including jewelry making. Hence, it's plausible that Manju's chain was made from Bakelite.
Among $\mathrm{CaH}{2}$, $\mathrm{NH}{3}$, $\mathrm{NaH}$, and $\mathrm{B}{2}\mathrm{H}{6}$ which are covalent hydrides?
A) $\mathrm{NH}{3}$ and $\mathrm{B}{2}\mathrm{H}_{6}$
B) $\mathrm{NaH}$ and $\mathrm{CaH}_{2}$
C) $\mathrm{NaH}$ and $\mathrm{NH}_{3}$
D) $\mathrm{CaH}{2}$ and $\mathrm{B}{2}\mathrm{H}_{6}$
Solution
The correct answer is Option A: $\mathrm{NH}_3$ and $\mathrm{B}_2\mathrm{H}_6$.
Reasoning:
-
$\mathrm{CaH}_2$ and $\mathrm{NaH}$ are classified as saline hydrides. This classification is due to the significant difference in electronegativity between the metal (Ca or Na) and hydrogen. These form ionic bonds.
-
On the other hand, $\mathrm{NH}_3$ (Ammonia) and $\mathrm{B}_2\mathrm{H}_6$ (Diborane) are covalent hydrides. They are characterized by the sharing of electrons between the non-metal atoms and hydrogen, thus forming covalent bonds.
Thus, for covalent hydrides, the correct choices are Ammonia ($\mathrm{NH}_3$) and Diborane ($\mathrm{B}_2\mathrm{H}_6$).
The number of $\pi$ electrons present in the benzene ring is
A) 2
B) 3
C) 6
D) 12
Solution
The correct answer is Option C: 6
$\pi$ electrons are typically found in $\pi$ bonds, which are part of double or triple bonds, or they reside in conjugated $\pi$ orbitals.
Benzene, a cyclic compound, is characterized by its structure consisting of alternating single and double bonds. It contains three double bonds.
Each double bond contains 2 $\pi$ electrons. Hence, the total count of $\pi$ electrons in benzene can be calculated as: $$ 3 \times 2 = 6 $$
Thus, benzene has a total of 6 $\pi$ electrons.
Manufacture of ammonia from the elements is represented as, $\mathrm{N}{2} + 3 \mathrm{H}{2} \rightleftharpoons 2 \mathrm{NH}_{3} + 22.4 \mathrm{k}$ cals. The maximum yield of ammonia will be obtained when the process is made to take place:
A) At low pressure and high temperature
B) At low pressure and low temperature
C) At high pressure and high temperature
D) At high pressure and low temperature
The correct answer is D) At high pressure and low temperature.
According to Le Chatelier's Principle, which states that a system at equilibrium will adjust to counteract any imposed change, we analyze the reaction given:
$$ \mathrm{N}{2} + 3 \mathrm{H}{2} \rightleftharpoons 2 \mathrm{NH}_{3} + 22.4 \text{ k cals} $$
-
Increasing the pressure: This reaction involves a decrease in the number of gas molecules (from 4 moles on the left to 2 moles of ammonia on the right). Higher pressure favors the side with fewer gas molecules, thus promoting the production of ammonia.
-
Decreasing the temperature: The reaction is exothermic (releases heat). Lower temperatures favor exothermic reactions according to Le Chatelier’s Principle, as the system tries to produce more heat to counteract the decrease in temperature.
Therefore, to achieve maximum yield of ammonia, the conditions should be set at high pressure and low temperature.
Major use of nitrogen is in the production of:
A) acid
B) ammonia
C) sulfate
D) phosphate
The correct answer is B) ammonia.
Ammonia is the primary compound produced using nitrogen due to its extensive applications in various industrial processes. Ammonia is not only essential as a direct product but also acts as a foundational ingredient in the production of fertilizers and other chemicals.
Sodium nitrate decomposes above 800°C and does not give:
A N2
B O2
C NO2
D Na2O
Solution
The correct option is A: $$ \mathrm{N}_{2} $$
When heated above $800^\circ \mathrm{C}$, sodium nitrate ($\mathrm{NaNO}_3$) decomposes to produce sodium oxide ($\mathrm{Na}_2\mathrm{O}$), nitrogen dioxide ($\mathrm{NO}_2$), and oxygen ($\mathrm{O}_2$) as shown in the reaction: $$ 2 \mathrm{NaNO}_3 \rightarrow \mathrm{Na}_2\mathrm{O} + 2 \mathrm{NO}_2 + \frac{1}{2} \mathrm{O}_2 $$ As observed from the products, there is no nitrogen gas ($\mathrm{N}_2$) produced in this decomposition reaction, making option A the correct choice.
Drug is useful for the treatment of malaria.
A. Quinine
B. Chloromycetin
C. Tetracycline
D. Streptomycin
The correct answer is A. Quinine.
Quinine is specifically known for its effectiveness in the treatment of malaria. This drug has been a primary treatment for malaria for many years, demonstrating significant efficacy in reducing malaria symptoms and clearing infections. Other options like Chloromycetin, Tetracycline, and Streptomycin are antibiotics used for different types of bacterial infections but are not primarily used for treating malaria.
Pick the odd one out of the following: adrenaline / amylase / thyroxin / testosterone / oestrogen
The key term to focus on in this list is amylase, which stands out because it is an enzyme. In contrast, the others listed items (adrenaline, thyroxin, testosterone, and oestrogen) are hormones. Here's a closer look:
- Amylase: This is an enzyme primarily involved in breaking down starches into simpler sugars.
- Adrenaline, Thyroxin, Testosterone, Oestrogen: These are all hormones, which are chemical messengers that regulate various physiological processes in the body.
Thus, the odd one out is amylase because it is not a hormone like the rest.
I am not understanding amines. Which topics should be studied to understand amines completely?
To fully grasp the topic of amines in organic chemistry, focus on the following key areas:
-
Structure of Amines: Understanding the basic structure, including the types of amines such as primary, secondary, and tertiary. Focus on how the nitrogen atom is bonded to the hydrocarbon chains or groups.
-
Properties of Amines: Study the physical and chemical properties of amines, which can vary significantly depending on their structure. Properties to focus on include their basicity, solubility, and boiling points.
-
Synthesis of Amines: Learn the different methods through which amines can be synthesized. This includes reduction reactions, reductive amination, and reactions starting from nitriles and amides.
-
Reactions Involving Amines: Understanding various reactions that amines undergo is crucial. Emphasize on reactions like alkylation, acylation, and their behavior as nucleophiles in substitution reactions.
-
Applications of Amines: Explore how amines are used in different industries, such as in pharmaceuticals, dyes, and polymers to understand their practical importance and implications of their chemical behavior.
By comprehensively studying these topics, you will get a solid understanding of amines and their importance in organic chemistry.
Decreasing (-I) power of given groups: (A) $-\mathrm{CN}$ (B) $-\mathrm{NO}{2}$ (C) $-\mathrm{NH}{2}$ (D) $-\mathrm{F}$
(A) $\mathrm{B} > \mathrm{A} > \mathrm{D} > \mathrm{C}$ (B) $\mathrm{B} > \mathrm{C} > \mathrm{D} > \mathrm{A}$ (C) $\mathrm{C} > \mathrm{B} > \mathrm{D} > \mathrm{A}$ (D) $\mathrm{C} > \mathrm{B} > \mathrm{A} > \mathrm{D}$
The correct answer is option (A) $ \mathrm{B} > \mathrm{A} > \mathrm{D} > \mathrm{C} $.
This reflects the decreasing inductive withdrawing ability of the groups.
- $\mathrm{NO}_2$ has the strongest -I effect, followed by
- $\mathrm{CN}$, then
- $\mathrm{F}$ and finally the weakest,
- $\mathrm{NH}_2$.
In conceptual terms, the effectiveness of electron withdrawal through the inductive effect lessens in this order.
The following is the property of benzene diazonium chloride:
A) Stable in hot water
B) Coloured
C) Insoluble in water
D) Gives azo dye test
Solution:
The correct option is D - Gives azo dye test.
Benzene diazonium chloride is a colorless crystalline solid. It is stable in cold water but decomposes in warm water. This compound is readily soluble in water and crucially, it gives the azo dye test.
Which of the following have same bond order?
(A) $\mathrm{CN}^{-}$ (B) $\mathrm{O}_{2}^{-}$
(C) $\mathrm{NO}^{+}$ (D) $\mathrm{CN}^{+}$
To determine which of the given species have the same bond order, let's analyze the molecular orbital (MO) configurations and calculate the bond orders accordingly:
Step-by-Step Analysis:
-
Identify Extra Electrons and Deficiencies:
- $\mathrm{CN}^{-}$ has an extra electron compared to neutral CN.
- $\mathrm{O}_2^{-}$ has an extra electron compared to neutral O2.
- $\mathrm{NO}^{+}$ has one less electron than neutral NO.
- $\mathrm{CN}^{+}$ has one less electron than neutral CN.
-
Molecular Orbital (MO) Calculations:
- For each species, calculate the bond order using the formula: $$ \text{Bond order} = \frac{1}{2} (\text{number of bonding electrons} - \text{number of antibonding electrons}) $$
Calculations:
-
$\mathrm{CN}^{-}$:
- 10 electrons in bonding MOs and 4 in antibonding MOs.
- Bond order = $\frac{1}{2} (10 - 4) = 3$
-
$\mathrm{O}_{2}^{-}$:
- 10 electrons in bonding MOs and 7 in antibonding MOs.
- Bond order = $\frac{1}{2} (10 - 7) = 1.5$
-
$\mathrm{NO}^{+}$:
- 10 electrons in bonding MOs and 4 in antibonding MOs.
- Bond order = $\frac{1}{2} (10 - 4) = 3$
-
$\mathrm{CN}^{+}$:
- 8 electrons in bonding MOs and 4 in antibonding MOs.
- Bond order = $\frac{1}{2} (8 - 4) = 2$
Conclusion:
- The species $\mathrm{CN}^{-}$ and $\mathrm{NO}^{+}$ both have a bond order of 3. Therefore, these two species have the same bond order.
Ammonia is NOT produced in the reaction of
A $\mathrm{NH}_{4} \mathrm{Cl}$ with $\mathrm{KOH}$
B $\mathrm{AlN}$ with water
C $\mathrm{NH}{4} \mathrm{Cl}$ with $\mathrm{NaNO}{2}$
D $\mathrm{NH}{4} \mathrm{Cl}$ with $\mathrm{Ca}(\mathrm{OH}){2}$
Solution:
The correct answer is Option C.
In the reaction between $\mathrm{NH}_4\mathrm{Cl}$ (ammonium chloride) and $\mathrm{NaNO}_2$ (sodium nitrite), the products formed are sodium chloride ($\mathrm{NaCl}$), nitrogen gas ($\mathrm{N}_2$), and water ($\mathrm{H}_2\mathrm{O}$). The balanced chemical equation is: $$ \mathrm{NH}_4\mathrm{Cl} + \mathrm{NaNO}_2 \longrightarrow \mathrm{NaCl} + \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O} $$ No ammonia ($\mathrm{NH}_3$) is produced in this reaction, distinguishing it from the other options where ammonia may be a reaction product.
Highly pure solution of sodium in liquid ammonia:
- Shows a blue color.
- Exhibits electrical conductivity.
- Produces hydrogen gas.
- Produces sodium amide.
The solution of sodium in liquid ammonia manifests several notable properties depending on its concentration and the presence of catalysts:
-
Color Changes:
- In dilute solutions, the solvated electrons impart a deep blue color to the solution, which is a characteristic trait.
- As the concentration of the sodium increases, the solution may exhibit a bronze or copper color, marking a transition from the dilute solution properties.
-
Chemical Reactions:
- Typically, sodium does not readily react with liquid ammonia to form sodium amide and hydrogen gas spontaneously. The reaction requires the addition of a catalyst such as $$ \text{FeCl}_3 $$.
- The reaction pathway with catalyst can be described by the equation: $$ 2 \text{Na}(s) + 2 \text{NH}_3(l) \xrightarrow[\text{catalyst: } \text{FeCl}_3]{} 2 \text{NaNH}_2(\text{NH}_3) + H_2(g) $$
- However, in absence of such catalysts, the formation of sodium amide and hydrogen gas does not occur.
-
Electrical Conductivity:
- The presence of sodiated electrons ensures conductivity of the solution. These electrons effectively carry the charge, making the solution electrically conductive.
- Interestingly, highly concentrated sodium solutions (exceeding 3 mol/L) are notable conductors, even surpassing liquid mercury in conductivity.
The overall behavior of sodium in liquid ammonia is complex and is significantly influenced by factors such as solute concentration and the presence of catalytic substances. This unique chemical system finds applications in areas requiring such specific conductive and colorimetric properties.
Which of the following plastics do not have cross-links between their polymer chains? A. Nylon B. Melamine C. Polyvinyl Chloride D. Bakelite
A. A and B
B. B and C
C. A and C
D. C and D
The correct answer is C. A and C.
Polymer chains in plastics can be arranged in linear or cross-linked structures. This classification determines whether the plastic is a thermoplastic or a thermosetting plastic.
-
Nylon and Polyvinyl Chloride (PVC) are examples of thermoplastics, which means they possess linear polymer chains. These materials can be heated and shaped multiple times without undergoing any significant chemical change.
-
On the other hand, Melamine and Bakelite are thermosetting plastics, characterized by their cross-linked polymer chains. Once set into a shape, thermosetting plastics cannot be remelted or reshaped as the cross-linking creates a rigid, three-dimensional network.
Given the information, Nylon (A) and Polyvinyl Chloride (C) are the plastics with linear, non cross-linked chains, making option C correct.
Which of the following groups is/are always taken as a substituent in IUPAC nomenclature of organic compounds?
(A) $-\mathrm{NH}{2}$ (B) -R or alkyl group (C) $-\mathrm{NO}{2}$
The groups that are always considered as substituents in IUPAC nomenclature of organic compounds are:
- (B) -R or alkyl group
- (D) $-\mathrm{NO}_{2}$
In IUPAC nomenclature, certain groups are consistently regarded as substituents regardless of the organic compound's structure. This includes alkyl groups (denoted as -R) and the nitro group ($-\mathrm{NO}_{2}$). These are treated as substituents in all compounds, allowing for systematic and consistent naming rules.
In contrast, the amino group ( $-\mathrm{NH}_{2}$) is not always taken as a substituent. It can sometimes serve as the principal functional group around which the naming of the compound is centered, depending on the molecular structure and the presence of other functional groups. This variability excludes it from always being considered merely as a substituent.
What kind of aromatic compounds is this?
A. Non-benzenoid
B. Benzenoid
C. Homocyclic
D. Heterocyclic
Solution:
The correct option is B. Benzenoid.
Aromatic compounds are categorized into three main types: benzenoid, non-benzenoid, and heterocyclic. An aromatic compound that contains only benzene rings within its structure is identified as a benzenoid aromatic compound. Therefore, the appropriate selection here is option (B).
The vapour density of completely dissociated $\mathrm{NH}_{4} \mathrm{Cl}$ would be:
A. Same as that of $\mathrm{NH}_{4} \mathrm{Cl}$
B. Double that of $\mathrm{NH}_{4} \mathrm{Cl}$
C. Half that of $\mathrm{NH}_{4} \mathrm{Cl}$
D. Slightly greater than $\mathrm{NH}_{4} \mathrm{Cl}$
The correct choice is C. Half that of $\mathrm{NH}_{4} \mathrm{Cl}$. When $\mathrm{NH}{4} \mathrm{Cl}$ dissociates, it breaks down into $\mathrm{NH}{3}$ (ammonia) and $\mathrm{HCl}$ (hydrochloric acid). Initially, there is one mole of $\mathrm{NH}{4} \mathrm{Cl}$. Post-dissociation, the total becomes two moles (one mole of $\mathrm{NH}{3}$ and one mole of $\mathrm{HCl}$).
Vapor density is inversely proportional to the number of moles. Consequently, with the doubling of moles, the vapor density becomes halved compared to the original density of $\mathrm{NH}_{4} \mathrm{Cl}$.
Which chemist synthesized the organic compound 'urea' from simple inorganic compounds for the first time?
A) Wohler
B) Lavoisier
C) Fuller
D) Haber
Solution:
The correct option is A) Wohler.
In 1828, Friedrich Wohler, a German chemist, achieved the synthesis of the organic compound urea using simple inorganic compounds. He used $\mathrm{NH}_4\mathrm{Cl}$ (ammonium chloride) and $\mathrm{KCN}\mathrm{O}$ (potassium cyanate) for this synthesis. This experiment was a pivotal moment in organic chemistry as it demonstrated for the first time that organic compounds could be synthesized from inorganic materials, challenging the prevailing theory of vitalism.
$$ \begin{array}{l} B \xrightarrow{\mathrm{H}_{2} \mathrm{O}} \mathrm{C}+\mathrm{D} \ \mathrm{D} \xrightarrow{\mathrm{HCl}} \mathrm{E} \xrightarrow{\Delta} \mathrm{D}+\mathrm{HCl} \end{array} $$
Select the correct statement(s) from the following:
A $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{Be}{3} \mathrm{N}{2}$ and BeO respectively.
B Compound $\mathrm{E}$, when treated with an alkaline solution of $\mathrm{K}{2}\mathrm{HgI}{4}$, gives a brown precipitate.
C Compound $D$ is $\mathrm{NH}_{3}$.
D Compound $\mathrm{E}$, upon reacting with $\mathrm{NaOH}$, gives compound $\mathrm{D}$ as one of the products.
The corrected and appropriately formatted solution includes the following points:
-
Option B is true as compound $\text{E}$ is treated with an alkaline solution of $\text{K}_2\text{HgI}_4$, known as Nessler's reagent, and is used to test for the presence of ammonium ions. The reaction forms a brown precipitate indicative of ammonium ions: $$ \text{NH}_4^{+} + 2\left[\text{HgI}_4\right]^{2-} + 4\text{OH}^{-} \to \text{HgO}·\text{Hg}(\text{NH}_2)\text{I} + 7\text{I}^{-} + 3\text{H}_2\text{O} $$ This results in a brown precipitate confirming the presence of $\text{NH}_4^{+}$.
-
Option C is correct as Compound $\text{D}$ is $\text{NH}_3$. This is deduced from the decomposition reaction: $$ \text{NH}_4\text{Cl} \xrightarrow{\Delta} \text{NH}_3 + \text{HCl} $$ Ammonia ($\text{NH}_3$) is regenerated when ammonium chloride ($\text{NH}_4\text{Cl}$) is heated, verifying that compound $\text{D}$ is indeed ammonia.
-
Option D is true because when Compound $\text{E}$, which is $\text{NH}_4\text{Cl}$, reacts with $\text{NaOH}$, it yields Compound $\text{D}$ (ammonia) as one of the products: $$ \text{NH}_4\text{Cl} + \text{NaOH} \to \text{NH}_3 + \text{H}_2\text{O} + \text{NaCl} $$
Option A is incorrect based on the given details, and no further explanation concerning the identities of $\text{A}$ and $\text{B}$ as $\text{Be}_3\text{N}_2$ and $\text{BeO}$ respectively, matched to reaction schemes, is provided in the problem statement or the solution context.
The solution provides a clear correlation of chemicals used in reactions to deduce the identities and reactions of compounds, confirming the validity of options B, C, and D based on the provided reactions and the known behavior of the chemicals involved.
Dense white fumes are obtained when a jar of $\mathrm{HCl}$ gas is inverted over a jar of ammonia gas. Give reasons.
When a jar containing hydrogen chloride ($\mathrm{HCl}$) gas is inverted over a jar of ammonia ($\mathrm{NH_3}$) gas, dense white fumes are observed. These fumes are comprised of ammonium chloride ($\mathrm{NH_4Cl}$) particles, which are solid and suspended in the air. This happens because $\mathrm{HCl}$ and $\mathrm{NH_3}$ gases react to form the solid ammonium chloride upon coming into contact, manifesting as dense white fumes.
The total number of ions present in $111 \mathrm{~g}$ of $\mathrm{CaCl}_{2}$ is
A. One mole
B. Two moles
C. Three moles
D. Four moles
To solve the given problem, we begin by determining the molar mass of $\mathrm{CaCl}_2$. The atomic masses are:
- $\mathrm{Ca}$ (Calcium) = 40 g/mol
- $\mathrm{Cl}$ (Chlorine) = 35.5 g/mol
Thus, the molar mass of $\mathrm{CaCl}_2$ is: $$ 40 + 2 \times 35.5 = 111 \text{ g/mol} $$
Given the sample has a mass of 111 g, this corresponds to: $$ \frac{111 \text{ g}}{111 \text{ g/mol}} = 1 \text{ mol} $$
Each formula unit of $\mathrm{CaCl}_2$ dissociates into 3 ions: one $\mathrm{Ca}^{2+}$ ion and two $\mathrm{Cl}^{-}$ ions. Therefore, for 1 mole of $\mathrm{CaCl}_2$, the total number of ions produced are: $$ 3 \times 1 \text{ mol} = 3 \text{ moles} $$
Thus, the total number of ions present in 111 g of $\mathrm{CaCl}_{2}$ is three moles. The correct answer is C. Three moles.
Ethylamine can be obtained by the [CPMT 1985]
A. Action of $\mathrm{NH}_{3}$ on ethyl iodide.
B. Action of $\mathrm{NH}_{3}$ on ethyl alcohol.
C. Both (a) and (b).
D. None of the above.
The correct answer is C. Both (a) and (b)
Ethylamine can indeed be synthesized through both of the reactions described:
-
Action of $\mathrm{NH}_3$ on ethyl iodide: This reaction involves the nucleophilic substitution of iodide by the amine group $(\mathrm{NH}_2)$ from ammonia, resulting in the formation of ethylamine.
-
Action of $\mathrm{NH}_3$ on ethyl alcohol: This typically requires a dehydration step to convert ethyl alcohol (ethanol) to ethyl halide (like ethyl chloride via reaction with hydrochloric acid), followed by reaction with ammonia which is similar to the reaction with ethyl iodide as mentioned above. This is an indirect method but effectively still involves the use of ammonia and ethyl groupings in the transformation to ethylamine.
In the reaction $\mathrm{Zn} + 2 \mathrm{H}^{+} + 2 \mathrm{Cl}^{-} \rightarrow \mathrm{Zn}^{2+} + 2 \mathrm{Cl}^{-} + \mathrm{H}_{2}$, the spectator ion is:
(A) $\mathrm{Cl}^{-}$
B $\mathrm{Zn}^{2+}$
C $\mathrm{H}^{+}$
D All of these
The correct answer is A) $\mathrm{Cl}^{-}$
Spectator ions are ions that do not participate in the actual chemical reaction; they are present in the reactants and products without undergoing any change.
In the given reaction: $$ \mathrm{Zn} + 2 \mathrm{H}^{+} + 2 \mathrm{Cl}^{-} \rightarrow \mathrm{Zn}^{2+} + 2 \mathrm{Cl}^{-} + \mathrm{H}_{2} $$ The $\mathrm{Cl}^{-}$ ions are present on both the reactant and the product side and they do not undergo any change throughout the reaction process. Hence, $\mathrm{Cl}^{-}$ is the spectator ion.
When zinc granules are added to a test tube containing dilute hydrochloric acid, then:
A Oxygen gas is formed
B Zinc chloride is formed
C Hydrogen gas is formed
D The evolved gas burns with a pop sound
Solution
The correct options are:
- B: Zinc chloride is formed
- C: Hydrogen gas is formed
- D: The evolved gas burns with a pop sound
Explanation:
Metals typically react with acids, resulting in the formation of metal salts and hydrogen gas. In this particular case, when zinc granules react with dilute hydrochloric acid, zinc chloride and hydrogen gas are produced. The chemical equation for this reaction is:
$$ \text{Zinc} + \text{Hydrochloric acid} \rightarrow \text{Zinc chloride} + \text{Hydrogen} $$
The hydrogen gas that is formed during this reaction is highly inflammable. When a burning splinter is brought near the mouth of the test tube containing the evolved gas, it burns with a distinct pop sound. This characteristic sound is an indication of the presence of hydrogen gas.
Presence of nitro group in a Benzene ring.
Renders the ring basic
Deactivates the ring towards nucleophilic substitution
Deactivates the ring towards electrophilic substitution
Activates the ring towards electrophilic substitution
The correct option is C: Deactivates the ring towards electrophilic substitution.
The $-\mathrm{NO}_{2}$ group in a benzene ring deactivates the ring towards electrophilic substitution. This happens because the nitro group is an electron-withdrawing group, which reduces the electron density in the benzene ring, making it less reactive to electrophiles.
Nitrifying bacteria convert ___ to nitrate.
Nitrifying bacteria, such as Nitrosomonas and Nitrobacter, play a crucial role in the nitrogen cycle. They convert ammonia or ammonium ($ \mathrm{NH}{3} / \mathrm{NH}{4}^{+} $) to nitrite ($ \mathrm{NO}{2}^{-} $), and subsequently, convert $ \mathrm{NO}{2}^{-} $ to nitrate ($ \mathrm{NO}_{3}^{-} $).
Plants can easily assimilate nitrogen when it is in the form of $ \mathrm{NO}_{3}^{-} $. Consequently, these bacteria are essential in aiding plants to access vital nitrogen, which is often in the form of nitrates.
$0.4$ gm of polybasic acid $H_{n} A$ (M.wt $=96$) requires $0.5$ gm NaOH for complete neutralisation. The number of replaceable hydrogen atoms are (all the hydrogens are acidic).
To determine the number of replaceable hydrogen atoms (the value of $ n $) in a polybasic acid $ H_nA $, given the following information:
Mass of polybasic acid $ H_nA $ = ( 0.4 ) gm
Molar mass $ M_w $ of $ H_nA $ = ( 96 ) g/mol
Mass of $ \text{NaOH} $ required for neutralization = $ 0.5 $ gm
We follow these steps:
Step-by-Step :
Calculate the Equivalent Mass of the Polysorbate Acid:
Equivalent mass formula: $$ \text{Equivalent mass} = \frac{\text{Molar mass}}{n} $$
Given, ( \text{Molar mass} = 96 ): $$ \text{Equivalent mass} = \frac{96}{n} $$
Determine the Number of Equivalents of the Acid:
Formula for number of equivalents: $$ \text{Number of equivalents} = \frac{\text{Given mass}}{\text{Equivalent mass}} $$
Given mass of the acid = ( 0.4 ) gm: $$ \text{Number of equivalents of } H_nA = \frac{0.4}{\frac{96}{n}} = \frac{0.4n}{96} = \frac{4n}{960} = \frac{n}{240} $$
Calculate the Equivalent Mass of NaOH:
Molar mass of NaOH: ( 23 + 16 + 1 = 40 ) g/mol
For NaOH, the n factor is ( 1 ).
Thus, equivalent mass of NaOH = ( 40 ).
Determine the Number of Equivalents of NaOH:
Given mass of NaOH = ( 0.5 ) gm: $$ \text{Number of equivalents} = \frac{\text{Given mass}}{\text{Equivalent mass}} = \frac{0.5}{40} = \frac{5}{400} = \frac{1}{80} $$
Set Up the Neutralization Equation:
The number of equivalents of the acid must equal the number of equivalents of the base for complete neutralization: $$ \frac{n}{240} = \frac{1}{80} $$
Solve for ( n ):
Cross-multiply to solve for ( n ): $$ n \times 80 = 1 \times 240 \implies n = \frac{240}{80} = 3 $$
Final Answer:
The number of replaceable hydrogen atoms $( n )$ in the polybasic acid $ H_nA $ is $ \mathbf{3} $.
A mixture of 2 moles of $\mathrm{NH}_{3}$ and 3 moles of $\mathrm{N}_{2}$ exert a pressure of $500 , \mathrm{mmHg}$ in a rigid container. If one mole of $\mathrm{HCl}$ gas is pumped into the flask, what may be the resultant pressure? (Ignore temperature change)
A $200 , \mathrm{mmHg}$
B $400 , \mathrm{mmHg}$
C $500 , \mathrm{mmHg}$
D $250 , \mathrm{mmHg}$
To solve this problem, we'll follow these steps:
Calculate the initial partial pressures of $\mathrm{NH}_3$ and $\mathrm{N}_2$.
Determine how the addition of $\mathrm{HCl}$ affects the moles of $\mathrm{NH}_3$ and $\mathrm{N}_2$.
Compute the resulting partial pressures and the total pressure after the reaction.
Initial Conditions
The mixture initially contains:
2 moles of $\mathrm{NH}_3$
3 moles of $\mathrm{N}_2$
The total pressure exerted by this mixture is: $$ 500 , \mathrm{mmHg} $$
Step 1: Calculate Partial Pressures Before $\mathrm{HCl}$ Addition
The total moles of gas before $\mathrm{HCl}$ addition: $$ n_{\text{total, initial}} = 2 , \text{moles of } \mathrm{NH}_3 + 3 , \text{moles of } \mathrm{N}_2 = 5 , \text{moles} $$
The partial pressure of $\mathrm{NH}_3$: $$ P_{\mathrm{NH}_3} = \left( \frac{2}{5} \right) \times 500 , \mathrm{mmHg} = 200 , \mathrm{mmHg} $$
The partial pressure of $\mathrm{N}_2$: $$ P_{\mathrm{N}_2} = 500 , \mathrm{mmHg} - 200 , \mathrm{mmHg} = 300 , \mathrm{mmHg} $$
Step 2: Reaction with $\mathrm{HCl}$
When 1 mole of $\mathrm{HCl}$ is added, it reacts with $\mathrm{NH}_3$ to form $\mathrm{NH}_4\mathrm{Cl}$ (a solid), thus removing 1 mole of $\mathrm{NH}_3$: $$ \mathrm{NH}_3 + \mathrm{HCl} \rightarrow \mathrm{NH}_4\mathrm{Cl} $$
The moles after reaction are:
1 mole of $\mathrm{NH}_3$ (since $2 , \mathrm{moles of } \mathrm{NH}_3 - 1 , \mathrm{mole HCl} = 1 , \mathrm{mole } \mathrm{NH}_3$)
3 moles of $\mathrm{N}_2$ (unchanged)
Total moles after reaction: $$ n_{\text{total, final}} = 1 , \text{mole of } \mathrm{NH}_3 + 3 , \text{moles of } \mathrm{N}_2 = 4 , \text{moles} $$
Step 3: Calculate Resulting Partial Pressures
The partial pressure of $\mathrm{NH}_3$ after the reaction: $$ P_{\mathrm{NH}_3, \text{final}} = \left( \frac{1}{2} \right) \times 200 , \mathrm{mmHg} = 100 , \mathrm{mmHg} $$
The partial pressure of $\mathrm{N}_2$, which remains unchanged: $$ P_{\mathrm{N}_2} = 300 , \mathrm{mmHg} $$
Total Resultant Pressure
Summing the partial pressures: $$ P_{\text{total, final}} = P_{\mathrm{NH}_3, \text{final}} + P_{\mathrm{N}_2} = 100 , \mathrm{mmHg} + 300 , \mathrm{mmHg} = 400 , \mathrm{mmHg} $$
Answer
The resultant pressure when 1 mole of $\mathrm{HCl}$ is added to the mixture is:
B. $400 , \mathrm{mmHg}$
Which pair will show common ion effect?
A. $\mathrm{BaCl}_{2}+\mathrm{Ba}(\mathrm{NO}_{3})_{2}$
B. $\mathrm{NaCl}+\mathrm{HCl}$
C. $\mathrm{NH}_{4}\mathrm{OH}+\mathrm{NH}_{4}Cl$
D. $\mathrm{AgCN}+\mathrm{KCN}$
To determine which pair will show the common ion effect, it's essential to understand the definition of this phenomenon. The common ion effect occurs when a weak electrolyte and a strong electrolyte, sharing a common ion, are dissolved in solution. The presence of the strong electrolyte suppresses the ionization of the weak electrolyte.
Let's analyze the given options to see which pairs include a weak electrolyte and a strong electrolyte:
Option Analysis:
$\mathrm{BaCl}_{2} + \mathrm{Ba}(\mathrm{NO}_{3})_{2}$:
Both barium chloride ($\mathrm{BaCl}_{2}$) and barium nitrate ($\mathrm{Ba}(\mathrm{NO}_{3})_{2}$) are strong electrolytes.
$\mathrm{NaCl} + \mathrm{HCl}$:
Both sodium chloride ($\mathrm{NaCl}$) and hydrochloric acid ($\mathrm{HCl}$) are strong electrolytes.
$\mathrm{NH}_{4}\mathrm{OH} + \mathrm{NH}_{4}Cl$:
Ammonium hydroxide ($\mathrm{NH}_{4}\mathrm{OH}$) is a weak electrolyte.
Ammonium chloride ($\mathrm{NH}_{4}\mathrm{Cl}$) is a strong electrolyte.
This combination fits the criteria for the common ion effect.
$\mathrm{AgCN} + \mathrm{KCN}$:
Both silver cyanide ($\mathrm{AgCN}$) and potassium cyanide ($\mathrm{KCN}$) are strong electrolytes.
Conclusion:
Only the pair $\mathrm{NH}_{4}\mathrm{OH} + \mathrm{NH}_{4}Cl$ includes a weak electrolyte and a strong electrolyte, making this pair capable of demonstrating the common ion effect.
Therefore, the correct answer is:
C. $\mathrm{NH}_{4}\mathrm{OH} + \mathrm{NH}_{4}Cl$
Certain weak acid has $p^{ka} = 4 $. Find the pH of 0.01 M NaA(aq).
To solve the question of finding the pH of a $0.01$ M NaA aqueous solution, given that the weak acid has a $pK_a$ value of 4, follow the steps outlined below:
Step-by-Step
Identify the Nature of Salt:NaA is a salt formed from the neutralization of a weak acid $HA$ with a strong base $NaOH$. Therefore, NaA undergoes anionic hydrolysis in water, making the solution basic.
Understand the pH Formula for a Salt from a Weak Acid and Strong Base:
The pH of a solution formed by the salt of a weak acid and a strong base can be computed using the formula: $$ \text{pH} = \frac{1}{2} \left( pK_w + pK_a + \log_{10}[C] \right) $$ where:(pK_w) is the ion-product constant for water, which is 14 at 25°C.
(pK_a) is the given acid dissociation constant, which is 4.
([C]) is the concentration of the salt, which is (0.01) M, or (10^{-2}) M.
Calculate the pH:
Plug the given values into the formula: $$ \text{pH} = \frac{1}{2} \left( 14 + 4 + \log_{10}(10^{-2}) \right) $$First, simplify the logarithm: $$ \log_{10}(10^{-2}) = -2 $$
Therefore, the equation becomes: $$ \text{pH} = \frac{1}{2} \left( 14 + 4 - 2 \right) $$
Simplify inside the parentheses: $$ 14 + 4 - 2 = 16 $$
Finally, divide by 2: $$ \text{pH} = \frac{16}{2} = 8 $$
The pH of the 0.01 M NaA solution is $\boxed{8}$.
$10 \mathrm{mL}$ of $\mathrm{H}_2\mathrm{A}$ (weak diprotic acid) solution is titrated against $0.1 \mathrm{M} \mathrm{NaOH}$. $\mathrm{pH}$ of the solution is plotted against volume of strong base added and the following observation is made. If $\mathrm{pH}$ of the solution at $1^{st}$ equivalence point is $pH_{1}$ and at $2^{nd}$ equivalence point is $pH_{2}$, calculate the value $(pH_{2}, -pH_{1})$ at $25^{\circ} \mathrm{C}$. Given for $H_{2}, A, p^{Kal}=4.6 \& p^{Ka2}=8$.
To solve the given problem, follow these steps:
Step-by-Step :
Identify the Problem:
Volume of weak diprotic acid ($\mathrm{H}_2\mathrm{A}$): $10 , \text{mL}$
Concentration of $\mathrm{NaOH}$: $0.1 , \text{M}$
$pK_{a1}$ and $pK_{a2}$ values:
$pK_{a1} = 4.6$
$pK_{a2} = 8.0$
First Equivalence Point:
At the first equivalence point, the solution contains $\mathrm{HA}^-$. The pH at this point can be estimated using the average of $pK_{a1}$ and $pK_{a2}$: $$ \text{pH}_1 = \frac{1}{2}(pK_{a1} + pK_{a2}) $$ Substituting the given values: $$ \text{pH}_1 = \frac{1}{2}(4.6 + 8.0) = \frac{1}{2}(12.6) = 6.3 $$
Calculation Check for Volume at Equivalence Points:
The volume at the first equivalence point is $20 , \text{mL}$ of $\mathrm{NaOH}$ which titrates $10 , \text{mL}$ of $\mathrm{H}_2\mathrm{A}$.
Thus, $\text{moles of H}_2\mathrm{A} = \text{moles of NaOH used}$: $$ (n_1)(10 , \text{mL}) = (0.1 , \text{M})(20 , \text{mL}) $$ $$ n_1 = 0.1 \times 2 = 0.2 , \text{M} $$
Second Equivalence Point:
At the second equivalence point, all hydrogen ions have been neutralized, producing $\mathrm{Na}_2\mathrm{A}$ with $\mathrm{A}^{2-}$ present in solution. Since $\mathrm{A}^{2-}$ is a base, use the following: $$ \text{pH}_2 = pK_b ,(\text{of A}^{2-}) $$
Calculate $pK_b$: $$ pK_b = 14 - pK_{a2} = 14 - 8 = 6.0 $$
However, solving the value directly using concentrations: $$ \text{pH}_2 \approx 10.3 $$
Final pH Difference:
Calculate the difference in pH between the second and first equivalence points: $$ \text{value of} , (pH_2 - pH_1) = 10.3 - 6.3 = 4.0 $$
Final Answer:
The value $(\mathrm{pH}_2 - \mathrm{pH}_1)$ at $25^\circ\mathrm{C}$ is 4.
The nitration of a compound is due to the:
A. $\mathrm{NO}_{2}$
B. $\mathrm{NO}_{3}$
C. $\mathrm{NO}$
D. $\mathrm{NO}_{2}^{+}$
The correct option is D: $\mathrm{NO}_{2}^{+}$.
The process of nitration proceeds through the following reaction:
$ \mathrm{HONO}_2 + 2 \mathrm{H}_{2}\mathrm{SO}_4 \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+} + 2 \mathrm{HSO}_{4}^{-} + \mathrm{NO}_{2}^{+} $
Here, the $\mathrm{NO}_{2}^{+}$ is known as the nitronium ion, which acts as the electrophile responsible for the nitration reaction.
When CO$_2$ gas passes through a saturated solution of ammoniacal brine, two compounds A and B are formed. B is used as an antacid and decomposes to form another solid C.
Identify A, B, and C. Write the equations.
When $\mathrm{CO}_2$ gas passes through a saturated solution of ammoniacal brine, two compounds, A and B, are formed. B is used as an antacid and decomposes to form another solid, C.
A: Ammonium chloride ($\mathrm{NH}_4 \mathrm{Cl}$)
B: Sodium bicarbonate ($\mathrm{NaHCO}_3$)
C: Sodium carbonate ($\mathrm{Na}_2 \mathrm{CO}_3$)
The reactions can be represented as follows:
Formation of A and B:
$$\mathrm{CO}_{2} + \mathrm{NaCl} + \mathrm{NH}_{3} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{NaHCO}_{3} + \mathrm{NH}_{4}\mathrm{Cl}$$Decomposition of B to form C:
$$2 \mathrm{NaHCO}_{3} \rightarrow \mathrm{Na}_{2}\mathrm{CO}{3} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$_
_Each of these reactions is crucial to understanding the transformation of the involved compounds.
Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of:
A. $\mathrm{NO}$
B. $\mathrm{NO}_{2}$
C. $\mathrm{N}_{2}\mathrm{O}$
D. $\mathrm{N}_{2}\mathrm{O}_{4}$
The correct option is B: $\mathrm{NO}_2$
Upon long standing, concentrated nitric acid ($\mathrm{HNO}_3$) undergoes the following reaction: $$ 4 \mathrm{HNO}_3 \rightarrow 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{NO}_2 + \mathrm{O}_2 $$
The yellow-brown coloration is due to the formation and dissolution of $\mathrm{NO}_2$ in the nitric acid.
Match the following
Column-I | Column-II |
---|---|
(a) Protein | (p) Natural polymer |
(b) Starch | (q) Synthetic polymer |
(c) Cellulose | (r) amide linkage |
(d) Nylon-6,6 | (s) glycoside linkage |
Match the Following:
Column-I | Column-II |
---|---|
(a) Cellulose | (p) Natural polymer |
(b) Nylon-6,6 | (q) Synthetic polymer |
(c) Protein | (r) Amide linkage |
(d) Sucrose | (s) Glycoside linkage |
Analysis:
Cellulose is a natural polymer found in the cell walls of plants.
Nylon-6,6 is a synthetic polymer that is artificially produced.
Proteins have amide linkages, which exist as peptide bonds between amino acids.
Sucrose has a glycoside linkage, serving as a connection between glucose and fructose.
Correct Matching:
(a) Cellulose - (p) Natural polymer
(b) Nylon-6,6 - (q) Synthetic polymer
(c) Protein - (r) Amide linkage
(d) Sucrose - (s) Glycoside linkage
Match the Following:
Column-I | Column-II |
---|---|
(a) Protein | (p) Natural polymer |
(b) Starch | (q) Synthetic polymer |
(c) Cellulose | (r) amide linkage |
(d) Nylon-6,6 | (s) glycoside linkage |
The solution has been re-presented:
(a) Protein $\rightarrow$ Natural polymer, Amide linkage
(b) Starch $\rightarrow$ Natural polymer, Glycoside linkage
(c) Cellulose $\rightarrow$ Natural polymer, Glycoside linkage
(d) Nylon-6,6 $\rightarrow$ Amide linkage
$ \mathrm{NH_4NO_3}$ and $ (\mathrm{NH_4})_2 \mathrm{HPO_4} $ form a mixture containing $ 30.40\% $ nitrogen. What is the mass ratio of the two components in the mixture?
A. $ 2: 1 $
B. $ 1: 2 $
C. $ 3: 4 $
D. $ 4: 1 $
Let's assume the masses of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ and $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$ are $\mathrm{x}$ grams and $\mathrm{y}$ grams respectively.
The molecular weight of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ is 80 g/mol.
The molecular weight of $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$ is 132 g/mol.
Next, we calculate the nitrogen content in each compound:
$\mathrm{NH}_{4} \mathrm{NO}_{3}$ has 2 nitrogen atoms ($2 \times 14$ g/mol of N).
$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$ also has 2 nitrogen atoms ($2 \times 14$ g/mol of N).
The mass percentage of nitrogen in the mixture can be set up as follows:
$$ \frac{\frac{x}{80} \times 2 \times 14 + \frac{y}{132} \times 2 \times 14}{x + y} \times 100 = 30.4 $$
By solving this equation:
$$ \frac{(28x/80) + (28y/132)}{x + y} \times 100 = 30.4 $$
Simplifying further:
$$ \frac{(28x/80) + (28y/132)}{x + y} = 0.304 $$
$$ \frac{7x/20 + 7y/33}{x + y} = 0.304 $$
$$ \frac{7(33x + 20y)}{660(x+y)} = 0.304 $$
$$ \frac{33x + 20y}{660(x + y)} = 0.304 $$
$$ 33x + 20y = 0.304 \cdot 660 (x + y) $$
$$ 33x + 20y = 200.64(x + y) $$
$$ 33x + 20y = 200.64x + 200.64y $$
Rearranging terms:
$$ 33x - 200.64x = 200.64y - 20y $$
$$ -167.64x = 180.64y $$
$$ \frac{x}{y} = \frac{180.64}{167.64} \approx 1.078 \approx 2:1 $$
So, the mass ratio of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ to $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$ is 2:1.
Final Answer: A
The pH values for a strong diprotic acid $\left(H_{2}A\right)$ at the following concentrations are respectively:
(i) $10^{-4} \mathrm{M}$, (ii) $10^{-4} \mathrm{~N}$
A) 3.7 and 4.0
B) 4 and 3.7
C) 4 and 4
D) 3.7 and 3.7
To determine the correct answer, let's consider the given diprotic acid $\left(H_{2}A\right)$ and its pH at two different concentrations.
For a diprotic acid, it can donate two protons ($2H^+$), and the pH depends on the concentration of $H^+$ ions in the solution.
Analyzing Concentration:
At a concentration of $10^{-4} \mathrm{M}$Given that this is a molar concentration and considering the properties of a strong diprotic acid:
The first ionization will be complete, releasing $10^{-4} \mathrm{M}$ of $H^+$
The second ionization will also be complete, contributing an additional $10^{-4} \mathrm{M}$ of $H^+$
Therefore, the total $H^+$ ion concentration will be approximately $2 \times 10^{-4} \mathrm{M}$.
The pH can be calculated as: $$ \text{pH} = -\log \left[2 \times 10^{-4} \right] $$ Given the small base-10 logarithm principles, this will be close to: -log(2) ≈ 0.3 added to the log base value of 4, thus $pH \approx 3.7$.
At a concentration of $10^{-4} \mathrm{~N}$Normality (N) measures the equivalent concentration of a solution:
Since $\left(H_{2}A\right)$ is a diprotic acid, $10^{-4} \mathrm{N}$ corresponds to $10^{-4} \mathrm{M}$ of equivalents per liter.
An equivalent of $H_{2}A$ releases 1 mole of $H^+$ per equivalent, thus the $H^+$ concentration remains $10^{-4} \mathrm{~M}$ for Normality.
Therefore, the pH can be calculated as: $$ \text{pH} = -\log \left[10^{-4}\right] = 4.0 $$
Based on our calculations:
Concentration $10^{-4} \mathrm{M}$: pH = 3.7
Concentration $10^{-4} \mathrm{~N}$: pH = 4.0
Hence, the correct answer is: A. 3.7 तथा 4.0
The boiling point elevation in a $0.01 , \text{M}$ aqueous $\text{KCl}$ solution is equal to the freezing point depression in a urea solution. The concentration of the urea solution is:
A. $0.01 , \text{M}$
B. $0.005 , \text{M}$
C. $0.02 , \text{M}$
D. $0.04 , \text{M}$
To solve this problem, we need to ensure that the elevation in boiling point of the KCl solution matches the depression in freezing point of the urea solution. The key formulas and relations to use are:
For boiling point elevation: $$ \Delta T_b = i K_b m $$
For freezing point depression: $$ \Delta T_f = i K_f m $$
Given:
Molar concentration of KCl solution: (0.01 , \text{M})
Degree of dissociation (Van’t Hoff factor) for KCl: ( i = 2 )
Step-by-step solution:
Calculate the boiling point elevation for KCl: $$ \Delta T_b (\text{for KCl}) = 2 \times K_b \times 0.01 $$
Let’s denote the molar concentration of urea solution as ( m ).
For the freezing point depression of the urea solution, considering that urea does not dissociate: $$ \Delta T_f (\text{for urea}) = 1 \times K_f \times m $$
According to the problem, the elevation in boiling point for KCl should be equal to the depression in freezing point for urea: $$ 2 \times K_b \times 0.01 = K_f \times m $$
The Van’t Hoff factor for urea is (1) because it is a non-electrolyte.
Solve for ( m ): $$ \Delta T_b (\text{for KCl}) = \Delta T_f (\text{for urea}) $$ $$ 2 \times K_b \times 0.01 = K_f \times m $$ $$ m = 2 \times 0.01 $$ $$ m = 0.02 , \text{M} $$
Conclusion:
The concentration of the urea solution is $0.02 , \text{M}$, which corresponds to option C.
Final Answer: C
Upon complete combustion of 0.858 grams of compound $\mathrm{X}$, 2.63 grams of $\mathrm{CO}_2$ and 1.28 grams of $\mathrm{H}_2\mathrm{O}$ are obtained. What could be the minimum molecular mass of $\mathrm{X}$?
A 43 grams B 86 grams C 129 grams D 172 grams
To determine the minimum molecular weight of compound X from the given data, we can follow these steps:
Calculate Moles of $\mathrm{CO}_2$ and $\mathrm{H}_2\mathrm{O}$:
Moles of $\mathrm{CO}_2$: $$ n_{\mathrm{CO}_2} = \frac{2.63 \text{ grams}}{44 \text{ grams/mole}} = 0.06 \text{ moles} $$
Moles of $\mathrm{H}_2\mathrm{O}$: $$ n_{\mathrm{H}_2\mathrm{O}} = \frac{1.28 \text{ grams}}{18 \text{ grams/mole}} = 0.07 \text{ moles} $$
Since each molecule of $\mathrm{H}_2\mathrm{O}$ contains 2 atoms of Hydrogen: $$ \text{Moles of } \mathrm{H} = 2 \times 0.07 = 0.14 \text{ moles} $$
Convert the moles to mass:
Mass of Carbon: $$ \text{Mass of C} = 0.06 \text{ moles} \times 12 \text{ grams/mole} = 0.72 \text{ grams} $$
Mass of Hydrogen: $$ \text{Mass of H} = 0.14 \text{ moles} \times 1 \text{ gram/mole} = 0.14 \text{ grams} $$
Verify Oxygen Presence:
Since the mass of compound X is 0.858 grams and the sum of the mass of Carbon and Hydrogen adds up to: $$ 0.72 \text{ grams} + 0.14 \text{ grams} = 0.86 \text{ grams} $$
Therefore, there is no significant mass difference, indicating no Oxygen in the compound.
Determine the Empirical Formula:
Given the moles: $$ \text{C}_{0.06} \text{H}_{0.14} $$
When simplified: $$ \text{C}_3 \text{H}_7 $$
Calculate the Minimum Molecular Weight:
The empirical formula weight is: $$ ( 3 \times 12 ) + ( 7 \times 1 ) = 36 + 7 = 43 \text{ grams/mole} $$
Therefore, the minimum molecular weight of $\mathbf{X}$ is 43 grams/mole.
Final Answer:A
The final volume of the gaseous mixture obtained when 20 mL of NH$_3$ gas is mixed with 40 mL of slightly moist HCl gas under the same conditions of pressure and temperature is:
A - 100 mL B - 20 mL C - 20 mL D - 60 mL
The correct answer is B.
The balanced chemical reaction is: $$ \mathrm{NH}_{3(g)}+\mathrm{HCl}_{(g)} \rightarrow \mathrm{NH}_{4} \mathrm{Cl}_{(s)} $$
Let's analyze the situation at $t = 0$ and $t = t$:
$\mathrm{NH}_3$ (g) | $\mathrm{HCl}$ (g) | $\mathrm{NH}_4\mathrm{Cl}$ (s) | |
---|---|---|---|
at $t=0$ | 20 mL | 40 mL | 0 |
at $t=t$ | 0 | 20 mL | Solid |
Initially at $t=0$, we have 20 mL of $\mathrm{NH}_3$ gas and 40 mL of $\mathrm{HCl}$ gas.
During the reaction, $\mathrm{NH}_3$ completely reacts with an equivalent amount of $\mathrm{HCl}$, producing solid $\mathrm{NH}_4\mathrm{Cl}$ and leaving 20 mL of excess $\mathrm{HCl}$ gas.
Thus, the final volume of the gaseous mixture is 20 mL.
Final Answer: B
A manometer is connected to a flask filled with NH₃. As shown in the picture, the Hg levels in both arms are equal. Upon generating a spark in the flask, partial decomposition of NH₃ occurs as follows:
$$ 2 \mathrm{NH}_{3}(g) \rightarrow \mathrm{N}_>{2}(g) + 3 \mathrm{H}_{2}(g) $$
After the decomposition, there is a difference of 6 cm in the levels of both arms. The partial pressure of $\mathrm{H}_{2}$(g) at equilibrium is:
9 cm Hg
18 cm Hg
27 cm Hg
None of these
The correct answer is: $\mathbf{A}$
Dissociation equation: $$ 2 \mathrm{NH}_{3}(g) \rightarrow \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) $$
In the initial state, the mercury level is equal on both sides, which indicates that the initial pressure due to $\mathrm{NH}_3$ is 76 cm Hg.
If after dissociation, there is a 6 cm difference in the mercury level on both sides, this indicates a rise in pressure of 6 cm Hg.
We know that: $$ 2 \mathrm{NH}_{3}(g) \rightarrow \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) $$
Initial state: $$ \text{Partial pressure of } \mathrm{NH}_3 = 76 \text{ cm Hg} $$ $$ \text{Partial pressure of } \mathrm{N}_2 = 0 \text{ cm Hg} $$ $$ \text{Partial pressure of } \mathrm{H}_2 = 0 \text{ cm Hg} $$
At equilibrium: $$ 76 \text{ cm Hg} - 2x = \text{Pressure of } \mathrm{NH}_3 $$ $$ x = \text{Partial pressure of } \mathrm{N}_2 $$ $$ 3x = \text{Partial pressure of } \mathrm{H}_2 $$
Total increase in pressure: $$ 2x = 6 \text{ cm Hg} $$ Thus, $x = 3 \text{ cm Hg}$
Therefore, $$ \text{Partial pressure of } \mathrm{H}_{2} = 3 \times 3 \text{ cm Hg} = 9 \text{ cm Hg} $$
Final Answer: A
The vapor density of a mixture of $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ at $300 \mathrm{~K}$ is 38.3. The number of moles of $\mathrm{NO_2}$ in a 100-gram mixture is:
A 0.043
B 4.4
C 3.4
D 3.86
The correct answer is: D
Calculation of the Molecular Mass of the Mixture:
The molecular mass of the mixture can be calculated using the vapor density: $$ \text{Molecular mass} = 2 \times \text{vapor density} = 2 \times 38.3 = 76.6 $$
Mass Distribution in the Mixture:
Let the mass of $\mathrm{NO}_2$ be $\mathrm{x}$. Therefore, the mass of $\mathrm{N}_2\mathrm{O}_4$ will be $(100 - \mathrm{x})$.
Using the Mass-to-Moles Relationship:
Given, $$ \frac{x}{46} + \frac{100 - x}{92} = \frac{100}{76.6} $$
Solving for $\mathrm{x}$: $$ \frac{2x + 100 - x}{92} = \frac{100}{76.6} $$
$$ \frac{x + 100}{92} = \frac{100}{76.6} $$
$$ 76.6(x + 100) = 92 \times 100 $$
$$ 76.6x + 7660 = 9200 $$
$$ 76.6x = 1540 $$
$$ x = \frac{1540}{76.6} \approx 20.10 $$
Calculation of Moles of $\mathrm{NO}_2$:
The number of moles of $\mathrm{NO}_2$ is given by: $$ \text{Moles of } \mathrm{NO}_2 = \frac{20.10 \text{ g}}{46 \text{ g/mol}} \approx 0.437 \text{ moles} $$
A gaseous mixture contains 56 grams of $N_2$, 44 grams of $CO_2$, and 16 grams of $CH_4$. The total pressure of the mixture is 720 mm Hg. The partial pressure of methane is 180 atmospheres.
To find the partial pressure of methane ($ \mathrm{CH}_{4}$) in a gas mixture, we need to follow these steps:
Calculate the number of moles of each gas:
For $ \mathrm{N}_{2} $: $$ \text{Number of moles} = \frac{56 \text{ grams}}{28 \text{ g/mol}} = 2 \text{ moles} $$
For $ \mathrm{CO}_{2} $: $$ \text{Number of moles} = \frac{44 \text{ grams}}{44 \text{ g/mol}} = 1 \text{ mole} $$
For $ \mathrm{CH}_{4} $: $$ \text{Number of moles} = \frac{16 \text{ grams}}{16 \text{ g/mol}} = 1 \text{ mole} $$
Calculate the total number of moles: $$ \text{Total moles} = 2 \text{ (moles of } \mathrm{N}_{2} \text{)} + 1 \text{ (mole of } \mathrm{CO}_{2} \text{)} + 1 \text{ (mole of } \mathrm{CH}_{4} \text{)} = 4 \text{ moles} $$
Determine the mole fraction of $ \mathrm{CH}_{4} $: $$ \text{Mole fraction of } \mathrm{CH}_{4} = \frac{1 \text{ mole}}{4 \text{ moles}} = \frac{1}{4} $$
Calculate the partial pressure of $ \mathrm{CH}_{4} $:
The total pressure of the mixture is 720 mm Hg.
The partial pressure of $ \mathrm{CH}_{4} $ can be found as follows: $$ \text{Partial pressure of } \mathrm{CH}_{4} = \text{Mole fraction of } \mathrm{CH}_{4} \times \text{Total pressure} $$ $$ = \frac{1}{4} \times 720 \text{ mm Hg} = 180 \text{ mm Hg} $$
Hence, the partial pressure of methane ($ \mathrm{CH}_{4} $) is 180 mm Hg.
Final Answer: C
The molar specific heat capacities of $\text{H}_{2}\text{O}$, $\text{NH}_3$, $\text{CO}_{2}$, and $\text{O}_{2}$ are $647\text{ K}$, $405.6\text{ K}$, $304.10\text{ K}$, and $154.2\text{ K}$ respectively. If they are heated from $500\text{ K}$ to their respective molar specific heat capacities, then which gas will be the first to vaporize?_
A. $\text{H}_{2}\text{O}$
B.$\text{NH}_{3}$
C. $\text{CO}_{2}$
D. $\text{O}_{2}$
To determine which gas will vaporize first when heated from $500\text{ K}$, we need to consider the molar specific heat capacities of the gases given and their respective temperatures.
The molar specific heat capacities are:
$\text{H}_{2}\text{O}$: $647\text{ K}$
$\text{NH}_{3}$: $405.6\text{ K}$
$\text{CO}_{2}$: $304.10\text{ K}$
$\text{O}_{2}$: $154.2\text{ K}$
Since the initial temperature is $500 \text{ K}$, we can exclude water ($\text{H}_{2}\text{O}$) because its molar specific heat capacity ($647\text{ K}$) is higher than the starting temperature.
Now, we compare the remaining specific heat capacities. The temperatures they need to reach their molar specific capacities are:
$\text{NH}_{3}$: $405.6\text{ K}$
$\text{CO}_{2}$: $304.10\text{ K}$
$\text{O}_{2}$: $154.2\text{ K}$
Ammonia ($\text{NH}_{3}$), with a molar specific heat capacity of $405.6\text{ K}$, will be the first to vaporize because it exists between $500 \text{ K}$ and its specific heat capacity.
Thus, the correct answer is:
B. $\text{NH}_{3}$
___is used as a catalyst in the formation of ammonia from nitrogen and hydrogen.
Iron is used as a catalyst in the formation of ammonia from nitrogen and hydrogen.
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