Alcohols, Phenols and Ethers - Class 12 Chemistry - Chapter 7 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Alcohols, Phenols and Ethers | NCERT | Chemistry | Class 12
List 1 | List 2 | |
---|---|---|
P. | Aniline | Sulpha drug |
Q. | TNT | Solvent in Friedel Craft |
R. | Sulphanilamide | Explosive |
S. | Nitrobenzene | Used in Azo dyes |
A) P-1, Q-2, R-3, S-4
B) P-4, Q-3, R-2, S-1
C) P-2, Q-3, R-4, S-1
D) P-4, Q-3, R-1, S-2
The correct answer is Option D $P,4; Q,3; R,1; S,2$.
Aniline is utilized in azo-coupling reactions to synthesize various dyes, aligning it with option 4 from List-2.
TNT (Trinitrotoluene) is well-known for its use as an explosive, which corresponds to option 3.
Sulphanilamide belongs to a class known as sulpha drugs, aligning with option 1.
Nitrobenzene, on the other hand, is used as a solvent in Friedel-Crafts reactions, matching option 2.
Thus, the mappings for each chemical in List-1 to their respective uses in List-2 are accurately reflected in Option D.
Which of the following compounds will give a positive iodoform test?
A. Isopropyl alcohol
B. Propionaldehyde
C. Ethylphenyl ketone
D. Benzyl alcohol
The correct option is A. Isopropyl alcohol
Isopropyl alcohol (chemical formula $ \left(\mathrm{CH}_3\right)_2\mathrm{CHOH} $) gives a positive iodoform test because it can be easily oxidized to acetone $ \mathrm{CH}_3\mathrm{COCH}_3 $, which is a methyl ketone. Methyl ketones are required for the iodoform reaction to occur, therefore making isopropyl alcohol a compound that will test positive.
How can we convert ethanol to but-1-yne?
To convert ethanol to but-1-yne, follow the outlined reaction steps below:
Convert ethanol to ethyl chloride: This is achieved through the reaction of ethanol (ethyl alcohol) with thionyl chloride (SOCl₂). The reaction can be represented as: $$ \mathrm{C}_2\mathrm{H}_5\mathrm{OH} + \mathrm{SOCl}_2 \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{Cl} + \mathrm{SO}_2 + \mathrm{HCl} $$ Here, ethanol reacts with thionyl chloride to form ethyl chloride along with sulfur dioxide and hydrogen chloride as by-products.
Reaction of ethyl chloride with sodium acetylide to form but-1-yne: Ethyl chloride reacts with sodium acetylide (a compound derived from acetylene) to form but-1-yne. The reaction proceeds as follows: $$ \mathrm{C}_2\mathrm{H}_5\mathrm{Cl} + \text{Sodium acetylide} \rightarrow \text{but-1-yne} + \mathrm{NaCl} $$ In this step, ethyl chloride reacts with sodium acetylide, yielding but-1-yne and sodium chloride as the final products.
These reactions collectively transform ethanol into but-1-yne through a two-step process involving an initial substitution followed by a coupling with sodium acetylide.
Why is bimolecular dehydration not suitable for the preparation of ethyl methyl ether?
Bimolecular dehydration is not a suitable method for synthesizing ethyl methyl ether due to the following reasons:
-
Bimolecular dehydration involves the dehydrating action of concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) on alcohols. This process generally works by abstracting a hydrogen atom from one alcohol molecule and a hydroxyl group (-OH) from another, effectively removing a molecule of water. For example, when ethanol undergoes bimolecular dehydration, diethyl ether is typically formed:
$$ 2\text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_2\text{H}_5\text{O}\text{C}_2\text{H}_5 + \text{H}_2\text{O} $$
-
This method is suitable primarily for the formation of symmetrical ethers (di[alkyl] ethers). This specificity arises because the reaction ideally requires two identical alkyl groups from similar alcohol molecules, thus limiting the formation to structurally symmetrical ethers.
-
Ethyl methyl ether (an asymmetrical ether) is more efficiently synthesized using alternative methods such as the reaction between sodium ethoxide and methanol or sodium methoxide and ethanol. These methods allow for more controlled synthesis of the desired asymmetrical ether:
$$ \text{NaOCH}_2\text{CH}_3 + \text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_2\text{H}_5\text{OCH}_3 + \text{NaOH} $$
-
Additionally, using bimolecular dehydration for a mixture of methanol and ethanol under the influence of sulfuric acid at high temperatures would lead to a mixture of ethers (not just ethyl methyl ether). Possible products include dimethyl ether, ethyl methyl ether, and diethyl ether. Each of these ethers forms from different combinations of the original alcohols, complicating the purification process to isolate the desired ether.
To summarize, for the synthesis of ethyl methyl ether specifically, targeted methods like reactions with sodium alkoxides provide more straightforward and predictable results compared to bimolecular dehydration, which tends to yield mixtures when applied to mixtures of different alcohols.
For the complete combustion of ethanol, $\mathrm{C}{2}\mathrm{H}{5}\mathrm{OH}(\mathrm{I}) + 3\mathrm{O}{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}{2} + 2\mathrm{H}{2}\mathrm{O}(\mathrm{I})$, the amount of heat produced, as measured in the bomb calorimeter, is $1364.47 , \mathrm{kJ} , \mathrm{mol}^{-1}$ at $25^{\circ} \mathrm{C}$. Assuming ideality, the enthalpy of combustion, $\Delta{\mathrm{c}}H$, for the reaction will be $\left[R=8.314 , \mathrm{JK}^{-1} , \mathrm{mol}^{-1}\right]$.
(A) $-1366.95 , \mathrm{kJ} , \mathrm{mol}^{-1}$ (B) $1361.95 , \mathrm{kJ} , \mathrm{mol}^{-1}$ (C) $1460.50 , \mathrm{kJ} , \mathrm{mol}^{-1}$ (D) $-1350.50 , \mathrm{kJ} , \mathrm{mol}^{-1}$
The correct option is (A) $-1366.95 , \mathrm{kJ} , \mathrm{mol}^{-1}$.
The given chemical equation for the combustion of ethanol is:
$$ \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l}) + 3 \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{g}) + 3 \mathrm{H}_2\mathrm{O}(\mathrm{l}) $$
The amount of heat produced in a bomb calorimeter, which corresponds to the change in internal energy ($\Delta U$), is:
$$ \Delta U = -1364.47 , \mathrm{kJ} , \mathrm{mol}^{-1} $$
To find the enthalpy of combustion ($\Delta_{\mathrm{c}} H$), we use the relation: $$ \Delta H = \Delta U + \Delta n_{\mathrm{g}} R T $$ where:
- $\Delta H$ is the enthalpy change,
- $\Delta U$ is the internal energy change,
- $\Delta n_{\mathrm{g}}$ is the change in moles of gas between products and reactants,
- $R$ is the universal gas constant ($8.314 , \mathrm{J} , \mathrm{K}^{-1} , \mathrm{mol}^{-1}$),
- $T$ is the temperature in Kelvin.
For the reaction, the change in the number of moles of gas ($\Delta n_{\mathrm{g}}$) is calculated as follows: $$ \Delta n_{\mathrm{g}} = (2 , \text{moles of } CO_2) - (3 , \text{moles of } O_2) = -1 $$
Converting the temperature from Celsius to Kelvin: $$ T = 25°C + 273.15 = 298.15 , \mathrm{K} $$
Calculating the enthalpy change: $$ \Delta H = (-1364.47 , \mathrm{kJ} , \mathrm{mol}^{-1}) + \left(-1 \times 8.314 , \mathrm{J} , \mathrm{K}^{-1} , \mathrm{mol}^{-1} \times 298.15 , \mathrm{K}/1000\right) $$ $$ \Delta H = -1364.47 , \mathrm{kJ} , \mathrm{mol}^{-1} - 2.477 , \mathrm{kJ} , \mathrm{mol}^{-1} \approx -1366.947 , \mathrm{kJ} , \mathrm{mol}^{-1} $$
Thus, the enthalpy of combustion, $\Delta_{\mathrm{c}} H$, for ethanol is approximately $-1366.95 , \mathrm{kJ} , \mathrm{mol}^{-1}$.
The correct IUPAC name of the compound given below is:
A. 3-ethyl-3-isopropyl-4-tertiarybutylhexane B. 3,4,4-triethyl-2,2,5-trimethylhexane C. 3,4-diethyl-4-isopropyl-2,2-dimethylhexane D. 3,3,4-triethyl-2,5,5-trimethylhexane
The correct IUPAC name of the compound is Option B: 3,4,4-triethyl-2,2,5-trimethylhexane.
For naming organic compounds, the longest continuous carbon chain is identified as the parent chain. This chain is then used as the baseline to determine the locants (positions) of substituents attached to it.
Key considerations for choosing the correct answer include finding the longest chain and properly identifying and positioning the substituents. In this case, the longest chain consists of 6 carbon atoms (hexane), and the substituents are ethyl and methyl groups located at positions 3, 4, 2, and 5 as described in Option B. Each position and type of substituent is carefully noted according to IUPAC nomenclature rules.
The molecular mass of $\mathrm{CH}{3}\mathrm{CH}{2}\mathrm{OH}$ is
A) $50 \mathrm{u}$
B) $35 \mathrm{u}$
C) $70 \mathrm{u}$
D) $46 \mathrm{u}$
Solution
The correct answer is $\mathbf{D)} , 46 \mathrm{u}$.
To determine the molecular mass of $\mathrm{CH}{3}\mathrm{CH}{2}\mathrm{OH}$, we sum the atomic masses of all its atoms:
- Carbon (C) has an atomic mass of $12 \mathrm{u}$. Since there are 2 carbon atoms, this contributes $2 \times 12 \mathrm{u} = 24 \mathrm{u}$.
- Hydrogen (H) has an atomic mass of $1 \mathrm{u}$. With 6 hydrogen atoms, this adds up to $6 \times 1 \mathrm{u} = 6 \mathrm{u}$.
- Oxygen (O) has an atomic mass of $16 \mathrm{u}$. There is 1 oxygen atom, contributing $1 \times 16 \mathrm{u} = 16 \mathrm{u}$.
Therefore, the molecular mass of $\mathrm{CH}{3}\mathrm{CH}{2}\mathrm{OH}$ is: $$ (2 \times 12 \mathrm{u}) + (6 \times 1 \mathrm{u}) + (1 \times 16 \mathrm{u}) = 24 \mathrm{u} + 6 \mathrm{u} + 16 \mathrm{u} = 46 \mathrm{u}. $$ This calculation confirms the correct choice: Option D, $46 \mathrm{u}$.
Dettol consists of:
A) cresol + chlorobenzene
B) chlorobenzene + terpineol
C) terpineol + chloroxylenol
D) cresol + terpineol
The correct answer is C) terpineol + chloroxylenol.
Dettol consists primarily of 4.8% w/v chloroxylenol, alongside terpineol and absolute alcohol.
An ester (A) with molecular formula $\mathrm{C}{9}\mathrm{H}{10}\mathrm{O}{2}$ was treated with excess of $\mathrm{CH}{3}-\mathrm{Mg} \mathrm{Br}$ and the complex so formed was treated with $\mathrm{H}{2}\mathrm{SO}{4}$ to give an olefin (B). Ozonolysis of (B) gave a ketone with molecular formula $\mathrm{C}{8}\mathrm{H}{8}\mathrm{O}$ which shows the iodoform test. The structure of (A) is -
(A) $\mathrm{C}{6}\mathrm{H}{5}\mathrm{COOC}{2}\mathrm{H}{5}$
(B) $\mathrm{CH}{3}\mathrm{COCH}{2}\mathrm{COC}{6}\mathrm{H}{5}$
(C) $\mathrm{P}\mathrm{CH}{3}\mathrm{O}-\mathrm{C}{6}\mathrm{H}{4}-\mathrm{COCH}{3}$
(D) $\mathrm{C}{6}\mathrm{H}{5}\mathrm{COOC}{6}\mathrm{H}{5}$
The solution to identify the correct structure of ester (A) can be explained through a series of chemical reactions and deductions based on the given information:
-
Structure Analysis:
- Ester (A) has the molecular formula $ \mathrm{C}{9}\mathrm{H}{10}\mathrm{O}_{2} $. This indicates the presence of 9 carbons, 10 hydrogens, and 2 oxygens.
-
Reaction with $\mathrm{CH}_{3}-\mathrm{Mg} \mathrm{Br}$:
- When reacted with excess methylmagnesium bromide ($\mathrm{CH}_3\mathrm{MgBr}$), the ester would generally form a tertiary alcohol after protonation with $\mathrm{H}_2\mathrm{SO}_4$.
-
Formation of Olefin (B):
- The treatment with $\mathrm{H}_2\mathrm{SO}_4$ leads to the elimination reaction, forming an olefin (B).
-
Ozonolysis of Olefin:
- Ozonolysis of the olefin, followed by reductive workup, typically breaks the double bond to form carbonyl compounds. Given this olefin results in a ketone $ \mathrm{C}{8}\mathrm{H}{8}\mathrm{O} $ that shows iodoform reaction, the presence of a $\mathrm{CH}_3\mathrm{CO}-$ group is inferred.
-
Iodoform Test:
- Only ketones (and some alcohols) with a $\mathrm{CH}_3\mathrm{CO}-$ group next to a methyl group ($\mathrm{-CH}_3$) can give a positive iodoform test. This directs towards a part of the molecule that must be $\mathrm{CH}_3\mathrm{CO}\mathrm{CH}_3$ post ozonolysis.
Given these details:
- Option (A) $\mathrm{C}{6}\mathrm{H}{5}\mathrm{COOC}{2}\mathrm{H}{5}$ (phenyl ethanoate) appears promising as the ethyl group ($\mathrm{C}_2\mathrm{H}_5$) can be converted into the required $\mathrm{CH}_3\mathrm{CO}\mathrm{CH}_3$ structure upon ozonolysis.
- The other options do not conveniently lead to a structure that simplifies into $\mathrm{CH}_3\mathrm{CO}\mathrm{CH}_3$ upon undergoing the prescribed reactions or ozonolysis.
Conclusion: The ester (A) with structure $\mathrm{C}_6\mathrm{H}_5\mathrm{COOC}_2\mathrm{H}_5$ (option A) is the correct answer as it fits well with the reactions and the molecular formulas provided.
Number of structural isomers possible for $\mathrm{C}{6}\mathrm{H}{14}$ and $\mathrm{C}{7}\mathrm{H}{16}$ are respectively?
A) 5 and 7
B) 5 and 9
C) 6 and 7
D) 4 and 8
The correct answer is Option B: 5 and 9.
Both $\mathrm{C}_6\mathrm{H}_{14}$ and $\mathrm{C}_7\mathrm{H}_{16}$ are alkanes, characterized by the general formula $\mathrm{C}_n\mathrm{H}_{2n+2}$. Each of these compounds can form multiple structural isomers due to differing configurations of carbon atoms.
Isomers of $\mathrm{C}_6\mathrm{H}_{14}$:
The structural isomers of hexane ($\mathrm{C}_6\mathrm{H}_{14}$) are as follows:
- n-hexane
- 2-methylpentane (isohexane)
- 3-methylpentane
- 2,2-dimethylbutane
- 2,3-dimethylbutane
These compounds total to 5 isomers.
Isomers of $\mathrm{C}_7\mathrm{H}_{16}$:
The structural isomers of heptane ($\mathrm{C}_7\mathrm{H}_{16}$) include:
- n-heptane
- 2-methylhexane
- 3-methylhexane
- 2,2-dimethylpentane
- 3,3-dimethylpentane
- 2,3-dimethylpentane
- 2,4-dimethylpentane
- 3-ethylpentane
- 2,2,3-trimethylbutane
These configurations total to 9 isomers.
Therefore, the number of structural isomers possible for $\mathrm{C}_6\mathrm{H}_{14}$ and $\mathrm{C}_7\mathrm{H}_{16}$ are respectively 5 and 9, as indicated in Option B.
What are polar and nonpolar solvents?
Polar solvents possess large dipole moments; this characteristic arises because they contain bonds between atoms that exhibit significant differences in electronegativities, such as oxygen and hydrogen.
Nonpolar solvents, on the other hand, feature bonds between atoms with comparable electronegativities, typical examples include carbon and hydrogen (often found in hydrocarbons like gasoline).
Two varieties of coffee worth Rs. 126 per $\mathrm{kg}$ and Rs. 130 per $\mathrm{kg}$ are mixed with a third variety in the ratio 1:1:2. If the resulting mixture is worth Rs. 172 per $\mathrm{kg}$, then find the price (Rs./kg) of the third variety of the coffee.
(A) 216 (B) 198 (C) 195 (D) 207 (E) Can't be determined
To determine the price per kilogram of the third variety of coffee, we calculate as follows:
The coffee varieties priced at Rs. 126 and Rs. 130 per kg are mixed in a 1:1 ratio, giving us: $$ \text{Average price} = \frac{126 + 130}{2} = 128 \text{ Rs./kg} $$ Now this average priced variety and the third variety are then mixed in a 2:2 ratio (simplified to 1:1) to result in a mixture priced at Rs. 172 per kg.
We apply the Alligation Rule, a method used to solve problems based on mixture and solution: $$ \text{Mean Price} = 172 \text{ Rs./kg} \ \text{Cheaper Price} = 128 \text{ Rs./kg} \ \text{Dearer Price (x)} = ? \ $$ The alligation formula calculates the difference between the mean price and each of the other price points: $$ \begin{array}{c|c} \text{Cheaper} & \text{Dearer} \ \hline 128 & x \ \hline & 172 \ \end{array} $$ $$ \text{Difference with Dearer} = x - 172 \ \text{Difference with Cheaper} = 172 - 128 = 44 \ $$ Using alligation, the differences must be equal because of the 1:1 ratio, so: $$ x - 172 = 44 \ x = 172 + 44 = 216 $$ Therefore, the price of the third coffee variety is Rs. 216 per kg, and the correct answer is (A) 216.
Tincture of iodine is a mixture of two materials: iodine and alcohol. One of the materials has a property that its solid form can be converted directly into vapors on heating by a process called X. Identify the material that sublimes and process X.
A) Volatile component: Iodine, X: Distillation
B) Volatile component: Alcohol, X: Fractional distillation
C) Volatile component: Alcohol, X: Melting
D) Volatile component: Iodine, X: Sublimation
Solution
The correct option is D
- Volatile Component: Iodine
- Process X: Sublimation
Key Points:
- Sublimation is the process where a substance converts directly from the solid state to the gaseous state without passing through the liquid state.
- A substance that can sublime, such as iodine in this scenario, is known as a sublimate and is also volatile as it tends to evaporate or transition to a gas easily.
- In a tincture of iodine, which is composed of iodine mixed with alcohol, iodine is identified as the substance that sublimes, making it the volatile component in this mixture.
Arrange Ethane, Ethene, and Ethyne in decreasing order of the bond strength between the two carbon atoms in the given hydrocarbons.
A) Ethene > Ethane > Ethyne
B) Ethane > Ethene > Ethyne
C) Ethyne = Ethene = Ethane
D) Ethyne > Ethene > Ethane
The correct answer is D) Ethyne > Ethene > Ethane.
Bond strength between carbon atoms typically increases with the number of electron pairs shared in the bond. The three compounds—Ethyne, Ethene, and Ethane—differ in the types of bonds between the carbon atoms:
- Ethyne (also known as acetylene) has a triple bond, sharing six electrons.
- Ethene (ethylene) features a double bond, with four electrons shared.
- Ethane contains a single bond, with only two electrons shared.
Given that more shared electrons generally result in a stronger bond, the decreasing order of bond strength among these hydrocarbons is:
$$ \text{Ethyne} > \text{Ethene} > \text{Ethane} $$
Each bond type (triple, double, and single) corresponds to higher electron sharing, and consequently, stronger bonds in Ethyne. Thus, option D accurately reflects the correct order based on bond strength.
(i) What is the major mononitration product of $\mathrm{Ph}-\mathrm{COO}-\mathrm{Ph}$ when it is reacted in a nitrating mixture? (ii) Which of the following are formed when $\mathrm{Ph}-\mathrm{COOH}$ reacts with $\mathrm{CH}_{3} \mathrm{MgI}$?
(i) $\mathrm{Ph}-\mathrm{COO}-\mathrm{Ph}-\mathrm{NO}_{2}$ - ortho product (ii) Phenol
B (i) $\mathrm{Ph}-\mathrm{COO}-\mathrm{Ph}-\mathrm{NO}_{2}-$ para product (ii) Benzene
C (i) $\mathrm{Ph}-\mathrm{COO}-\mathrm{Ph}-\mathrm{NO}{2}$ - para product (ii) $\mathrm{Ph}-\mathrm{CH}{3}$
D (i) $\mathrm{Ph}-\mathrm{COO}-\mathrm{Ph}-\mathrm{NO}_{2}$ - ortho product (ii) Methane
Solution
The correct answer is option D:
-
For part (i), the major mononitration product of $\mathbf{Ph-COO-Ph}$ under a nitrating mixture is ortho-nitration yielding $\mathbf{Ph-COO-Ph-NO_2}$ (ortho product). This occurs because of the directing effects of the functional groups within the molecule.
-
For part (ii), when $\mathbf{Ph-COOH}$ reacts with $\mathbf{CH_3MgI}$, $\mathbf{Methane (CH_4)}$ is formed. This happens via the reaction of the acid $\mathbf{Ph-COOH}$ with the Grignard reagent $\mathbf{CH_3MgI}$, which typically leads to the formation of a new carbon-carbon bond, but in this specific reaction, it results in the formation of methane through decarboxylation.
Hence, the full selections for option D are correct: (i) $\mathbf{Ph-COO-Ph-NO_2}$ - ortho product (ii) Methane
The suitable reagent for the following conversion? i.e. $\mathrm{CH}{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}{3} \rightarrow ^{\mathrm{R}}$ trans-alkene:
(A) $\mathrm{Cr}{2} \mathrm{O}{3} / \mathrm{Al}{2} \mathrm{O}{3}$
(B) $\mathrm{H}{2} / \mathrm{Pd} / \mathrm{CaCO}{3}$
(C) $\mathrm{Na} / \mathrm{NH}_{3}(\mathrm{I})$
(D) $\mathrm{Na} /$ dry ether
Solution
The correct option is (C) $\mathrm{Na} / \mathrm{NH}_{3}(\mathrm{l})$.
Using the reagent $\mathrm{Na} / \mathrm{NH}_{3}(\mathrm{l})$, which is utilized in the Birch reduction, will successfully convert an alkyne into a trans-alkene. This specific reduction selectively reduces alkynes to their corresponding trans-alkenes by adding hydrogen atoms to opposite sides of the triple bond.
Which of the following compounds is used as an antifreeze in automobile radiators?
A) Methyl alcohol
B) Glycol
C) Nitrophenol
D) Ethyl alcohol
Correct Option: B) Glycol
Glycol, specifically ethylene glycol, is utilized as an antifreeze in automobile radiators due to its high boiling point and its excellent miscibility with water ($\text{H}_2\text{O}$). These properties ensure that the radiator fluid does not freeze under low temperatures and effectively transfers heat.
When ethyl iodide and propyl iodide react with $\mathrm{Na}$ in the presence of ether, they form:
A) One alkane
B) Two alkanes
C) Four alkanes
D) Three alkanes
The correct option is D) Three alkanes.
The given reaction involves ethyl iodide ($\mathrm{C}{2}\mathrm{H}{5}\mathrm{I}$) and propyl iodide ($\mathrm{C}{3}\mathrm{H}{7}\mathrm{I}$) reacting with sodium ($\mathrm{Na}$) in the presence of ether to undergo a coupling reaction. This type of reaction is called the Wurtz reaction, which generally leads to the formation of new alkanes by forming new carbon-carbon bonds between alkyl halides.
In this specific scenario, three different alkanes can be formed due to the possible combinations of the two alkyl groups:
-
Pentane ($\mathrm{C}{2}\mathrm{H}{5}-\mathrm{C}{3}\mathrm{H}{7}$) can form when one molecule of ethyl iodide reacts with one molecule of propyl iodide: $$ \mathrm{C}{2}\mathrm{H}{5}\mathrm{I} + 2 \mathrm{Na} + \mathrm{C}{3}\mathrm{H}{7}\mathrm{I} \xrightarrow[\text{Ether}]{\text{Dry}} \mathrm{C}{2}\mathrm{H}{5}-\mathrm{C}{3}\mathrm{H}{7} + 2 \mathrm{NaI} $$
-
Butane ($\mathrm{C}{2}\mathrm{H}{5}-\mathrm{C}{2}\mathrm{H}{5}$) results from the coupling of two ethyl iodide molecules: $$ \mathrm{C}{2}\mathrm{H}{5}\mathrm{I} + 2 \mathrm{Na} + \mathrm{C}{2}\mathrm{H}{5}\mathrm{I} \xrightarrow[\text{Ether}]{\text{Dry}} \mathrm{C}{2}\mathrm{H}{5}-\mathrm{C}{2}\mathrm{H}{5} + 2 \mathrm{NaI} $$
-
Hexane ($\mathrm{C}{3}\mathrm{H}{7}-\mathrm{C}{3}\mathrm{H}{7}$) is created when two propyl iodide molecules react together: $$ \mathrm{C}{3}\mathrm{H}{7}\mathrm{I} + 2 \mathrm{Na} + \mathrm{C}{3}\mathrm{H}{7}\mathrm{I} \xrightarrow[\text{Ether}]{\text{Dry}} \mathrm{C}{3}\mathrm{H}{7}-\mathrm{C}{3}\mathrm{H}{7} + 2 \mathrm{NaI} $$
Thus, the reaction between ethyl iodide and propyl iodide with sodium in the presence of ether can produce three different alkanes: pentane, butane, and hexane.
The monomer of polyester is:
A) ester
B) rubber
C) vinyl chloride
D) starch
The correct answer is A) ester.
Polyester is derived from the term "poly+ester," indicating that the polymer is formed from repeating units of a chemical compound known as ester.
Study the following information carefully and answer the questions given below it. Nine friends $L, M, N, O, P, Q, R, S$, and I are sitting in a circle facing the center. T sits fifth to the right of $R$. $N$ is not an immediate neighbor of either $R$ or T. M sits between $S$ and P. N sits fourth to the left of P. O sits second to the right of Q. S is not an immediate neighbor of T. If $S: Q$, then $N:$?
A $\mathrm{R}$
B $\mathrm{O}$
C $\mathrm{L}$
D $\mathrm{T}$
E None of these
The answer to the question is Option C: L.
Here's how the seating arrangement and logic work out based on the clues provided:
- Starting with $R$, $T$ is seated fifth to the right of $R$.
- $N$ is not adjacent to $R$ or to $T$.
- $M$ is seated between $S$ and $P$.
- Taking count from $P$, $N$ is seated fourth to the left of $P$.
- With $Q$ as a reference, $O$ sits second to the right of $Q$.
- $S$ is not directly next to $T$.
Given the pattern of seating, if the position held by $S$ relative to $Q$ is observed (perhaps being opposite each other), and the question is about a similar relational position for $N$, it turns out that $N$ and $L$ hold corresponding positions. This pairing of positions leads us to conclude that the answer is L.
Reduction of $\mathrm{CH}{3}\mathrm{CH}{2}\mathrm{COCH}{3}$ to $\mathrm{CH}{3}\mathrm{CH}{2}\mathrm{CH}{2}\mathrm{CH}_{3}$ can be carried out with:
A. $\mathrm{H}_{2}/\mathrm{Pd}$
B. $\mathrm{Zn}-\mathrm{Hg}/\mathrm{HCl}$ C. $\mathrm{NaBH}_{4}$
D. $\mathrm{LiAlH}_{4}$
Solution
The correct answer is Option B: $\mathrm{Zn}-\mathrm{Hg}/\mathrm{HCl}$. This process utilizes the Clemmensen reduction, which specifically employs the reagents $\mathrm{Zn}-\mathrm{Hg}/\mathrm{HCl}$ to reduce the ketone group in $\mathrm{CH}{3}\mathrm{CH}{2}\mathrm{COCH}{3}$ to a methylene group, resulting in $\mathrm{CH}{3}\mathrm{CH}{2}\mathrm{CH}{2}\mathrm{CH}_{3}$. This method is effective in reducing carbonyl groups to methylene groups, particularly in cases where the ketone is within a non-sensitive chemical environment.
Assertion: Phenol forms 2, 4, 6-tribromophenol on treatment with $\mathrm{Br}_{2}$ in carbon disulphide at $273 \mathrm{~K}$. Reason: Bromine polarizes in carbon disulphide.
(a) Assertion and reason both are correct, and reason is a correct explanation of assertion. (b) Assertion and reason are both wrong statements. (c) Assertion is a correct statement, but the reason is a wrong statement. (d) Assertion is a wrong statement, but the reason is a correct statement. (e) Both assertion and reason are correct statements, but the reason is not a correct explanation of the assertion.
Answer: (b) Assertion and reason both are wrong statements.
Correct Assertion (A): Phenol forms 2, 4, 6-tribromophenol upon treatment with bromine in water, not in carbon disulfide. In phenols, the polarization of bromine can occur even in the absence of a Lewis acid, facilitating a reaction that typically leads to bromination in the ortho and para positions of the phenol ring.
Correct Reason (R): In water, phenol transforms into the phenoxide ion, which is a more reactive species. This ion significantly activates the ring towards the electrophilic substitution reaction, making the bromination process more effective.
When we make a binary solution of $\mathrm{HNO}{3}$ in $\mathrm{H}{2} \mathrm{O}$, deviation from Raoult's law is observed. Which of the following option(s) is/are correct?
A $\Delta \mathrm{H}_{\text {mixing}} < 0$
B $\Delta \mathrm{V}_{\text {mixing}} < 0$
C $\mathrm{P}{\mathrm{H}{2} \mathrm{O}} < \mathrm{P}{\mathrm{H}{2} \mathrm{O}}^{\mathrm{O}} \mathrm{X}{\mathrm{H}{2} \mathrm{O}}$
D $\Delta \mathrm{S}_{\text {mixing}} < 0$
The question involves understanding how a solution of $\mathrm{HNO}_3$ in $\mathrm{H}_2\mathrm{O}$ displays deviations from Raoult's Law. Given the options, let's explore which are correct based on the properties of mixing and deviations:
-
Option A: $$\Delta \mathrm{H}_{\text{mixing}} < 0$$
- This option is correct. The enthalpy of mixing, $\Delta \mathrm{H}_{\text{mixing}}$, is less than zero. This is indicative of an exothermic process, meaning heat is released when $\mathrm{HNO}_3$ and $\mathrm{H}_2\mathrm{O}$ mix. This is consistent with the extensive hydration of $\mathrm{H^+}$ ions, a process which typically releases heat.
-
Option B: $$\Delta \mathrm{V}_{\text{mixing}} < 0$$
- This option is also correct. The volume of mixing, $\Delta \mathrm{V}_{\text{mixing}}$, being negative suggests that the total volume of the solution after mixing is less than the sum of the volumes of pure $\mathrm{HNO}_3$ and $\mathrm{H}_2\mathrm{O}$ before mixing. This typically occurs due to the close packing of molecules and significant interactions between $\mathrm{HNO}_3$ and water molecules.
-
Option C: $$\mathrm{P}_{\mathrm{H}2\mathrm{O}} < \mathrm{P}{\mathrm{H}2\mathrm{O}}^\circ \mathrm{X}{\mathrm{H}_2\mathrm{O}}$$
- This option is correct and illustrates a deviation from Raoult's Law, where $\mathrm{P}_{\mathrm{H}2\mathrm{O}}$ represents the partial vapor pressure of water in the solution, $\mathrm{P}{\mathrm{H}2\mathrm{O}}^\circ$ is the vapor pressure of pure water, and $\mathrm{X}{\mathrm{H}_2\mathrm{O}}$ is the mole fraction of water in the solution. The inequality shows that the vapor pressure of water in the mixture is less than what Raoult's Law predicts, indicating a negative deviation which is common in solutions with strong molecular interactions.
In light of this analysis, options A, B, and C are correct due to the negative deviation from ideality, characterized by exothermic enthalpy, decreased volume on mixing, and lower vapor pressure of water than expected from Raoult's Law.
Note that Option D is not provided as correct in the solution, suggesting that the entropy of mixing, $\Delta \mathrm{S}_{\text{mixing}}$, is not less than zero but rather is expected to be positive, reflecting increased disorder upon forming the solution.
How to find the total number of isomers in a given compound?
To determine the total number of isomers, especially in organic compounds, understanding the distinction between structural and stereo isomers is crucial. However, no direct formula exists for calculating structural isomers.
Stereoisomers Calculation
When it comes to stereoisomers, which include both geometric and optical isomers, the calculation depends primarily on the number of chiral carbons in the molecule. Chiral carbons are carbons bonded to four different groups.
Consider the following scenarios:
-
Asymmetric Ends:
- If the molecule has asymmetric ends, then the number of stereoisomers can be determined as: $$ 2^n $$ where $n$ is the number of chiral carbons. All these isomers are optically active.
-
Symmetric Ends:
-
Case 2a: n is even
- Number of optically active forms: $$ 2^{n-1} $$
- Number of meso forms (achiral compounds): $$ 2^{0.5n - 1} $$
- Total Number of Stereoisomers (sum of optically active and meso forms): $$ 2^{n-1} + 2^{0.5n - 1} $$
-
Case 2b: n is odd
- Number of optically active forms: $$ 2^{n-1} $$
- Number of meso forms: $$ 2^{0.5n - 0.5} $$
- Total Number of Stereoisomers: $$ 2^{n-1} + 2^{0.5n - 0.5} $$
- Note: For odd $n$, the number of optically active forms will be the total minus meso forms.
-
This approach helps us systematically determine the possible isomers focusing on how symmetry and chirality impact isomer count in molecules.
Which of the following is not a usage of ethanol?
A) used as preservatives
B) solvent for medicines
C) used as fuel
D) manufacturing of paint
The correct answer is D) manufacturing of paint.
Ethanol is used in numerous applications, but there are some common purposes it serves including:
- Preservative purposes
- Acting as a solvent for medicines
- Being used as a fuel
Additionally, ethanol is utilized as a solvent in the manufacturing of varnishes and perfumes. However, it is specifically methanol, another type of alcohol, which is highly toxic and used in the manufacturing of paint, not ethanol. Thus, ethanol's use in paint manufacturing is not typical, making option D the incorrect choice in this context.
What are polar protic and polar aprotic solvents?
Polar and Protic vs. Aprotic Solvents
Polar Solvent
A polar solvent is characterized by a non-zero net dipole moment. This net dipole moment arises due to the presence of partial charges, which occur when two elements with different electronegativities form a covalent bond.
Protic Solvent
A protic solvent is any solvent that contains a labile H+ (a hydrogen ion that can be easily donated). Typically, these solvents include hydroxyl or amine groups. These solvents readily donate protons to reagents.
Polar Protic Solvent
Combining the definitions, a polar protic solvent is one that not only has a net dipole moment but also has the capability of releasing H+ ions.
Examples include:
Water
Acetic acid
Methanol
Note: Every protic solvent is a polar solvent, but not every polar solvent is a protic solvent.
Polar Aprotic Solvent
A polar aprotic solvent consists of polar molecules but does not contain labile H+ ions. These solvents do not have acidic hydrogens and are particularly useful in discussing reaction mechanisms.
Examples include:
Dichloromethane (DCM)
Tetrahydrofuran (THF)
Dimethylsulfoxide (DMSO)
Acetone
What is the specific volume of castor oil of density 956 $\mathrm{kgm}^{-3}$?
A. $1.05 \times 10^{-3} \mathrm{m}^{3}\mathrm{~kgm}^{-3}$
B. $1.05 \times 10^{-3} \mathrm{m}^{3}\mathrm{~kg}^{-1}$
C. $1.25 \times 10^{-3} \mathrm{~kg}^{-1}$
D. $9.56 \times 10^{-3} \mathrm{kgm}^{-3}$
To determine the specific volume of castor oil given its density, you need to understand the relationship between density and specific volume. Specific volume is defined as the volume per unit mass, or more simply, it is the reciprocal of density.
Here’s the step-by-step solution:
Identify the given density:
The density of castor oil is provided as: $$ \rho = 956 , \text{kg/m}^3 $$
Calculate the specific volume:
Specific volume ( v ) is given by the reciprocal of density: $$ v = \frac{1}{\rho} $$
Hence, $$ v = \frac{1}{956 , \text{kg/m}^3} $$
Perform the division:
Calculate the value: $$ v \approx 1.046 \times 10^{-3} , \text{m}^3/\text{kg} $$
Round to the given precision:
For comparison with the provided options, round the result appropriately: $$ v \approx 1.05 \times 10^{-3} , \text{m}^3/\text{kg} $$
So, the specific volume of castor oil with a density of 956 $\text{kg/m}^3$ is approximately $1.05 \times 10^{-3} , \text{m}^3/\text{kg}$.
Therefore, the correct answer is:
Option B: $1.05 \times 10^{-3} , \text{m}^3/\text{kg}$
The compound that is the most difficult to protonate is:
water ($\mathrm{H}_{2}\mathrm{O}$)
methyl alcohol ($\mathrm{H}_{3}\mathrm{C}-\mathrm{OH}$)
dimethyl ether $\mathrm{H}{3}\mathrm{C}-\mathrm{O}-\mathrm{CH}{3}$
phenol ($\mathrm{Ph}-\mathrm{OH}$)
The compound that is the most difficult to protonate is option D: phenol ($\mathrm{Ph}-\mathrm{OH}$).
Explanation:
In phenol:
The lone pair of electrons on the oxygen atom is involved in resonance with the aromatic ring.
This delocalization imparts a partial positive charge on the oxygen.
As a result, the incoming proton ($\mathrm{H}^+$) finds it challenging to attack the oxygen.
Therefore, phenol is the compound that is most difficult to protonate.
Chloroform is prepared by the use of bleaching powder. The organic compound taken is:
A) $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$ B) $\mathrm{CH}_{3}\mathrm{CHO}$ C) $\mathrm{CCl}_{3}\mathrm{CHO}$ D) $\mathrm{CH}_{3}\mathrm{COOH}$
A. A,B,C
B. B,C,D
C. A,C,D
D. A,B,D
The question asks: "Chloroform is prepared by the use of bleaching powder. The organic compound taken is:"
Given options: A) $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$ (Ethyl Alcohol) B) $\mathrm{CH}_{3}\mathrm{CHO}$ (Acetaldehyde) C) $\mathrm{CCl}_{3}\mathrm{CHO}$ (Trichloroacetaldehyde) D) $\mathrm{CH}_{3}\mathrm{COOH}$ (Acetic Acid)
The correct options are A, C, and D.
Explanation:
The formation of chloroform from bleaching powder involves several steps and specific organic compounds. Here is the detailed breakdown:
Bleaching powder reaction: $$ \mathrm{Ca(OCl)_2 + H_2O \rightarrow Ca(OH)_2 + 2 Cl_2} $$
In this reaction, bleaching powder ($\mathrm{Ca(OCl)_2}$) mixed with water releases chlorine gas.
Oxidation of Ethyl Alcohol: Ethyl alcohol ($ \mathrm{C_2H_5OH} $) reacts with the liberated chlorine gas. Chlorine acts as an oxidizing agent and converts ethyl alcohol into acetaldehyde ($ \mathrm{CH_3CHO} $).
Formation of Trichloroacetaldehyde: Acetaldehyde ($ \mathrm{CH_3CHO} $) further reacts with more chlorine, forming trichloroacetaldehyde ($ \mathrm{CCl_3CHO} $).
Reaction with Calcium Hydroxide: Trichloroacetaldehyde ($ \mathrm{CCl_3CHO} $) finally reacts with calcium hydroxide ($ \mathrm{Ca(OH)_2} $) to produce chloroform ($ \mathrm{CHCl_3} $) and calcium formate ($ \mathrm{HCOOCH_2Ca} $).
$$ 2 \mathrm{CHCl_3} + \mathrm{HCOOCH_2Ca} $$
Therefore, the essential organic compounds involved in this process are:
Ethyl Alcohol ($ \mathrm{C_2H_5OH} $)
Trichloroacetaldehyde ($ \mathrm{CCl_3CHO} $)
Acetic Acid ($ \mathrm{CH_3COOH} $)
Correct answer: A) $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$, C) $\mathrm{CCl}_{3}\mathrm{CHO}$, and D) $\mathrm{CH}_{3}\mathrm{COOH}$
So, options A, C, and D are indeed correct for the preparation of chloroform using bleaching powder.
The strength of H₂O₂ is expressed by various terms such as molarity, normality % (w/v), volume strength, etc. The meaning of "$10$ V H₂O₂" is that 1 volume of H₂O₂ upon decomposition gives $10$ volumes of O₂ at 1 atmospheric pressure and $273$ K temperature, or 1 liter of H₂O₂ provides $10$ liters of O₂ at 1 atmospheric pressure and $273$ K temperature. The decomposition of H₂O₂ occurs as follows: H₂O₂ (aqueous) $\rightarrow$ H₂O (liquid) $+\frac{1}{2}$ O₂ (gas).
H₂O₂ can act both as an oxidizing agent and a reducing agent. When it acts as an oxidizing agent, it converts into H₂O, whereas as a reducing agent, it releases O₂. In both conditions, the value of the factor $ \mathrm{n} $ is $2$. The normality of the H₂O solution $ = 2 \times $ the molarity of the H₂O₂ solution. A $40$ gram sample of $\text{Ba(MnO₄)₂}$ with impurities (a.d.r. $= 375$) reacts with $125$ mL of "33.6 V" H₂O₂ in an acidic medium. The % purity of the sample is:
A $28.12%$
B $70.31%$
C $85%$
D None of these
The correct answer is: B
The solution terminology can be detailed as follows:
Step 1: Equating the milliequivalents of $\mathrm{Ba}\left(\mathrm{MnO}_4\right)_2$ with $\mathrm{H}_2\mathrm{O}_2$: $$ \text{Milliequivalents of } \mathrm{Ba}\left(\mathrm{MnO}_4\right)_2 = \text{Milliequivalents of } \mathrm{H}_2 \mathrm{O}_2 $$
Step 2: Calculating Molarity (M) of the H₂O₂ solution: $$ \text{Given the solution is } "33.6 V" \implies M = \frac{33.6}{11.2} = 3 $$ Therefore, we have: $$ 33.6 \text{ times of } 11.2 \text{ (volume strength)} \Rightarrow 3 \text{ (molarity)} $$
Step 3: Applying the formula for milliequivalents: $$ \frac{w}{375} \times 1000 = 3 \times 125 \times 2 $$ Where:
$w$ is the weight in grams.
$375$ is the molar mass of $\mathrm{Ba}\left(\mathrm{MnO}_4\right)_2$.
$3$ is the molarity of $\mathrm{H}_2 \mathrm{O}_2$.
$125 \text{ ml}$ is the volume of $\mathrm{H}_2 \mathrm{O}_2$ used.
Solving for $w$: $$ \frac{w}{375} \times 1000 = 3 \times 125 \times 2 $$ $$ w = 28.125 \text{ grams} $$
Step 4: Determining the % purity of the sample: $$ % \text{ purity } = \frac{w}{40} \times 100 $$ Where $40 \text{ grams}$ is the sample weight.
Calculating: $$ % \text{ purity } = \frac{28.125}{40} \times 100 = 70.31 % $$
Thus, the correct % purity of the sample is: B $\boldsymbol{70.31%}$.
The following equilibrium is established in the solution of $ \text{Al(OH)}_3 $.
$$ \text{Al(OH)}_3\text{(s)} \rightleftharpoons \text{Al}^{3+}\text{(aq.)} + 3\text{OH}^-\text{(aq.)}, \ K_{sp} $$
$$ \text{Al(OH)}_3\text{(s)} + \text{OH}^-\text{(aq.)} \rightleftharpoons \text{Al(OH)}_4^-\text{(aq.)}, \ K_c $$
For minimum solubility, which of the following relations is correct?
A. $ [\text{OH}^-] = \left( \frac{K_{sp}}{K_c} \right)^{1/3} $
B. $ [\text{OH}^-] = \left( \frac{K_c}{K_{sp}} \right)^{1/4} $
C. $ [\text{OH}^-] = \sqrt{\left( \frac{K_{sp}}{K_c} \right)^{1/4}} $
D. None of the above
The correct answer is D.
In the solution, $\mathrm{Al}(\mathrm{OH})_{3}$ exists in equilibrium among its dissociated forms $\mathrm{Al}^{3+}$ (aq) and $\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}$.
The solubility, $S$, can be expressed as: $$ S = \left[\mathrm{Al}^{3+}(aq)\right] + \left[\mathrm{Al}(\mathrm{OH})_{4}(aq)\right] $$
Given the respective solubility product ($K_{sp}$) and equilibrium constant ($K_c$), we write: $$ S = \frac{K_{sp}}{\left[\mathrm{OH}^{-}\right]^3} + K_c \cdot \left[\mathrm{OH}^{-}\right] $$
To find the minimum solubility, we set the derivative of $S$ with respect to $\left[\mathrm{OH}^{-}\right]$ to zero: $$ \frac{dS}{d\left[\mathrm{OH}^{-}\right]} = 0 $$
This leads us to: $$
\frac{3 \cdot K_{sp}}{\left[\mathrm{OH}^{-}\right]^4} + K_c = 0 $$
Solving for $\left[\mathrm{OH}^{-}\right]$, we get: $$ \left[\mathrm{OH}^{-}\right] = \left( \frac{3 K_{sp}}{K_c} \right)^{1/4} $$
Final Answer: D
Water and chlorobenzene are two immiscible liquids. Under a reduced pressure of $7.7 \times 10^{4}$ Pa, this mixture boils at $89^\circ$C. The vapor pressure of pure water at $89^\circ$C is $7 \times 10^{4}$ Pa. The mass percentage of chlorobenzene in the distillate is:
A. 50
B. 60
C. 78.3
D. 38.46
The correct answer is D.
Given:
The reduced pressure is $7.7 \times 10^4$ Pa.
The mixture boils at $89^\circ$C.
The vapor pressure of pure water at $89^\circ$C is $7 \times 10^4$ Pa.
To determine the mass percentage of chlorobenzene in the distillate, we'll use the formula:
$$ \frac{W_A}{W_B} = \frac{P_A^{\circ}}{P_B^{\circ}} \times \frac{M_A}{M_B} $$
Where:
$W_A$ and $W_B$ are the masses of water and chlorobenzene, respectively.
$P_A^{\circ}$ and $P_B^{\circ}$ are the vapor pressures of pure water and chlorobenzene, respectively.
$M_A$ and $M_B$ are the molar masses of water (18 g/mol) and chlorobenzene (112.5 g/mol), respectively.
We know:
$$ P_A^{\circ} = 7 \times 10^4 , \text{Pa} $$ $$ P_B^{\circ} = 7.7 \times 10^4 - 7 \times 10^4 = 0.7 \times 10^4 , \text{Pa} $$
Substitute these values into the formula:
$$ \frac{W_A}{W_B} = \frac{7 \times 10^4}{0.7 \times 10^4} \times \frac{112.5}{18} $$
Simplify:
$$ \frac{W_A}{W_B} = \frac{7}{0.7} \times \frac{112.5}{18} $$ $$ \frac{W_A}{W_B} \approx 10 \times 6.25 = 62.5 $$
This simplifies further:
$$ \frac{W_A}{W_B} = 0.625 $$
To find the mass percentage of chlorobenzene in the mixture:
$$ % , \text{chlorobenzene} = \frac{W_B}{W_A + W_B} \times 100 = \frac{0.625}{1.625} \times 100 $$
Calculate the percentage:
$$ % , \text{chlorobenzene} \approx 38.46 $$
Thus, the mass percentage of chlorobenzene in the distillate is 38.46%.
Final Answer: D
How much ethyl alcohol should be mixed in 1.0 liter of water so that the solution does not freeze at $-4^\circ F$? ( $K_f = 1.86^\circ \text{C/m}$)
A. <20 g B. <10.75 g C. 494.5 g D. >494.5 g
The correct answer is D, which requires more than 494.5 g of ethanol.
To find the amount of ethyl alcohol ($\mathrm{C_2H_5OH}$) needed to prevent the solution from freezing at $-4^\circ \text{F}$, we follow these steps:
Convert the temperature from Fahrenheit to Celsius: $$ -4^\circ \text{F} = -20^\circ \text{C} $$ This implies that the freezing point depression $\Delta T_f$ is: $$ \Delta T_f = 20^\circ \text{C} $$
Use the formula for freezing point depression: $$ \Delta T_f = K_f \cdot m $$ Where:
$ \Delta T_f = 20^\circ \text{C} $
$ K_f = 1.86^\circ \text{C/m} $
Solve for molality ($m$): $$ m = \frac{\Delta T_f}{K_f} = \frac{20}{1.86} \approx 10.75 \text{ m} $$
Calculate the mass of ethyl alcohol needed: $$ \text{Mass of } \mathrm{C_2H_5OH} = 10.75 \times 46 = 494.5 \text{ g} $$
Thus, the mass of $\mathrm{C_2H_5OH}$ required is greater than 494.5 g, which makes the final answer: D.
The correct order of osmotic pressures for equimolar solutions of Urea, $\mathrm{BaCl_{2}}$, and $\mathrm{AlCl_{3}}$ is:
A. $\mathrm{AlCl_{3} > BaCl_{2} >}$ Urea
B. $\mathrm{BaCl_{2} > AlCl_{2} >}$ Urea
C. Urea $> \mathrm{BaCl_{2} > \mathrm{AlCl_{3}}}$
D. $\mathrm{BaCl_{2} >}$ Urea $> \mathrm{AlCl_{3}}$
The correct answer is Option A.
For equal molar solutions of urea, $\mathrm{BaCl}_{2}$, and $\mathrm{AlCl}_{3}$, the order of osmotic pressures is: $$ \mathrm{AlCl}_{3} > \mathrm{BaCl}_{2} > \text{urea} $$
This is because osmotic pressure ($ \Pi $) is directly proportional to the number of particles (ions or molecules) in the solution, as given by the formula: $$ \Pi = iCRT $$ where:
$i$ is the van 't Hoff factor,
$C$ is the molar concentration,
$R$ is the gas constant,
$T$ is the temperature.
For equimolar solutions:
Urea does not dissociate in water, so $i = 1$.
$\mathrm{BaCl}_{2}$ dissociates into 3 ions ($\mathrm{Ba}^{2+}$ and $2 \mathrm{Cl}^{-}$), so $i = 3$.
$\mathrm{AlCl}_{3}$ dissociates into 4 ions ($\mathrm{Al}^{3+}$ and $3 \mathrm{Cl}^{-}$), so $i = 4$.
Thus, osmotic pressure follows the order: $$ \mathrm{AlCl}_{3} (i=4) > \mathrm{BaCl}_{2} (i=3) > \text{urea} (i=1) $$
Final Answer: A
A $0.10 \, \mathrm{M}$ solution of a monoprotic acid ($d = 1.01$ g/cc) dissociates to $5\%$. What is the freezing point of the solution? The molecular weight of the acid is $300$ and $K_{f(H_{2}O)} = 1.86\^{\circ} \mathrm{C}/\mathrm{m}$
A $-0.189^{\circ} \mathrm{C}$
B $-0.194^{\circ} \mathrm{C}$
C $-0.199^{\circ} \mathrm{C}$
D None of these
The correct answer is: C
Calculate the mass of $1$ liter of the solution:
Given density: $1.01$ grams/mL
Hence, $1$ liter (or $1000$ mL) of the solution weighs ( 1010 ) grams.
Calculate the mass of the solvent:
Molecular weight of acid: $300 , \mathrm{g/mol}$
The amount of acid in $1$ liter of $0.10 , \mathrm{M}$ solution: $$ 0.10 \times 300 = 30 \text{ grams} $$
Therefore, mass of solvent (water) in the solution: $$ 1010 , \text{grams} - 30 , \text{grams} = 980 , \text{grams} $$
Convert the mass of solvent from grams to kilograms: $$ 980 , \text{grams} = 0.980 , \text{kg} $$
Calculate the molality ( m ): $$ m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.10 , \text{moles}}{0.980 , \text{kg}} = 0.102 , \text{m} $$
Calculate the van 't Hoff factor, ( i ):
Given the acid is $5%$ dissociated, ( \alpha = 0.05 )
Since it is a monoprotic acid (( i = 1 + \alpha )): $$ i = 1 + 0.05 = 1.05 $$
Apply the freezing point depression formula:
The formula for depression of freezing point is: $$ \Delta T_f = K_f \cdot m \cdot i $$
Using values given: $$ \Delta T_f = 1.86 , ^\circ \mathrm{C/m} \times 0.102 , \mathrm{m} \times 1.05 $$ $$ \Delta T_f = 0.199 , ^\circ \mathrm{C} $$
Calculate the final freezing point:
Since the freezing point of water is $0^\circ \mathrm{C}$: $$ T_f = 0 - 0.199 = -0.199 , ^\circ \mathrm{C} $$
Therefore, the freezing point of the solution is (\boxed{-0.199 , ^\circ \mathrm{C}}).
The density of an aqueous solution with 56.0% by mass of 1-propanol (CH₃CH₂CH₂OH) is 0.8975 grams/cm³. The mole percent of 1-propanol in the solution is:
A) 0.292
B) 0.227
C) 0.241
D) 0.276
The correct answer is D.
Mass of the solution: ( 100 \text{ grams} )
Mass of the solute (1-propanol): ( 56 \text{ grams} )
Molar mass of 1-propanol (( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} )): ( 60 \text{ g/mol} )
Moles of solute: $$ \frac{56}{60} = 0.93 \text{ moles} $$
Mass of the solvent (water): ( 44 \text{ grams} )
Molar mass of water (( \text{H}_2\text{O} )): ( 18 \text{ g/mol} )
Moles of solvent: $$ \frac{44}{18} = 2.44 \text{ moles} $$
Mole fraction of 1-propanol: $$ X_{\text{sol solute}} = \frac{0.93}{0.93 + 2.44} = \frac{0.93}{3.37} = 0.276 $$
Thus, the moles percentage of 1-propanol in the solution is 0.276. Therefore, the final answer is D.
Statement: The enthalpy of neutralization of strong acids and strong bases is always the same.
Reason: The enthalpy of neutralization is equal to the heat of formation of water.
If both the statement and the reason are true, and the reason explains the statement correctly:
If the statement is true but the reason is false:
If both the statement and the reason are true, but the reason does not explain the statement correctly:
If both the statement and the reason are false:
The correct answer is: $\mathbf{A}$
The enthalpy of neutralization for strong acids and strong bases is the same and it is 13.7 kilocalories. The main reason for this is that it is released during the formation of water from the combination of $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$.
$$ \begin{align*} \mathrm{H}^{+} + \mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}, \quad \Delta \mathrm{H} = -13.7 \text{ kilocalories} \end{align*} $$
Thus, both the statement and the reason are correct and they explain each other.
Final Answer: $\mathbf{A}$
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