Some Basic Concepts of Chemistry - Class 11 Chemistry - Chapter 1 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Some Basic Concepts of Chemistry | NCERT | Chemistry | Class 11
For the reaction: $$ \mathrm{Al} + \mathrm{O}{2} \rightarrow \mathrm{Al}{2}\mathrm{O}_{3} $$ (The equation is not balanced).
$54 , \mathrm{g}$ of $\mathrm{Al}$ is reacted with $67.4$ liters at STP. $76.5 , \mathrm{g}$ of $\mathrm{Al}{2}\mathrm{O}{3}$ was obtained. The percentage yield of the product obtained is:
Balanced Chemical Equation: $$ 4 \mathrm{Al} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2\mathrm{O}_3 $$
From stoichiometry, $108 , \mathrm{g}$ of $\mathrm{Al}$ reacts with $67.4 , \mathrm{liters}$ of $\mathrm{O}_2$ at STP to yield $204 , \mathrm{g}$ of $\mathrm{Al}_2\mathrm{O}_3$. Therefore, $54 , \mathrm{g}$ of $\mathrm{Al}$ would require half the amount of $\mathrm{O}_2$ and produce half as much $\mathrm{Al}_2\mathrm{O}_3$:
Required $\mathrm{O}_2 = 33.7 , \mathrm{liters}$
Theoretical yield of $\mathrm{Al}_2\mathrm{O}_3 = 102 , \mathrm{g}$
Given that actual experimental yield of $\mathrm{Al}_2\mathrm{O}_3$ is $76.5 , \mathrm{g}$, we calculate the percentage yield using: $$ \text{Percentage Yield} = \left(\frac{\text{Experimental Yield}}{\text{Theoretical Yield}}\right) \times 100 = \left(\frac{76.5}{102}\right) \times 100 $$ $$ = 75% $$
Excess Oxygen: Amount of unused $\mathrm{O}_2$ is $67.4 , \mathrm{liters} - 33.7 , \mathrm{liters} = 33.7 , \mathrm{liters}$.
Thus, the percentage yield of $\mathrm{Al}_2\mathrm{O}_3$ produced is 75%.
Four 1-litre flasks are separately filled with $\mathrm{SO}{2}$, $\mathrm{SO}{3}$, $\mathrm{Ar}$, and $\mathrm{CH}_{4}$ gases at the same temperature and pressure. The ratio of the total number of atoms of these gases present in the different flasks would be:
(A) 1:1:1:1 (B) 3:4:1:5 (C) 4:3:1:5 (D) None
The correct option is (B) 3:4:1:5.
Avogadro's Law states that: Equal volumes of gases at the same temperature and pressure contain the same number of molecules.
Given the gases:
$\mathrm{SO}_2$ (Sulfur dioxide)
$\mathrm{SO}_3$ (Sulfur trioxide)
$\mathrm{Ar}$ (Argon)
$\mathrm{CH}_4$ (Methane)
Under the same conditions of temperature and pressure, we consider equal volumes for each gas. By Avogadro's Law, this implies that the numbers of molecules in each flask are equal. Therefore, the molecule ratio is:
$$ \text{Molecule ratio} = 1 : 1 : 1 : 1 $$
Now, let's count the number of atoms per molecule of each gas:
For $\mathrm{SO}_2$, each molecule consists of 1 sulfur atom and 2 oxygen atoms, totaling 3 atoms.
For $\mathrm{SO}_3$, each molecule consists of 1 sulfur atom and 3 oxygen atoms, totaling 4 atoms.
For $\mathrm{Ar}$, each molecule (being an inert noble gas) consists of one atom itself, thus 1 atom.
For $\mathrm{CH}_4$, each molecule consists of 1 carbon atom and 4 hydrogen atoms, totaling 5 atoms.
Now, since the ratios of the molecules are equal, the ratio of total atoms in each flask will depend on the number of atoms per molecule. Therefore, the ratio of total atoms for $\mathrm{SO}_2$, $\mathrm{SO}_3$, $\mathrm{Ar}$, and $\mathrm{CH}_4$ is:
$$ \text{Ratio of atoms} = 3 : 4 : 1 : 5 $$
Thus, the correct answer is option B, where the ratio of the total number of atoms of these gases present in the different flasks is 3:4:1:5.
In which of the following does the given amount of chlorine gas exert the least pressure in a vessel of $1 \mathrm{~L}$ capacity at $273 \mathrm{~K}$?
A) $0.0355 \mathrm{~g}$
B) $0.071 \mathrm{~g}$
C) $0.01 \mathrm{~mol}$
D) $0.02 \mathrm{~mol}$
The correct option is A) $0.0355 \mathrm{~g}$.
To determine which quantity of chlorine gas exerts the least pressure, we can consider the mole concept. Using the ideal gas law, pressure is directly proportional to the number of moles. Thus, fewer moles mean lower pressure, provided volume and temperature are constant. We need to convert the mass of chlorine into moles to compare effectively.
First, we need to know the molar mass of chlorine (Cl$_2$):
- The atomic mass of chlorine (Cl) is approximately $35.5 , \text{g/mol}$.
- As chlorine is diatomic in its natural state, its molar mass is $35.5 \times 2 = 71 , \text{g/mol}$.
We calculate the moles for options A and B:
- For option A: $$ \text{Moles} = \frac{0.0355 , \text{g}}{71 , \text{g/mol}} = 0.0005 , \text{mol} $$
- For option B: $$ \text{Moles} = \frac{0.071 , \text{g}}{71 , \text{g/mol}} = 0.001 , \text{mol} $$
Comparing moles in all options:
- A) $0.0355 , \text{g}$ corresponds to $0.0005 , \text{mol}$
- B) $0.071 , \text{g}$ corresponds to $0.001 , \text{mol}$
- C) $0.01 , \text{mol}$
- D) $0.02 , \text{mol}$
Option A ($0.0005 , \text{mol}$) has the smallest number of moles, indicating it will exert the least pressure in a $1 , \mathrm{L}$ vessel at $273 , \mathrm{K}$.
The full form of IUPAC is
A. International Union of Physics and Chemistry
B. International Union of Planning, Action, and Chemistry
C. Indian Union of Pure and Applied Chemistry
D. International Union of Pure and Applied Chemistry
The correct answer is D. International Union of Pure and Applied Chemistry.
IUPAC, abbreviated from the International Union of Pure and Applied Chemistry, is an internationally recognized authority in chemical nomenclature, symbols, and standards. Though primarily known for its standardized chemistry nomenclature, IUPAC is also influential in fields like biology and physics, producing various publications and guidelines essential for scientific research and communication.
Calculate the number of:
(a) Particles in 0.1 moles of any substance,
(b) Hydrogen atoms in 0.1 mole of $\mathrm{H}_{2} \mathrm{SO}_{4}$
(c) Molecules in one $\mathrm{kg}$ of calcium chloride.
Let's break down the calculations into three parts:
(a) Number of particles in 0.1 moles of any substance
For one mole of any substance, the number of particles is given by Avogadro's number, which is:
$$ 6.023 \times 10^{23} $$
Therefore, for 0.1 moles:
$$ \text{Number of particles} = 6.023 \times 10^{23} \times 0.1 = 6.023 \times 10^{22} $$
(b) Number of Hydrogen atoms in 0.1 mole of $\mathrm{H}_{2}\mathrm{SO}_{4}$
One mole of sulfuric acid ($\mathrm{H}_{2}\mathrm{SO}_{4}$) contains 2 moles of hydrogen atoms. Hence, in terms of atoms:
$$ \text{Hydrogen atoms in 1 mole of } \mathrm{H}_{2}\mathrm{SO}_{4} = 2 \times 6.023 \times 10^{23} $$
For 0.1 moles of $\mathrm{H}_{2}\mathrm{SO}_{4}$:
$$ \text{Hydrogen atoms} = 2 \times 6.023 \times 10^{23} \times 0.1 = 1.2046 \times 10^{23} $$
(c) Number of molecules in 1 kg of calcium chloride ($\mathrm{CaCl}_{2}$)
The molar mass of $\mathrm{CaCl}_{2}$ is approximately $111 \text{ g/mol}$. Therefore:
$$ \text{Number of molecules in } 111 \text{ g of calcium chloride} = 6.023 \times 10^{23} $$
Given that we have 1 kg (1000 g) of calcium chloride, the number of molecules is:
$$ \text{Number of molecules in 1000 g} = \frac{1000 \text{ g}}{111 \text{ g/mol}} \times 6.023 \times 10^{23} $$
$$ = 9.009 \times 6.023 \times 10^{23} = 5.408 \times 10^{24} $$
Thus, the final results are:
Number of particles in 0.1 moles of any substance: $$6.023 \times 10^{22}$$
Number of hydrogen atoms in 0.1 mole of $\mathrm{H}_{2}\mathrm{SO}_{4}$: $$1.2046 \times 10^{23}$$
Number of molecules in 1 kg of calcium chloride: $$5.408 \times 10^{24}$$
Hydrogen gas is prepared in the laboratory by reacting dilute $\mathrm{HCl}$ with granulated zinc. Following reaction takes place:
$$ \mathrm{Zn} + 2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2} + \mathrm{H}_{2} $$
Calculate the volume of hydrogen gas liberated at STP when $32.65 \mathrm{~g}$ of zinc reacts with $\mathrm{HCl}$. $0.1 \mathrm{~mol}$ of a gas occupies $22.7 \mathrm{~L}$ volume at STP; atomic mass of $\mathrm{Zn} = 65.3 \mathrm{u}$
Given:
Mass of $\mathrm{Zn}$ = $32.65 \ \mathrm{g}$
1 mole of gas occupies $22.7 \ \mathrm{L}$ at STP
Atomic mass of $\mathrm{Zn}$ = $65.3 \ \mathrm{u}$
The reaction is:
$$ \mathrm{Zn} + 2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2} + \mathrm{H}_{2} $$
From the equation, it is clear that:
$65.3 \ \mathrm{g}$ of $\mathrm{Zn}$ produces $22.7 \ \mathrm{L}$ of $\mathrm{H}_{2}$ at STP.
Therefore, to find the volume of hydrogen gas produced when $32.65 \ \mathrm{g}$ of zinc reacts with $\mathrm{HCl}$, we use the ratio derived from the reaction:
$$ \text{Volume of } \mathrm{H}_{2} = \frac{22.7 \times 32.65}{65.3} $$
Calculating the volume:
$$ \text{Volume of } \mathrm{H}_{2} = 11.35 \ \mathrm{L} $$
Thus, $32.65 \ \mathrm{g}$ of zinc when reacting with $\mathrm{HCl}$ will produce $11.35 \ \mathrm{L}$ of $\mathrm{H}_{2}$ at STP.
In a sample of a pure compound, 9.81 grams of Zn, $1.8 \times 10^{23}$ chromium atoms, and 0.60 moles of oxygen atoms are present. The simplest formula of the compound is:
A. $ \mathrm{ZnCr}_{2}\mathrm{O}_{7} $
B. $ \mathrm{ZnCr}_{2}\mathrm{O}_{4} $
C. $ \mathrm{ZnCrO}_{4} $
D. $ \mathrm{ZnCrO}_{6} $
The correct answer is B.
First, we calculate the moles of each element:
Moles of Zn: $$ \text{Moles of } \mathrm{Zn} = \frac{9.81 , \text{grams}}{65 , \text{g/mol}} = 0.15 , \text{moles} $$
Moles of Cr: $$ \text{Moles of } \mathrm{Cr} = \frac{1.8 \times 10^{23} , \text{atoms}}{6.02 \times 10^{23} , \text{atoms/mol}} = 0.3 , \text{moles} $$
Moles of O: $$ \text{Moles of } \mathrm{O} = 0.6 , \text{moles} $$
Combining these, we get the formula: $$ \mathrm{Zn}_{0.15} \mathrm{Cr}_{0.3} \mathrm{O}_{0.6} $$
To get the simplest whole number ratio, we divide each by 0.15: $$ \mathrm{Zn}: \frac{0.15}{0.15} = 1 \ \mathrm{Cr}: \frac{0.3}{0.15} = 2 \ \mathrm{O}: \frac{0.6}{0.15} = 4 $$
Therefore, the simplest formula is: $$ \boxed{\mathrm{ZnCr}_{2}\mathrm{O}_{4}} $$
Final Answer: B
A metal $\mathrm{M}$ (equivalent mass $\mathrm{E}$) forms an oxide with the formula $ M_{x} O_{y} $. What is the equation to obtain the atomic mass of the metal?
(A) $ \frac{2E y}{x} $
(B) $ xyE $
(C) $ \frac{E}{y} $
(D) $ \frac{y}{E} $
The correct answer is: A
Given that a metal $\mathrm{M}$ (with equivalent mass $\mathrm{E}$) forms an oxide with the formula $M_{x}O_{y}$, we need to find the atomic mass of the metal.
Let's consider the atomic mass of metal $M$ to be $m$. Then, $16 \mathrm{y}$ grams of oxygen would react with $\mathrm{xm}$ grams of the metal.
Therefore, the mass of the metal that reacts with $8$ grams of oxygen would be: $$ \frac{x m \times 8}{16 y} = \frac{x m}{2 y} $$
Since this mass is equivalent to $\mathrm{E}$, we equate: $$ \frac{x m}{2 y} = E $$
Solving for $m$, we get: $$ m = \frac{2 E y}{x} $$
Thus, the atomic mass of the metal, $m$, is given by: $$ \boxed{\frac{2 E y}{x}} $$
Final Answer: A
If the value of $u $ at 27°C is $ 30R^{1/2} $, then calculate the molar mass of the gas (in kg):
A. 1
B. 2
C. 4
D. 0.001
Given:
The value of $u$ at 27°C is 30$R^{1/2}$.
To find:
The molar mass of the gas in kilograms.
We start with the rms speed formula: $$ u_{\text{rms}} = \sqrt{\frac{3RT}{M}} $$
Substituting the given value: $$ \sqrt{30^2 R} = \sqrt{\frac{3RT}{M}} $$
Squaring both sides: $$ 30 \times 30 R = \frac{3R \times 300}{M} $$
Solving for $M$: $$ M = \frac{3 \times 300}{30 \times 30} = 1 \text{ gram} = 0.001 \text{ kilograms} $$
Final Answer: D