Grade 11 : Chemistry - Basic Concepts (Set 5) | Concept of Maths

Question 1

1. A robotic probe measures the density of a liquid on a distant moon to be $1.20 \text{ g/cm}^3$. If the volume of the sample is recorded as $2.5 \text{ cm}^3$, what is the mass of the sample reported with the correct number of significant figures?

A) $3 \text{ g}$

B) $3.00 \text{ g}$

C) $3.0 \text{ g}$

D) $3.000 \text{ g}$

Answer

Correct Answer: C

In multiplication, the result must have the same number of significant figures as the measurement with the fewest significant figures. $\text{Density} = 1.20$ (3 sig figs) $\text{Volume} = 2.5$ (2 sig figs) $\text{Mass} = \text{Density} \times \text{Volume} = 1.20 \times 2.5 = 3.0 \text{ g}$ The result must be reported to 2 significant figures.

Question 2

2. In a combustion analysis, a $0.45 \text{ g}$ sample of a pure hydrocarbon produces $1.32 \text{ g}$ of $\text{CO}_2$ and $0.81 \text{ g}$ of $\text{H}_2\text{O}$. What is the empirical formula of the hydrocarbon?

A) $\text{CH}_2$

B) $\text{CH}_3$

C) $\text{C}_2\text{H}_5$

D) $\text{C}_3\text{H}_8$

Answer

Correct Answer: B

1. Moles of $\text{C} = \frac{1.32}{44.01} \approx 0.03 \text{ mol}$ 2. Moles of $\text{H} = 2 \times \frac{0.81}{18.02} \approx 0.09 \text{ mol}$ 3. Ratio $\text{C}:\text{H} = 0.03 : 0.09 = 1:3$ Empirical Formula = $\text{CH}_3$.

Question 3

3. A hypothetical reaction occurs: $2\text{A} + 5\text{B} \rightarrow 3\text{C} + 4\text{D}$. If $0.4$ moles of $\text{A}$ and $0.8$ moles of $\text{B}$ are reacted, which reagent is limiting and how many moles of $\text{C}$ are formed?

A) $\text{B}$ is limiting; $0.48 \text{ mol C}$

B) $\text{A}$ is limiting; $0.60 \text{ mol C}$

Question 4

4. An isotope of a newly discovered element $\text{X}$ has two naturally occurring isotopes: $^{70}\text{X}$ (abundance $60\%$) and $^{72}\text{X}$ (abundance $40\%$). What is the average atomic mass of element $\text{X}$?

A) $71.0 \text{ u}$

B) $70.5 \text{ u}$

C) $71.2 \text{ u}$

D) $70.8 \text{ u}$

Answer

Correct Answer: D

Average atomic mass $= \sum (\text{mass} \times \text{abundance})$ $= (70 \times 0.60) + (72 \times 0.40)$ $= 42.0 + 28.8 = 70.8 \text{ u}$.

Question 5

5. Calculate the molarity of $\text{Cl}^-$ ions when $200 \text{ mL}$ of $0.2 \text{ M } \text{MgCl}_2$ solution is mixed with $300 \text{ mL}$ of $0.1 \text{ M } \text{NaCl}$ solution.

A) $0.14 \text{ M}$

B) $0.22 \text{ M}$

C) $0.18 \text{ M}$

D) $0.25 \text{ M}$

Answer

Correct Answer: B

Moles of $\text{Cl}^-$ from $\text{MgCl}_2 = 2 \times (0.2 \text{ L} \times 0.2 \text{ M}) = 0.08 \text{ mol}$ Moles of $\text{Cl}^-$ from $\text{NaCl} = 1 \times (0.3 \text{ L} \times 0.1 \text{ M}) = 0.03 \text{ mol}$ Total moles of $\text{Cl}^- = 0.08 + 0.03 = 0.11 \text{ mol}$ Total Volume $= 0.2 + 0.3 = 0.5 \text{ L}$ Molarity $= \frac{0.11}{0.5} = 0.22 \text{ M}$.

Question 6

6. How many oxygen atoms are present in $88 \text{ g}$ of $\text{CO}_2$? (Take $N_A = 6.022 \times 10^{23}$)

A) $1.204 \times 10^{24}$

B) $6.022 \times 10^{23}$

C) $2.408 \times 10^{24}$

D) $3.011 \times 10^{23}$

Answer

Correct Answer: C

Moles of $\text{CO}_2 = \frac{88}{44} = 2 \text{ mol}$. Each molecule of $\text{CO}_2$ has $2$ Oxygen atoms. Total $\text{O}$ atoms $= 2 \times 2 \times N_A = 4 \times 6.022 \times 10^{23} = 2.408 \times 10^{24}$ atoms.

Question 7

7. Which of the following pairs of compounds illustrates the Law of Multiple Proportions?

A) $\text{SO}_2$ and $\text{SO}_3$

B) $\text{H}_2\text{O}$ and $\text{D}_2\text{O}$

C) $\text{NaCl}$ and $\text{NaBr}$

D) $\text{MgO}$ and $\text{Mg(OH)}_2$

Answer

Correct Answer: A

The Law of Multiple Proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. $\text{SO}_2$ and $\text{SO}_3$ consist of the same two elements ($\text{S}$ and $\text{O}$) in different ratios.

Question 8

8. A solution is prepared by dissolving $10 \text{ g}$ of a non-volatile solute in $81 \text{ g}$ of water. The mole fraction of the solute is $0.1$. What is the molar mass of the solute?

A) $18 \text{ g/mol}$

B) $40 \text{ g/mol}$

C) $60 \text{ g/mol}$

D) $20 \text{ g/mol}$

Answer

Correct Answer: D

$n_{\text{water}} = \frac{81}{18} = 4.5 \text{ mol}$. Let $n_{\text{solute}}$ be $x$. Mole fraction $X_{\text{solute}} = \frac{x}{x + 4.5} = 0.1$ $x = 0.1x + 0.45 \Rightarrow 0.9x = 0.45 \Rightarrow x = 0.5 \text{ mol}$. Molar mass $= \frac{\text{Mass}}{\text{moles}} = \frac{10}{0.5} = 20 \text{ g/mol}$.

Question 9

9. What is the mass of $11.2 \text{ L}$ of $\text{N}_2$ gas at STP?

A) $28 \text{ g}$

B) $7 \text{ g}$

C) $14 \text{ g}$

D) $22.4 \text{ g}$

Answer

Correct Answer: C

1 mole of any gas at STP occupies $22.4 \text{ L}$. Moles of $\text{N}_2 = \frac{11.2}{22.4} = 0.5 \text{ mol}$. Molar mass of $\text{N}_2 = 28 \text{ g/mol}$. Mass $= 0.5 \times 28 = 14 \text{ g}$.

Question 10

10. In the reaction $\text{X} + 2\text{Y} \rightarrow \text{Z}$, if you start with $5$ moles of $\text{X}$ and $8$ moles of $\text{Y}$, and the reaction has a $75\%$ yield, how many moles of $\text{Z}$ are actually produced?

A) $4 \text{ mol}$

B) $3 \text{ mol}$

C) $3.75 \text{ mol}$

D) $5 \text{ mol}$

Answer

Correct Answer: B

1. Limiting Reagent: $\text{Y}$ is limiting ($8/2 = 4$ vs $5/1 = 5$). 2. Theoretical yield of $\text{Z} = \frac{1}{2} \times \text{moles of Y} = 4 \text{ moles}$. 3. Actual yield $= 75\% \text{ of } 4 = 0.75 \times 4 = 3 \text{ moles}$.

Accepted Answer

Correct Answer: A

Required ratio $\text{B}/\text{A} = 5/2 = 2.5$. Given ratio $\text{B}/\text{A} = 0.8/0.4 = 2.0$. Since $2.0 < 2.5$, $\text{B}$ is the limiting reagent. Moles of $\text{C}$ formed $= \frac{3}{5} \times \text{moles of B} = 0.6 \times 0.8 = 0.48 \text{ mol}$.

Grade 11 : Chemistry - Basic Concepts (Set 5) | Concept of Maths