Grade 11 Chemistry Quiz - Basic Concepts (Set 7) | Concept of Maths

Question 1

1. An analytical chemist analyzes a mineral sample from a remote crater. The sample contains $70.0\%$ Iron (Fe) and $30.0\%$ Oxygen (O) by mass. What is the empirical formula of this mineral? ($A_r: \text{Fe} = 56, \text{O} = 16$)

A) $\text{FeO}$

B) $\text{Fe}_3\text{O}_4$

C) $\text{Fe}_2\text{O}_3$

D) $\text{Fe}_2\text{O}$

Answer

Correct Answer: C

Solution: Step 1: Moles of $\text{Fe} = \frac{70.0}{56} = 1.25\text{ mol}$. Step 2: Moles of $\text{O} = \frac{30.0}{16} = 1.875\text{ mol}$. Step 3: Divide by smallest value ($1.25$): $\text{Fe}: \frac{1.25}{1.25} = 1$; $\text{O}: \frac{1.875}{1.25} = 1.5$. Step 4: Multiply by $2$ to get whole numbers: $\text{Fe}_2\text{O}_3$.

Question 2

2. A laboratory stock of concentrated $\text{H}_2\text{SO}_4$ is labeled as $98\%$ by mass with a density of $1.84\text{ g/mL}$. Calculate the Molarity ($M$) of this solution. ($M_m: \text{H}_2\text{SO}_4 = 98\text{ g/mol}$)

A) $18.4\text{ M}$

B) $10.0\text{ M}$

C) $1.84\text{ M}$

D) $9.8\text{ M}$

Answer

Correct Answer: A

Solution: Density $= 1.84\text{ g/mL}$ means $1000\text{ mL}$ ($1\text{ L}$) weighs $1840\text{ g}$. Mass of $\text{H}_2\text{SO}_4$ in $1\text{ L} = 98\% \text{ of } 1840\text{ g} = 0.98 \times 1840 = 1803.2\text{ g}$. Molarity $M = \frac{\text{mass}}{\text{molar mass} \times \text{volume (L)}} = \frac{1803.2}{98 \times 1} = 18.4\text{ M}$.

Question 3

3. In the industrial synthesis of Ammonia: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$, a reactor is charged with $50.0\text{ kg}$ of $\text{N}_2$ and $10.0\text{ kg}$ of $\text{H}_2$. Identify the limiting reagent and the maximum mass of $\text{NH}_3$ produced.

Question 4

4. A synthetic element 'Z' has two stable isotopes: $^{200}\text{Z}$ (abundance $40\%$) and $^{202}\text{Z}$ (abundance $60\%$). What is the average atomic mass of element Z?

A) $200.0\text{ u}$

B) $200.8\text{ u}$

C) $201.0\text{ u}$

D) $201.2\text{ u}$

Answer

Correct Answer: D

Solution: Average Mass $= \frac{(200 \times 40) + (202 \times 60)}{100}$ $= \frac{8000 + 12120}{100} = \frac{20120}{100} = 201.2\text{ u}$.

Question 5

5. Solve the following calculation and express the result with the correct number of significant figures: $\frac{2.5 \times 10^{-2} \times 20.0}{0.050}$

A) $10$

B) $10.$

C) $10.0$

D) $1.0 \times 10^1$

Answer

Correct Answer: B

Solution: $2.5 \times 10^{-2}$ has 2 sig figs. $20.0$ has 3. $0.050$ has 2. Value $= \frac{0.025 \times 20.0}{0.050} = \frac{0.5}{0.050} = 10$. The result must have the least number of sig figs used in multiplication/division, which is 2. Thus, $10$ with 2 sig figs is expressed as $1.0 \times 10^1$ or $10.$ (with a decimal). Option B is $10.$ representing 2 sig figs.

Question 6

6. A combustion chamber contains $20\text{ mL}$ of Methane ($\text{CH}_4$) and $50\text{ mL}$ of Oxygen ($\text{O}_2$). After complete combustion and cooling to room temperature, what is the total volume of the remaining gas mixture?

A) $70\text{ mL}$

B) $20\text{ mL}$

C) $30\text{ mL}$

D) $50\text{ mL}$

Answer

Correct Answer: C

Solution: Reaction: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$. $20\text{ mL}$ of $\text{CH}_4$ requires $2 \times 20 = 40\text{ mL}$ of $\text{O}_2$. We have $50\text{ mL}$ of $\text{O}_2$, so remaining $\text{O}_2 = 50 - 40 = 10\text{ mL}$. $\text{CO}_2$ produced $= 20\text{ mL}$ (1:1 ratio with $\text{CH}_4$). Total gas (Water is liquid at room temp) $= 10\text{ mL}$ of $\text{O}_2 + 20\text{ mL}$ of $\text{CO}_2 = 30\text{ mL}$.

Question 7

7. Calculate the mole fraction of Ethanol ($\text{C}_2\text{H}_5\text{OH}$) in a solution containing $46\text{ g}$ of ethanol and $54\text{ g}$ of water. ($M_m: \text{Ethanol}=46, \text{Water}=18$)

A) $0.25$

B) $0.50$

C) $0.75$

D) $1.00$

Answer

Correct Answer: A

Solution: Moles of Ethanol ($n_A$) $= \frac{46}{46} = 1\text{ mol}$. Moles of Water ($n_B$) $= \frac{54}{18} = 3\text{ mol}$. Mole fraction $X_A = \frac{n_A}{n_A + n_B} = \frac{1}{1+3} = \frac{1}{4} = 0.25$.

Question 8

8. How many total atoms are present in $4.4\text{ g}$ of Carbon dioxide ($\text{CO}_2$)? ($N_A = 6.022 \times 10^{23}$)

A) $6.022 \times 10^{22}$

B) $1.204 \times 10^{23}$

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

D) $1.806 \times 10^{23}$

Answer

Correct Answer: D

Solution: Moles of $\text{CO}_2 = \frac{4.4}{44} = 0.1\text{ mol}$. 1 molecule of $\text{CO}_2$ has 3 atoms (1 C, 2 O). Total atoms $= 0.1 \times 3 \times N_A = 0.3 \times 6.022 \times 10^{23} = 1.806 \times 10^{23}$.

Question 9

9. A pharmaceutical production line attempts to produce $\text{CaO}$ by heating $100\text{ g}$ of $\text{CaCO}_3$. If the actual mass of $\text{CaO}$ obtained is $42\text{ g}$, what is the percentage yield? ($M_m: \text{CaCO}_3=100, \text{CaO}=56$)

A) $42\%$

B) $75\%$

C) $56\%$

D) $84\%$

Answer

Correct Answer: B

Solution: Theoretical yield: $100\text{ g}$ of $\text{CaCO}_3$ gives $56\text{ g}$ of $\text{CaO}$ (from stoichiometry). Actual yield $= 42\text{ g}$. $\%\text{ Yield} = \frac{\text{Actual}}{\text{Theoretical}} \times 100 = \frac{42}{56} \times 100 = 0.75 \times 100 = 75\%$.

Question 10

10. An element X forms two oxides. In the first, $50\%$ of X is present. In the second, $40\%$ of X is present. This data illustrates which chemical law?

A) Law of Multiple Proportions

B) Law of Definite Proportions

C) Law of Conservation of Mass

D) Gay-Lussac's Law

Answer

Correct Answer: A

Solution: In Oxide 1: $50\%$ X, $50\%$ O. Ratio X:O $= 1:1$. In Oxide 2: $40\%$ X, $60\%$ O. Ratio X:O $= 40:60 = 1:1.5$. For a fixed mass of X (1 part), the masses of Oxygen that combine are in the ratio $1 : 1.5$, which is $2 : 3$ (a simple whole number ratio). This is the definition of the Law of Multiple Proportions.

Accepted Answer

Correct Answer: B

Solution: Moles of $\text{N}_2 = \frac{50000\text{ g}}{28} \approx 1785.7\text{ mol}$. Moles of $\text{H}_2 = \frac{10000\text{ g}}{2} = 5000\text{ mol}$. Required $\text{H}_2$ for all $\text{N}_2 = 1785.7 \times 3 = 5357.1\text{ mol}$. Since we only have $5000\text{ mol}$, $\text{H}_2$ is limiting. Mass of $\text{NH}_3 = \frac{2}{3} \times 5000 \times 17\text{ g} = 56666\text{ g} \approx 56.6\text{ kg}$.

Grade 11 : Chemistry - Basic Concepts (Set 7) | Concept of Maths

Grade 11 : Chemistry - Basic Concepts (Set 7) | Concept of Maths

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4. A synthetic element 'Z' has two stable isotopes: $^{200}\text{Z}$ (abundance $40\%$) and $^{202}\text{Z}$ (abundance $60\%$). What is the average atomic mass of element Z?

5. Solve the following calculation and express the result with the correct number of significant figures: $\frac{2.5 \times 10^{-2} \times 20.0}{0.050}$

6. A combustion chamber contains $20\text{ mL}$ of Methane ($\text{CH}_4$) and $50\text{ mL}$ of Oxygen ($\text{O}_2$). After complete combustion and cooling to room temperature, what is the total volume of the remaining gas mixture?

7. Calculate the mole fraction of Ethanol ($\text{C}_2\text{H}_5\text{OH}$) in a solution containing $46\text{ g}$ of ethanol and $54\text{ g}$ of water. ($M_m: \text{Ethanol}=46, \text{Water}=18$)

8. How many total atoms are present in $4.4\text{ g}$ of Carbon dioxide ($\text{CO}_2$)? ($N_A = 6.022 \times 10^{23}$)

9. A pharmaceutical production line attempts to produce $\text{CaO}$ by heating $100\text{ g}$ of $\text{CaCO}_3$. If the actual mass of $\text{CaO}$ obtained is $42\text{ g}$, what is the percentage yield? ($M_m: \text{CaCO}_3=100, \text{CaO}=56$)

10. An element X forms two oxides. In the first, $50\%$ of X is present. In the second, $40\%$ of X is present. This data illustrates which chemical law?

Grade 11 Chemistry Quiz - Basic Concepts (Set 7) | Concept of Maths

1. An analytical chemist analyzes a mineral sample from a remote crater. The sample contains $70.0\%$ Iron (Fe) and $30.0\%$ Oxygen (O) by mass. What is the empirical formula of this mineral? ($A_r: \text{Fe} = 56, \text{O} = 16$)

2. A laboratory stock of concentrated $\text{H}_2\text{SO}_4$ is labeled as $98\%$ by mass with a density of $1.84\text{ g/mL}$. Calculate the Molarity ($M$) of this solution. ($M_m: \text{H}_2\text{SO}_4 = 98\text{ g/mol}$)

3. In the industrial synthesis of Ammonia: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$, a reactor is charged with $50.0\text{ kg}$ of $\text{N}_2$ and $10.0\text{ kg}$ of $\text{H}_2$. Identify the limiting reagent and the maximum mass of $\text{NH}_3$ produced.

Grade 11 : Chemistry - Basic Concepts (Set 7) | Concept of Maths

Grade 11 : Chemistry - Basic Concepts (Set 7) | Concept of Maths

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4. A synthetic element 'Z' has two stable isotopes: $^{200}\text{Z}$ (abundance $40\%$) and $^{202}\text{Z}$ (abundance $60\%$). What is the average atomic mass of element Z?

5. Solve the following calculation and express the result with the correct number of significant figures: $\frac{2.5 \times 10^{-2} \times 20.0}{0.050}$

6. A combustion chamber contains $20\text{ mL}$ of Methane ($\text{CH}_4$) and $50\text{ mL}$ of Oxygen ($\text{O}_2$). After complete combustion and cooling to room temperature, what is the total volume of the remaining gas mixture?

7. Calculate the mole fraction of Ethanol ($\text{C}_2\text{H}_5\text{OH}$) in a solution containing $46\text{ g}$ of ethanol and $54\text{ g}$ of water. ($M_m: \text{Ethanol}=46, \text{Water}=18$)

8. How many total atoms are present in $4.4\text{ g}$ of Carbon dioxide ($\text{CO}_2$)? ($N_A = 6.022 \times 10^{23}$)

9. A pharmaceutical production line attempts to produce $\text{CaO}$ by heating $100\text{ g}$ of $\text{CaCO}_3$. If the actual mass of $\text{CaO}$ obtained is $42\text{ g}$, what is the percentage yield? ($M_m: \text{CaCO}_3=100, \text{CaO}=56$)

10. An element X forms two oxides. In the first, $50\%$ of X is present. In the second, $40\%$ of X is present. This data illustrates which chemical law?