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? ($A_r: Fe = 56, O = 16$)

A) FeOB) $Fe_3O_4$C) $Fe_2O_3$D) $Fe_2O$

Answer

✓ Correct Answer: Option C

Explanation: Solution: Step 1: Moles of $Fe = \frac{70.0}{56} = 1.25 mol$. Step 2: Moles of $O = \frac{30.0}{16} = 1.875 mol$. Step 3: Divide by smallest value (1.25): $Fe: \frac{1.25}{1.25} = 1$; $O: \frac{1.875}{1.25} = 1.5$. Step 4: Multiply by 2 to get whole numbers: $Fe_2O_3$.

Question 2

2. A laboratory stock of concentrated $H_2SO_4$ is labeled as 98% by mass with a density of 1.84 g/mL. Calculate the Molarity (M) of this solution. ($M_m: H_2SO_4 = 98 g/mol$)

18.4 M

Question 3

3. In the industrial synthesis of Ammonia: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, a reactor is charged with 50.0 kg of $N_2$ and 10.0 kg of $H_2$. Identify the limiting reagent and the maximum mass of $NH_3$ produced.

A) $N_2$ is limiting; $60.7kg NH_3$B) $H_2$ is limiting; $56.6kg NH_3$C) $H_2$ is limiting; $60.7kg NH_3$D) $N_2$ is limiting; $112.3kg NH_3$

Answer

✓ Correct Answer: Option B

Explanation: Solution: Moles of $N_2 = \frac{50000g}{28} \approx 1785.7 mol$. Moles of $H_2 = \frac{10000g}{2} = 5000 mol$. Required $H_2$ for all $N_2 = 1785.7 \times 3 = 5357.1 mol$. Since we only have 5000 mol, $H_2$ is limiting. Mass of $NH_3 = \frac{2}{3} \times 5000 \times 17g = 56666g \approx 56.6kg$.

Question 4

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

A) 200.0 uB) 200.8 uC) 201.0 uD) 201.2 u

Answer

✓ Correct Answer: Option D

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

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) 10B) 10.C) 10.0D) $1.0 \times 10^1$

Answer

✓ Correct Answer: Option B

Explanation: 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 mL of Methane ($CH_4$) and 50 mL of Oxygen ($O_2$). After complete combustion and cooling to room temperature, what is the total volume of the remaining gas mixture?

A) 70 mLB) 20 mLC) 30 mLD) 50 mL

Answer

✓ Correct Answer: Option C

Explanation: Solution: Reaction: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$. 20 mL of $CH_4$ requires $2 \times 20 = 40 mL$ of $O_2$. We have 50 mL of $O_2$, so remaining $O_2 = 50 - 40 = 10 mL$. $CO_2$ produced = 20 mL (1:1 ratio with $CH_4$). Total gas (Water is liquid at room temp) = 10 mL of $O_2 + 20 mL$ of $CO_2 = 30 mL$.

Question 7

7. Calculate the mole fraction of Ethanol ($C_2H_5OH$) in a solution containing 46 g of ethanol and 54 g of water. ($M_m: Ethanol=46, Water=18$)

A) 0.25B) 0.50C) 0.75D) 1.00

Answer

✓ Correct Answer: Option A

Explanation: Solution: Moles of Ethanol ($n_A$) $= \frac{46}{46} = 1 mol$. Moles of Water ($n_B$) $= \frac{54}{18} = 3 mol$. 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 g of Carbon dioxide ($CO_2$)? ($N_A = 6.022 \times 10^{23}$)

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: Option D

Explanation: Solution: Moles of $CO_2 = \frac{4.4}{44} = 0.1 mol$. 1 molecule of $CO_2$ 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 CaO by heating 100 g of $CaCO_3$. If the actual mass of CaO obtained is 42 g, what is the percentage yield? ($M_m: CaCO_3=100, CaO=56$)

A) 42%B) 75%C) 56%D) 84%

Answer

✓ Correct Answer: Option B

Explanation: Solution: Theoretical yield: 100 g of $CaCO_3$ gives 56 g of CaO (from stoichiometry). Actual yield = 42g. $% Yield = \frac{Actual}{Theoretical} \times 100 = \frac{42}{56} \times 100 = 0.75 \times 100 = 75%$.

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 ProportionsB) Law of Definite ProportionsC) Law of Conservation of MassD) Gay-Lussac's Law

Answer

✓ Correct Answer: Option A

Explanation: 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: Option A

Explanation: Solution: Density = 1.84 g/mL means 1000 mL (1 L) weighs 1840 g. Mass of $H_2SO_4$ in $1 L = 98% of 1840g = 0.98 \times 1840 = 1803.2g$. Molarity $M = \frac{mass}{molar mass \times volume (L)} = \frac{1803.2}{98 \times 1} = 18.4m$.

Grade 11 Chemistry Quiz - Basic Concepts (Set 7)

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? ($A_r: Fe = 56, O = 16$)

2. A laboratory stock of concentrated $H_2SO_4$ is labeled as 98% by mass with a density of 1.84 g/mL. Calculate the Molarity (M) of this solution. ($M_m: H_2SO_4 = 98 g/mol$)

3. In the industrial synthesis of Ammonia: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, a reactor is charged with 50.0 kg of $N_2$ and 10.0 kg of $H_2$. Identify the limiting reagent and the maximum mass of $NH_3$ produced.

4. A synthetic element 'Z' has two stable isotopes: $^{200}Z$ (abundance 40%) and $^{202}Z$ (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 mL of Methane ($CH_4$) and 50 mL of Oxygen ($O_2$). 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 ($C_2H_5OH$) in a solution containing 46 g of ethanol and 54 g of water. ($M_m: Ethanol=46, Water=18$)

8. How many total atoms are present in 4.4 g of Carbon dioxide ($CO_2$)? ($N_A = 6.022 \times 10^{23}$)

9. A pharmaceutical production line attempts to produce CaO by heating 100 g of $CaCO_3$. If the actual mass of CaO obtained is 42 g, what is the percentage yield? ($M_m: CaCO_3=100, CaO=56$)

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?