Grade 11 : Chemistry - Basic Concepts (Set 4) | Concept of Maths

Question 1

1. A closed vessel contains 4 molecules of $A_2$ (white spheres) and 9 molecules of $B_2$ (black spheres). They react to form product molecules of $AB_3$ according to the balanced equation. Which of the following describes the contents of the vessel after the reaction is complete?

A) $A_2$ is the limiting reagent; 1 molecule of $B_2$ remains.

B) $B_2$ is the limiting reagent; 1 molecule of $A_2$ remains.

C) $B_2$ is the limiting reagent; 3 molecules of $A_2$ remain.

D) Both reagents are consumed completely.

Answer

Correct Answer: B

The balanced equation is: $A_2 + 3B_2 \rightarrow 2AB_3$. Stoichiometric ratio of $A_2 : B_2 = 1 : 3$. For 4 molecules of $A_2$, we need $4 \times 3 = 12$ molecules of $B_2$. Since we only have 9 molecules of $B_2$, $B_2$ is the limiting reagent. Molecules of $A_2$ consumed = $9 / 3 = 3$ molecules. Remaining $A_2$ = $4 - 3 = 1$ molecule.

Question 2

2. A technician fills a pipette with $50 \text{ mL}$ of a liquid solvent $X$. The density of the liquid is $1.2 \text{ g/mL}$ and its molar mass is $60 \text{ g/mol}$. How many molecules of $X$ are present in the pipette?

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

B) $1.2 \times 10^{24}$

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

D) $1.0 \times 10^{22}$

Answer

Correct Answer: C

Step 1: Calculate mass of liquid: $m = \text{Density} \times \text{Volume} = 1.2 \text{ g/mL} \times 50 \text{ mL} = 60 \text{ g}$. Step 2: Calculate moles: $n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{60 \text{ g}}{60 \text{ g/mol}} = 1 \text{ mol}$. Step 3: Calculate molecules: $N = n \times N_A = 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$ molecules.

Question 3

3. A pie chart representing the mass percentage composition of a compound shows Carbon at $40\%$, Hydrogen at $6.7\%$, and Oxygen at $53.3\%$. What is the empirical formula of this compound?

Question 4

4. A student performs three measurements of the mass of a standard $15.00 \text{ g}$ weight. The results are $10.25 \text{ g}$, $10.26 \text{ g}$, and $10.24 \text{ g}$. Which statement best describes these results?

A) Accurate and precise

B) Precise but not accurate

C) Accurate but not precise

D) Neither accurate nor precise

Answer

Correct Answer: B

The measurements ($10.24, 10.25, 10.26$) are very close to each other, indicating high **precision**. However, they are very far from the true value of $15.00 \text{ g}$, indicating low **accuracy**.

Question 5

5. In a laboratory experiment, $2 \text{ moles}$ of cane sugar are dissolved in $500 \text{ mL}$ of pure water. Assuming the density of water is $1.0 \text{ g/mL}$, what is the molality ($m$) of the resulting solution?

A) $2.0 \text{ m}$

B) $1.0 \text{ m}$

C) $4.0 \text{ m}$

D) $0.5 \text{ m}$

Answer

Correct Answer: C

Molality ($m$) = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$. Mass of solvent (water) = $500 \text{ mL} \times 1.0 \text{ g/mL} = 500 \text{ g} = 0.5 \text{ kg}$. Molality = $\frac{2 \text{ moles}}{0.5 \text{ kg}} = 4.0 \text{ mol/kg} = 4.0 \text{ m}$.

Question 6

6. Element $X$ and $Y$ form two compounds. In Compound A, $0.3 \text{ g}$ of $X$ reacts with $0.4 \text{ g}$ of $Y$. In Compound B, $18 \text{ g}$ of $X$ reacts with $48 \text{ g}$ of $Y$. This data illustrates which 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

In Compound A, ratio $X:Y = 0.3 : 0.4 = 3 : 4$. In Compound B, ratio $X:Y = 18 : 48 = 3 : 8$. Comparing the mass of $Y$ that combines with a fixed mass of $X$ (3 units): the ratio is $4 : 8$ or $1 : 2$. Since this is a simple whole number ratio, it illustrates the Law of Multiple Proportions.

Question 7

7. A graph of Pressure ($P$) vs Temperature ($T$ in Kelvin) for a fixed volume $V$ of an ideal gas shows a straight line with slope $k$. If the number of moles of the gas is doubled while keeping the volume constant, how will the slope of the graph change?

A) The slope remains $k$.

B) The slope becomes $2k$.

C) The slope becomes $k/2$.

D) The slope becomes $k^2$.

Answer

Correct Answer: B

From the Ideal Gas Equation: $PV = nRT \implies P = (\frac{nR}{V})T$. The slope of the $P$ vs $T$ graph is $m = \frac{nR}{V}$. If $n$ is doubled to $2n$, the new slope $m' = \frac{2nR}{V} = 2m$. Thus, the slope becomes $2k$.

Question 8

8. A student calculates the result of the following expression using a calculator: $\frac{0.0025 \times 1.20}{0.0010}$. How many significant figures should the final answer contain?

A) 1

B) 2

C) 3

D) 4

Answer

Correct Answer: B

In multiplication and division, the result must have the same number of significant figures as the term with the least number of significant figures. - $0.0025$ has 2 significant figures. - $1.20$ has 3 significant figures. - $0.0010$ has 2 significant figures. The least number of significant figures is 2.

Question 9

9. A volumetric flask contains $100 \text{ mL}$ of $0.1 \text{ M } HCl$. If distilled water is added to the flask until the total volume reaches the $500 \text{ mL}$ mark, what is the new molarity of the solution?

A) $0.05 \text{ M}$

B) $0.025 \text{ M}$

C) $0.02 \text{ M}$

D) $0.5 \text{ M}$

Answer

Correct Answer: C

Using the dilution formula $M_1V_1 = M_2V_2$: $0.1 \text{ M} \times 100 \text{ mL} = M_2 \times 500 \text{ mL}$ $M_2 = \frac{10}{500} = 0.02 \text{ M}$.

Question 10

10. Based on the molecular structure of Urea ($NH_2CONH_2$), what is the mass percentage of Nitrogen in the compound? (At. masses: $N=14, C=12, O=16, H=1$)

A) $23.33\%$

B) $46.67\%$

C) $28.00\%$

D) $35.12\%$

Answer

Correct Answer: B

Molar mass of Urea = $(14 \times 2) + 12 + 16 + (1 \times 4) = 28 + 12 + 16 + 4 = 60 \text{ g/mol}$. Mass of Nitrogen in 1 mole = $2 \times 14 = 28 \text{ g}$. $\% N = \frac{28}{60} \times 100 = 46.666... \approx 46.67\%$.

Accepted Answer

Correct Answer: A

Assume $100 \text{ g}$ of substance: Moles of $C = \frac{40}{12} = 3.33$ Moles of $H = \frac{6.7}{1} = 6.7$ Moles of $O = \frac{53.3}{16} = 3.33$ Simplest ratio: $C = \frac{3.33}{3.33} = 1$; $H = \frac{6.7}{3.33} \approx 2$; $O = \frac{3.33}{3.33} = 1$. Formula: $CH_2O$.

Grade 11 : Chemistry - Basic Concepts (Set 4) | Concept of Maths