Grade 11 : Chemistry - Basic Concepts (Set 3) | Concept of Maths

Question 1

1. A sealed glass tube contains $5.5\text{ g}$ of substance $X$ and $2.5\text{ g}$ of substance $Y$. After a chemical reaction, $4.0\text{ g}$ of a new gas $Z$ is formed along with a solid residue $R$. According to the Law of Conservation of Mass, what is the mass of the residue $R$?

A) $8.0\text{ g}$

B) $1.5\text{ g}$

C) $4.0\text{ g}$

D) $5.5\text{ g}$

Answer

Correct Answer: C

The Law of Conservation of Mass states that total mass of reactants = total mass of products. $Mass_{X} + Mass_{Y} = Mass_{Z} + Mass_{R}$ $5.5\text{ g} + 2.5\text{ g} = 4.0\text{ g} + Mass_{R}$ $8.0\text{ g} = 4.0\text{ g} + Mass_{R}$ $Mass_{R} = 8.0 - 4.0 = 4.0\text{ g}$.

Question 2

2. An analyst has $1\text{ g}$ samples of four different gases: $H_2, He, O_2,$ and $N_2$. Which sample contains the greatest number of molecules?

A) $H_2$

B) $He$

C) $O_2$

D) $N_2$

Answer

Correct Answer: A

Number of molecules is proportional to the number of moles ($n = \frac{m}{M}$). For a fixed mass $m = 1\text{ g}$, the gas with the smallest molar mass ($M$) will have the largest number of moles. $M(H_2) = 2\text{ g/mol}$, $M(He) = 4\text{ g/mol}$, $M(O_2) = 32\text{ g/mol}$, $M(N_2) = 28\text{ g/mol}$. Since $H_2$ has the smallest molar mass, it has the most molecules.

Question 3

3. A flask contains $6.022 \times 10^{23}$ molecules of $CO_2$. If $3.011 \times 10^{23}$ molecules are removed, how many total Oxygen atoms remain in the flask?

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

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

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

D) $1.505 \times 10^{23}$

Question 4

4. What is the result of the following calculation expressed to the correct number of significant figures? $$12.11 + 18.0 + 1.012$$

A) $31.122$

B) $31.12$

C) $31.1$

D) $31.1$

Answer

Correct Answer: D

In addition, the result must be reported to the least number of decimal places present in the terms. $12.11$ (2 decimal places), $18.0$ (1 decimal place), $1.012$ (3 decimal places). Sum = $31.122$. Rounding to 1 decimal place gives $31.1$.

Question 5

5. In the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, if you start with $2.0\text{ moles}$ of $N_2$ and $3.0\text{ moles}$ of $H_2$, which is the limiting reagent?

A) $H_2$

B) $N_2$

C) $NH_3$

D) None

Answer

Correct Answer: A

According to stoichiometry, $1\text{ mole}$ of $N_2$ requires $3\text{ moles}$ of $H_2$. So, $2.0\text{ moles}$ of $N_2$ would require $2.0 \times 3 = 6.0\text{ moles}$ of $H_2$. Since we only have $3.0\text{ moles}$ of $H_2$, $H_2$ is the limiting reagent.

Question 6

6. A solution is prepared by dissolving $10\text{ g}$ of $NaOH$ in enough water to make $250\text{ mL}$ of solution. What is the molarity of this solution? (Molar mass of $NaOH = 40\text{ g/mol}$)

A) $0.25\text{ M}$

B) $0.5\text{ M}$

C) $1.0\text{ M}$

D) $2.0\text{ M}$

Answer

Correct Answer: C

Moles of $NaOH = \frac{10}{40} = 0.25\text{ mol}$. Volume in Liters = $\frac{250}{1000} = 0.25\text{ L}$. Molarity $(M) = \frac{\text{moles}}{\text{volume in L}} = \frac{0.25}{0.25} = 1.0\text{ M}$.

Question 7

7. Two oxides of a metal contain $50\%$ and $40\%$ of the metal by mass respectively. This data illustrates which law?

A) Law of Definite Proportions

B) Law of Multiple Proportions

C) Law of Conservation of Mass

D) Avogadro's Law

Answer

Correct Answer: B

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. This is the Law of Multiple Proportions.

Question 8

8. A sample of $O_2$ gas contains $1.204 \times 10^{24}$ atoms of Oxygen. How many moles of $O_2$ molecules are present?

A) $2.0\text{ mol}$

B) $1.0\text{ mol}$

C) $0.5\text{ mol}$

D) $4.0\text{ mol}$

Answer

Correct Answer: B

Number of Oxygen atoms = $1.204 \times 10^{24}$. Since $O_2$ is diatomic, number of $O_2$ molecules = $\frac{1.204 \times 10^{24}}{2} = 6.022 \times 10^{23}$. $6.022 \times 10^{23}$ molecules = $1.0\text{ mole}$.

Question 9

9. If the density of a solution is $1.2\text{ g/mL}$, what is the mass of $50\text{ mL}$ of this solution?

A) $41.6\text{ g}$

B) $50.0\text{ g}$

C) $1.2\text{ g}$

D) $60.0\text{ g}$

Answer

Correct Answer: D

$Mass = Density \times Volume$. $Mass = 1.2\text{ g/mL} \times 50\text{ mL} = 60.0\text{ g}$.

Question 10

10. The empirical formula of a compound is $CH_2$ and its molar mass is $42\text{ g/mol}$. What is its molecular formula?

A) $C_3H_6$

B) $C_2H_4$

C) $C_4H_8$

D) $CH_2$

Answer

Correct Answer: A

Empirical formula mass $(CH_2) = 12 + (2 \times 1) = 14\text{ g/mol}$. $n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{42}{14} = 3$. Molecular formula = $3 \times (CH_2) = C_3H_6$.

Accepted Answer

Correct Answer: B

Remaining molecules of $CO_2 = 6.022 \times 10^{23} - 3.011 \times 10^{23} = 3.011 \times 10^{23}$ molecules. Each molecule of $CO_2$ contains 2 Oxygen atoms. Total O atoms = $2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23}$.

Grade 11 : Chemistry - Basic Concepts (Set 3) | Concept of Maths