Question 1

1. A sealed glass tube contains 5.5 g of substance X and 2.5 g of substance Y. After a chemical reaction, 4.0 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 gB) 1.5 gC) 4.0 gD) 5.5 g

Answer

✓ Correct Answer: Option C

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

Question 2

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

A) $H_2$ He

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) $1.505 \times 10^{23}$

Answer

✓ Correct Answer: Option B

Explanation: 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 = $2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23}$.

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) 31.122B) 31.12C) 31.1D) 31.1

Answer

✓ Correct Answer: Option D

Explanation: 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 moles of $N_2$ and 3.0 moles of $H_2$, which is the limiting reagent?

A) $H_2$B) $N_2$C) $NH_3$D) None

Answer

✓ Correct Answer: Option A

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

Question 6

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

A) 0.25 MB) 0.5 MC) 1.0 MD) 2.0 M

Answer

✓ Correct Answer: Option C

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

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 ProportionsB) Law of Multiple ProportionsC) Law of Conservation of MassD) Avogadro's Law

Answer

✓ Correct Answer: Option B

Explanation: 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 molB) 1.0 molC) 0.5 molD) 4.0 mol

Answer

✓ Correct Answer: Option B

Explanation: 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 mole.

Question 9

9. If the density of a solution is 1.2 g/mL, what is the mass of 50 mL of this solution?

A) 41.6 gB) 50.0 gC) 1.2 gD) 60.0 g

Answer

✓ Correct Answer: Option D

Explanation: $Mass = Density \times Volume$. $Mass = 1.2 g/mL \times 50 mL = 60.0g$.

Question 10

10. The empirical formula of a compound is $CH_2$ and its molar mass is 42 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: Option A

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

Accepted Answer

✓ Correct Answer: Option A

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

Grade 11 : Chemistry - Basic Concepts (Set 3)

1. A sealed glass tube contains 5.5 g of substance X and 2.5 g of substance Y. After a chemical reaction, 4.0 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?

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

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?

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

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

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

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

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?

9. If the density of a solution is 1.2 g/mL, what is the mass of 50 mL of this solution?

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