# 2022-may-Chemistry_paper_1__TZ1_HL Detailed Solution

### Question-1 :2022-may-Chemistry_paper_1__TZ1_HL

Topic:

Given: $2.67 \mathrm{~g}$ of lead (II) carbonate is decomposed by heating until constant mass.

$$\mathrm{PbCO}_3(\mathrm{~s}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$$

Calculate: What is the final mass of solid?

A. $0.44 \mathrm{~g}$
B. $2.23 \mathrm{~g}$
C. $2.67 \mathrm{~g}$
D. $3.11 \mathrm{~g}$

Solution:

To solve this problem, we need to use the balanced chemical equation to determine the theoretical yield of $\mathrm{PbO}$. From the equation, we know that one mole of $\mathrm{PbCO_3}$ will produce one mole of $\mathrm{PbO}$. The molar mass of $\mathrm{PbCO_3}$ is $267.2\mathrm{~g/mol}$ (Pb: $207.2\mathrm{~g/mol}$, C: $12.0\mathrm{~g/mol}$, O: $48.0\mathrm{~g/mol}$).

The number of moles of $\mathrm{PbCO_3}$ is:

$$n=\frac{m}{M}=\frac{2.67 \mathrm{~g}}{267.2 \mathrm{~g} / \mathrm{mol}}=0.01 \mathrm{~mol}$$



According to the balanced chemical equation, 1 mole of $\mathrm{PbCO_3}$ will produce 1 mole of $\mathrm{PbO}$, so the number of moles of $\mathrm{PbO}$ produced is also 0.01 mol. The molar mass of $\mathrm{PbO}$ is $223.2\mathrm{~g/mol}$ (Pb: $207.2\mathrm{~g/mol}$, O: $16.0\mathrm{~g/mol}$), so the mass of $\mathrm{PbO}$ produced is:

${}_{}$

$$m_{\mathrm{PbO}}=n \cdot M_{\mathrm{PbO}}=0.01 \mathrm{~mol} \cdot 223.2 \mathrm{~g} / \mathrm{mol}=2.23 \mathrm{~g}$$

Therefore, the final mass of solid is $\colorbox{yellow}{$\mathrm{(B)~2.23g}$}$

### Question-2 :2022-may-Chemistry_paper_1__TZ1_HL

Topic:

Given: $0.02 \mathrm{~mol}$ of zinc is added to $10.0 \mathrm{~cm}^3$ of $1.0 \mathrm{~mol} \mathrm{dm}^{-3}$ hydrochloric acid.

$$\mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_2(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g})$$

Calculate: How many moles of hydrogen are produced?
A. $0.005$
B. $0.01$
C. $0.02$
D. $0.04$

Solution:

According to the balanced chemical equation, one mole of $\mathrm{Zn}$ reacts with two moles of $\mathrm{HCl}$ to produce one mole of $\mathrm{H_2}$ and one mole of $\mathrm{ZnCl_2}$. Therefore, in this reaction, the number of moles of hydrogen produced will be half of the number of moles of $\mathrm{Zn}$ added.

The number of moles of $\mathrm{Zn}$ added is given as $0.02~\mathrm{mol}$. Therefore, the number of moles of hydrogen produced will be:

$$\text { Number of moles of } \mathrm{H}_2=\frac{1}{2} \times 0.02 \mathrm{~mol}=0.01 \mathrm{~mol}$$

Since the volume of the hydrochloric acid is given in cubic centimeters ($10.0~\mathrm{cm^3}$), we need to convert it to cubic decimeters (liters) to use the molarity given in units of moles per liter. One cubic decimeter is equal to 1000 cubic centimeters, so $10.0~\mathrm{cm^3} = 0.01~\mathrm{dm^3}$.

The molarity of the hydrochloric acid is given as $1.0~\mathrm{moldm^{-3}}$. Therefore, the number of moles of hydrochloric acid in $10.0\mathrm{cm^3}$ is:

$$\text { Number of moles of } \mathrm{HCl}=\text { Molarity } \times \text { Volume }=1.0 \mathrm{~mol} \mathrm{dm}^{-3} \times 0.01 \mathrm{dm}^3=0.01 \mathrm{~mol}$$

Since the stoichiometry of the balanced chemical equation tells us that two moles of hydrochloric acid react with one mole of hydrogen, and we have $0.01~\mathrm{mol}$ of hydrochloric acid, the maximum number of moles of hydrogen that could be produced is $0.005~\mathrm{mol}$.

Therefore, the correct answer is $\colorbox{yellow}{$\mathrm{(A)~0.005}$}$

### Question-3 :2022-may-Chemistry_paper_1__TZ1_HL

Topic:

Given: $8.8 \mathrm{~g}$ of an oxide of nitrogen contains $3.2 \mathrm{~g}$ of oxygen.

Calculate: What is the empirical formula of the compound?

A. $\mathrm{N}_2 \mathrm{O}_5$
B. $\mathrm{N}_2 \mathrm{O}$
C. $\mathrm{NO}_2$
D. $\mathrm{NO}$

Solution:

To find the empirical formula of the compound, we need to determine the simplest whole-number ratio of the atoms present in the compound.

We are given that the compound contains $8.8 \mathrm{~g}$ in total, and $3.2 \mathrm{~g}$ of oxygen. To find the mass of nitrogen, we can subtract the mass of oxygen from the total mass:

$$8.8 \mathrm{~g}-3.2 \mathrm{~g}=5.6 \mathrm{~g}$$

Therefore, the mass of nitrogen in the compound is $5.6 \mathrm{~g}$.

Next, we need to convert the masses of nitrogen and oxygen to moles. The molar mass of nitrogen is $14.01 \mathrm{~g/mol}$, and the molar mass of oxygen is $16.00 \mathrm{~g/mol}$, so:

\begin{aligned} \text { moles of nitrogen } & =\frac{5.6 \mathrm{~g}}{14.01 \mathrm{~g} / \mathrm{mol}} \approx 0.399 \mathrm{~mol} \\ \text { moles of oxygen } & =\frac{3.2 \mathrm{~g}}{16.00 \mathrm{~g} / \mathrm{mol}}=0.200 \mathrm{~mol} \end{aligned}

The ratio of nitrogen to oxygen in the compound is:

$$\frac{0.399 \mathrm{~mol}}{0.200 \mathrm{~mol}} \approx 2: 1$$

This means that the empirical formula of the compound is $\mathrm{N}_2\mathrm{O}$ (two nitrogen atoms for every one oxygen atom). Therefore, the correct answer is $\colorbox{yellow}{$\mathrm{(B)~\mathrm{N}_2 \mathrm{O}}$}$.

### Question-4 :2022-may-Chemistry_paper_1__TZ1_HL

Topic:

Given: Naturally occurring gallium consists of the isotopes ${ }^{71} \mathrm{Ga}$ and ${ }^{69} \mathrm{Ga}$.

$$M_{\mathrm{r}}(\mathrm{Ga})=69.72$$

Calculate: What is the approximate percentage abundance of ${ }^{69} \mathrm{Ga}$ ?

A. $40 \%$
B. $50 \%$
C. $60 \%$
D. $75 \%$

Solution:

### Question-5 :2022-may-Chemistry_paper_1__TZ1_HL

Topic:

Given: The graph shows the first six ionization energies of an element.

Discuss: In which group is the element?

A. $13$
B. $14$
C. $15$
D. $16$