2022-may-Chemistry_paper_1__TZ1_SL Detailed Solution

Solution:

To determine which sample contains the fewest moles of $\mathrm{HCl}$, we need to calculate the number of moles of $\mathrm{HCl}$ in each sample.

For sample A, we have:

Volume of $\mathrm{HCl}(\mathrm{aq})=10.0 \mathrm{~cm}^3=10^{-3} \mathrm{dm}^3$
Concentration of $\mathrm{HCl}(\mathrm{aq})=0.1 \mathrm{~mol} \mathrm{dm}^{-3}$

Using the formula for concentration, we can calculate the number of moles of $\mathrm{HCl}$:

$$
\text { Number of moles of } \mathrm{HCl}(\mathrm{aq})=\text { Concentration } \times \text { Volume }=0.1 \mathrm{~mol} \mathrm{dm}^{-3} \times 10^{-3} \mathrm{dm}^3=1.0 \times 10^{-5} \mathrm{~mol}
$$

For sample B, we are given the number of molecules of $\mathrm{HCl(g)}$, but we can convert this to moles using Avogadro’s number:

$$
\text { Number of moles of } \mathrm{HCl}(\mathrm{g})=\frac{\text { Number of molecules of } \mathrm{HCl}(\mathrm{g})}{N_{\mathrm{A}}}=\frac{6.02 \times 10^{24}}{6.02 \times 10^{23}}=10 \mathrm{~mol}
$$

For sample C, we are given the mass of $\mathrm{HCl(g)}$, but we can convert this to moles using the molar mass of $\mathrm{HCl}$:

$$
\text { Number of moles of } \mathrm{HCl}(\mathrm{g})=\frac{\text { Mass of } \mathrm{HCl}(\mathrm{g})}{\text { Molar mass of } \mathrm{HCl}}=\frac{0.365 \mathrm{~g}}{36.46 \mathrm{~g} \mathrm{~mol}^{-1}} \approx 0.01 \mathrm{~mol}
$$

For sample D, we are given the volume of $\mathrm{HCl(g)}$ at STP, but we can convert this to moles using the molar volume of an ideal gas at STP:

$$
\text { Number of moles of } \mathrm{HCl}(\mathrm{g})=\frac{\text { Volume of } \mathrm{HCl}(\mathrm{g}) \text { at STP }}{\text { Molar volume of gas at STP }}=\frac{2.27 \mathrm{dm}^3}{22.7 \mathrm{dm}^3 \mathrm{~mol}^{-1}}=0.1 \mathrm{~mol}
$$