9701 · 2.2

The mole and the Avogadro constant — practice questions

Practice and worked examples for 9701 The mole and the Avogadro constant. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Calculate the amount of substance (in moles) and the number of formula units in 25.0 g of calcium carbonate, CaCO₃. (Relative atomic masses: Ca = 40.1, C = 12.0, O = 16.0)

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Calculate the molar mass ( $M_r$ ) of CaCO₃: $M_r = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ g mol^{-1}
Calculate the amount in moles ( $n$ ): $n = \frac{m}{M_r} = \frac{25.0 \text{ g}}{100.1 \mathrm{g mol}^{-1}} = 0.24975...$ mol To 3 significant figures, $n = 0.250$ mol.
Calculate the number of formula units: Number of units = $n \times L$ Number of units = $0.250 \text{ mol} \times (6.02 \times 10^{23} \mathrm{mol}^{-1}) = 1.505 \times 10^{23}$ To 3 significant figures, this is $1.51 \times 10^{23}$ formula units.

Worked example 2

A student dissolves 4.38 g of anhydrous sodium carbonate, Na₂CO₃, in water and makes the solution up to 250 $\mathrm{cm}^{3} in a volumetric flask$ . Calculate the concentration of the solution in mol dm^{-3}. (Relative atomic masses: Na = 23.0, C = 12.0, O = 16.0)

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Calculate the molar mass ( $M_r$ ) of Na₂CO₃: $M_r = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0$ g mol^{-1}
Calculate the amount in moles ( $n$ ) of Na₂CO₃: $n = \frac{m}{M_r} = \frac{4.38 \text{ g}}{106.0 \mathrm{g mol}^{-1}} = 0.04132...$ mol
**Convert the volume to $\mathrm{dm}^{3}:**$ $$V = \frac{250 \mathrm{cm}^{3}}{1000} = 0.250 $\mathrm{dm}^{3}$
Calculate the concentration ( $c$ ): $c = \frac{n}{V} = \frac{0.04132... \text{ mol}}{0.250 \mathrm{dm}^{3}} = 0.16528...$ mol dm^{-3} To 3 significant figures, the concentration is $0.165$ mol dm^{-3}.

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