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WAEC Chemistry 2023 Paper

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Question : 31

Total: 50

If

100^3

of a saturated solution of sodium tetraoxosulphate

(\mathrm{VI})

30^{\circ}\,\mathrm{C}

contains 10.5 g of the salt, what would be its solubility at this temperature?

[ (\mathrm{Na}_2 \mathrm{SO}_4=142] .

2.15 mol dm-3
0.74 mol dm-3
0.60 mol dm-3
0.57 mol dm-3

Solution:

Concept:

Solubility is the number of moles of solute present in

1\,\mathrm{dm^3}

of a saturated solution.

Explanation:

Molar mass of sodium tetraoxosulphate(VI),

\mathrm{Na_2SO_4}

:

= 2(23) + 32 + 4(16) = 142\,\mathrm{g\,mol^{-1}}

Moles of salt in the solution:

\text{Moles} = \frac{10.5\,\mathrm{g}}{142\,\mathrm{g\,mol^{-1}}} = 0.0739\,\mathrm{mol}

Convert volume of solution:

100\,\mathrm{cm^3} = \frac{100}{1000} = 0.1\,\mathrm{dm^3}

Calculate solubility:

\text{Solubility} = \frac{0.0739\,\mathrm{mol}}{0.1\,\mathrm{dm^3}} = 0.739\,\mathrm{mol\,dm^{-3}} \approx 0.74\,\mathrm{mol\,dm^{-3}}

Answer:

The correct option is B:

0.74\,\mathrm{mol\,dm^{-3}}

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