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WAEC Chemistry 2017 Paper

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Question : 34

Total: 50

If 0.2g of a salt is required to saturate 200

\mathrm{cm}^{3}

of water at room temperature, what is the solubility of the salt?

0.2 $\mathrm{g\,dm}^{-3}$
1.0 $\mathrm{g\,dm}^{-3}$
2.0 $\mathrm{g\,dm}^{-3}$
5.0 $\mathrm{g\,dm}^{-3}$

Solution:

Solubility in $\mathrm{g\,dm}^{-3}$ = $\frac{\text{mass of salt in gram}}{\text{volume of liquid in}\; \mathrm{dm}^{3}}$

200 $\mathrm{cm}^{3}$ = $\;\frac{200}{1000} \mathrm{dm}^{3} = 0.2 \mathrm{dm}^{3}$

Solubility = $\;\frac{0.2\,\mathrm{g}}{0.2\,\mathrm{dm}^{3}}$

= 1.0 $\mathrm{g\,dm}^{-3}$

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