If 0.2g of a salt is required to saturate 200{cm}³ of water at room temperature, what is the solubility of the salt?

Correct Answer is : 1.0g dm−3{g\,dm}^{-3}gdm−3

Correct Answer is : 1.0g dm−3{g\,dm}^{-3}gdm−3

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WAEC Chemistry 2017 Paper

Question : 34

Total: 50

\mathrm{cm}^{3}

Solution:

Concept:

Solubility is the mass of solute required to saturate

1\,\mathrm{dm}^3

of solvent.

Explanation:

First convert the volume of water from

\mathrm{cm}^3

\mathrm{dm}^3

Since

1000\,\mathrm{cm}^3 = 1\,\mathrm{dm}^3

, we get:

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

Solubility is calculated as mass of salt divided by volume of water in

\mathrm{dm}^3

\text{Solubility} = \frac{0.2\,\mathrm{g}}{0.2\,\mathrm{dm}^3} = 1.0\,\mathrm{g\,dm}^{-3}

Thus, the salt has a solubility of

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

Answer:

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

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