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WAEC Chemistry 2019 Paper

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Question : 17

Total: 50

What is the solubility of a salt

X

at 25ºC if the saturated solution of the salt contains 0.28g in 100

cm^3

of the solution in mol

dm^3

?

0.05
0.10
2.60
2.80

Solution:

Concept:

Solubility in

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

is the number of moles of solute dissolved in

1\,\mathrm{dm^3}

of saturated solution.

Explanation:

Convert the given volume of solution to

\mathrm{dm^3}

:

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

.

Calculate the moles of salt

X

using its molar mass (

56.0\,\mathrm{g\,mol^{-1}}

):

\text{Moles of } X = \frac{0.28\,\mathrm{g}}{56.0\,\mathrm{g\,mol^{-1}}} = 0.005\,\mathrm{mol}

.

Now, find the solubility:

\text{Solubility} = \frac{\text{moles}}{\text{volume in } \mathrm{dm^3}} = \frac{0.005\,\mathrm{mol}}{0.1\,\mathrm{dm^3}} = 0.05\,\mathrm{mol\,dm^{-3}}

.

Alternatively, use the direct formula:

\frac{\text{mass}}{\text{molar mass}} \times \frac{1000}{\text{volume in } \mathrm{cm^3}} = \frac{0.28}{56.0} \times \frac{1000}{100} = 0.05\,\mathrm{mol\,dm^{-3}}

.

Answer:

Option A:

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

.

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