At 25^{}{C} evaporation of a 100\,{cm}³ solution of {K}₂{CO}₃ to dryness gave 14 g of the salt. What is the solubility of {K}₂{CO}₃ at 25^{}{C} ? [{K}₂{CO}₃=138]

Correct Answer is : 1.01 mol dn−3\;\;1.01\,{mol}\,{dn}^{-3}1.01moldn−3

Correct Answer is : 1.01 mol dn−3\;\;1.01\,{mol}\,{dn}^{-3}1.01moldn−3

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WAEC Chemistry 2021 Paper

Question : 30

Total: 50

25^{\circ}\text{C}

100\,\text{cm}^3

\mathrm{K}_2\mathrm{CO}_3

\mathrm{K}_2\mathrm{CO}_3

25^{\circ}\text{C}

[\mathrm{K}_2\mathrm{CO}_3=138]

Solution:

Concept:

Solubility in

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

is the number of moles of solute present in

1\,\mathrm{dm}^3

of solution.

Explanation:

Convert the volume of solution from

\mathrm{cm}^3

\mathrm{dm}^3

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

Calculate the moles of

\mathrm{K}_2\mathrm{CO}_3

\text{moles} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{14}{138} = 0.10145\,\mathrm{mol}

Find the solubility:

\text{solubility} = \dfrac{\text{moles}}{\text{volume in } \mathrm{dm}^3} = \dfrac{0.10145}{0.1} = 1.0145\,\mathrm{mol\,dm}^{-3}

This rounds to

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

Answer:

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

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