If 50\,{cm}³ of a saturated solution of {KNO}₃ at 40^{}\,{C} contained 5.05 g of the salt, its solubility at the same temperature would be [{KNO}₃=101]

Correct Answer is : 1.0 mol dm−3\;\; 1.0\,{mol}\,{dm}^{-3}1.0moldm−3.

Correct Answer is : 1.0 mol dm−3\;\; 1.0\,{mol}\,{dm}^{-3}1.0moldm−3.

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WAEC Chemistry 2022 Paper

Question : 30

Total: 50

50\,\mathrm{cm}^3

\mathrm{KNO}_3

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

[\mathrm{KNO}_3=101]

Solution:

Concept:

Solubility in

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

is the number of moles of solute dissolved in

1\,\mathrm{dm^3}

of solution.

Explanation:

Given: volume

V = 50\,\mathrm{cm^3}

and mass of

\mathrm{KNO_3} = 5.05\,\mathrm{g}

Molar mass of

\mathrm{KNO_3} = 39 + 14 + (3 \times 16) = 101\,\mathrm{g\,mol^{-1}}

Convert volume:

50\,\mathrm{cm^3} = 50 \times \frac{1}{1000} = 0.05\,\mathrm{dm^3}

Moles of

\mathrm{KNO_3} = \frac{5.05}{101} = 0.05\,\mathrm{mol}

Solubility

= \frac{\text{moles}}{\text{volume}} = \frac{0.05}{0.05} = 1.0\,\mathrm{mol\,dm^{-3}}

Answer:

\;1.0\,\mathrm{mol\,dm^{-3}}

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