If 50.0\text{ cm}^3 of a saturated solution of potassium chloride (\text{KCl}) contains 14… — JAMB UTME

If $50.0\text{ cm}^3$ of a saturated solution of potassium chloride ( $\text{KCl}$ ) contains $14.9\text{ g}$ of the salt at $298\text{ K}$ , what is the solubility of potassium chloride at this temperature in $\text{mol dm}^{-3}$ ? (Molar mass of $\text{KCl} = 74.5\text{ g mol}^{-1}$ )

Answer: 4 mol dm⁻³

Answer

The solubility of potassium chloride at

298\text{ K}

4.0\text{ mol dm}^{-3}

First, the mass concentration is determined by scaling the

14.9\text{ g}

50.0\text{ cm}^3

1000\text{ cm}^3

, giving

298.0\text{ g dm}^{-3}

. Dividing

298.0\text{ g dm}^{-3}

by the molar mass of

\text{KCl}

(

74.5\text{ g mol}^{-1}

) yields

4.0\text{ mol dm}^{-3}

Step-by-Step Solution

Calculate the mass of solute present per cubic decimetre (1000 cm³) of saturated solution.

Mass concentration = 298.0 g dm⁻³

Solubility is expressed relative to 1 dm³ of solution volume.

Divide the mass concentration in g dm⁻³ by the molar mass of KCl.

Solubility = 4.0 mol dm⁻³

Molar solubility equals mass concentration divided by molar mass (M).

Key Concept

Solubility Determination in Moles per Decimetre Cubed

Estimated Time:1m 30s

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