The solubility of a salt Y (molar mass = 160\text{ g mol}^{-1}) in water is 2.5\text{ mol … — JAMB UTME

The solubility of a salt $Y$ (molar mass = $160\text{ g mol}^{-1}$ ) in water is $2.5\text{ mol dm}^{-3}$ at $70^\circ\text{C}$ and $1.2\text{ mol dm}^{-3}$ at $30^\circ\text{C}$ . If $300\text{ cm}^3$ of a saturated solution of $Y$ at $70^\circ\text{C}$ is cooled to $30^\circ\text{C}$ , what mass of salt $Y$ will crystallize out of solution?

$62.4\text{ g}$ Answer
B
$120.0\text{ g}$
C
$208.0\text{ g}$
D
$0.39\text{ g}$

Answer

The mass of salt

Y

that crystallizes out of solution is

62.4\text{ g}

The mass of salt precipitated is determined by taking the difference in molar solubility (

2.5 - 1.2 = 1.3\text{ mol dm}^{-3}

), multiplying by the volume fraction (

300/1000 = 0.3\text{ dm}^3

) to find the number of moles (

0.39\text{ mol}

), and then multiplying by the molar mass (

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

) to obtain

62.4\text{ g}

Step-by-Step Solution

Determine the change in molar solubility upon cooling

\Delta S = 2.5\text{ mol dm}^{-3} - 1.2\text{ mol dm}^{-3} = 1.3\text{ mol dm}^{-3}

Solubility decreases as temperature drops, causing the excess solute to precipitate.

Calculate the amount in moles precipitated in

300\text{ cm}^3

of solution

n = 1.3\text{ mol dm}^{-3} \times \frac{300\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.39\text{ mol}

The solution volume is

300\text{ cm}^3

(

0.3\text{ dm}^3

), so the moles precipitated must be scaled from

1\text{ dm}^3

Convert the moles precipitated to mass in grams

\text{Mass} = 0.39\text{ mol} \times 160\text{ g mol}^{-1} = 62.4\text{ g}

Mass is found by multiplying the mole quantity by the given molar mass of the salt.

Key Concept

Solubility and Crystallization Calculations

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