The solubility of a solute X in water is 1.5\text{ mol/dm}^3 at 80^\circ\text{C} and 0.5\t… — JAMB UTME

The solubility of a solute $X$ in water is $1.5\text{ mol/dm}^3$ at $80^\circ\text{C}$ and $0.5\text{ mol/dm}^3$ at $30^\circ\text{C}$ . If $250\text{ cm}^3$ of a saturated solution of $X$ at $80^\circ\text{C}$ is cooled to $30^\circ\text{C}$ , what mass of $X$ will crystallize out of the solution? [Molar mass of $X = 160\text{ g/mol}$ ]

$40.0\text{ g}$ Answer
B
$160.0\text{ g}$
C
$0.25\text{ g}$
D
$60.0\text{ g}$

Answer

The mass of salt

X

that crystallizes out of solution is

40.0\text{ g}

Cooling

1\text{ dm}^3

of saturated solution from

80^\circ\text{C}

30^\circ\text{C}

precipitates

1.0\text{ mol}

of solute. For a

250\text{ cm}^3

(

0.25\text{ dm}^3

) solution,

0.25\text{ mol}

precipitates. Multiplying

0.25\text{ mol}

by the molar mass (

160\text{ g/mol}

) yields

40.0\text{ g}

Step-by-Step Solution

Calculate the difference in solubility between

80^\circ\text{C}

and

30^\circ\text{C}

per

\text{dm}^3

\Delta S = 1.5\text{ mol/dm}^3 - 0.5\text{ mol/dm}^3 = 1.0\text{ mol/dm}^3

Solubility decreases upon cooling, causing the excess solute to precipitate.

Scale the amount of precipitated solute to the specified volume of

250\text{ cm}^3

(

0.25\text{ dm}^3

n_{\text{precipitated}} = 1.0\text{ mol/dm}^3 \times 0.25\text{ dm}^3 = 0.25\text{ mol}

The solution volume is

250\text{ cm}^3

, which is a quarter of a cubic decimeter.

Convert the moles of precipitated solute to mass in grams using its molar mass.

\text{Mass} = 0.25\text{ mol} \times 160\text{ g/mol} = 40.0\text{ g}

Mass is obtained by multiplying the chemical amount in moles by the molar mass.

Key Concept

Calculation of mass crystallized from solubility curves and temperature changes

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