The solubility of a salt X is 0.80\text{ mol dm}^{-3} at 60^\circ\text{C} and 0.30\text{ m… — JAMB UTME

The solubility of a salt $X$ is $0.80\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $0.30\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . Calculate the mass of salt $X$ (in grams) that will crystallize out of solution when $1.0\text{ dm}^3$ of its saturated solution is cooled from $60^\circ\text{C}$ to $20^\circ\text{C}$ . (Molar mass of salt $X = 100\text{ g mol}^{-1}$ )

Answer: 50 g

Answer

The mass of salt

X

that crystallizes out of the solution is

50\text{ g}

60^\circ\text{C}

1.0\text{ dm}^3

of saturated solution contains

0.80\text{ mol}

of salt

X

. When cooled to

20^\circ\text{C}

, the solution can only hold

0.30\text{ mol}

. The excess amount that crystallizes out is

0.80\text{ mol} - 0.30\text{ mol} = 0.50\text{ mol}

. Converting this amount to mass yields

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

Step-by-Step Solution

Calculate the difference in molar solubility between the two temperatures

0.80\text{ mol dm}^{-3} - 0.30\text{ mol dm}^{-3} = 0.50\text{ mol dm}^{-3}

This difference represents the amount of solute in moles per cubic decimeter that can no longer remain dissolved when cooled to

20^\circ\text{C}

Convert the precipitated moles into mass in grams for

1.0\text{ dm}^3

of solution

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

Multiplying the precipitated amount in moles by the molar mass gives the total mass in grams that crystallizes out.

Key Concept

Solubility and crystallization calculations upon cooling

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