The table below shows the solubility of anhydrous copper(II) tetraoxosulfate(VI), \text{CuSO}_4, in water at different temperatures: | Temperature (^\circ\text{C}) | Solubility (\text{g} of \text{CuSO}_4 per 100\text{ g} of \text{H}_2\text{O}) | |---|---| | 20 | 20.0 | | 40 | 29.0 | | 80 | 55.0 | A

The table below shows the solubility of anhydrous copper(II) tetraoxosulfate(VI), \text{Cu… — JAMB UTME

The table below shows the solubility of anhydrous copper(II) tetraoxosulfate(VI), $\text{CuSO}_4$ , in water at different temperatures:

Temperature ( $^\circ\text{C}$ )	Solubility ( $\text{g}$ of $\text{CuSO}_4$ per $100\text{ g}$ of $\text{H}_2\text{O}$ )
$20$	$20.0$
$40$	$29.0$
$80$	$55.0$

A saturated solution of copper(II) tetraoxosulfate(VI) in $200.0\text{ g}$ of water at $80^\circ\text{C}$ is cooled to $20^\circ\text{C}$ . Given that the molar mass of $\text{CuSO}_4 = 160\text{ g mol}^{-1}$ and $\text{H}_2\text{O} = 18\text{ g mol}^{-1}$ , what is the exact mass of hydrated copper(II) tetraoxosulfate(VI) pentahydrate crystals, $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$ , that will deposit from the solution?

$123.24\text{ g}$ Answer
B
$109.38\text{ g}$
C
$70.00\text{ g}$
D
$44.80\text{ g}$

Answer

The mass of copper(II) tetraoxosulfate(VI) pentahydrate crystals deposited is

123.24\text{ g}

When a hydrated salt crystallizes, it removes both solute and water of crystallization from the saturated solution. Taking into account that

m

grams of

\text{CuSO}_4\cdot 5\text{H}_2\text{O}

contains

0.64m

grams of

\text{CuSO}_4

and

0.36m

grams of

\text{H}_2\text{O}

, setting up the solubility ratio at

20^\circ\text{C}

(110.0 - 0.64m) / (200.0 - 0.36m) = 0.20

gives

m = 123.24\text{ g}

Step-by-Step Solution

Calculate the initial mass of dissolved anhydrous

\text{CuSO}_4

80^\circ\text{C}

200.0\text{ g}

of water, mass of dissolved

\text{CuSO}_4 = 2 \times 55.0\text{ g} = 110.0\text{ g}

Solubility at

80^\circ\text{C}

55.0\text{ g}

per

100\text{ g}

of water.

Determine the molar masses of the anhydrous salt, water, and hydrate.

Molar mass of

\text{CuSO}_4 = 160\text{ g mol}^{-1}

;

\text{H}_2\text{O} = 18\text{ g mol}^{-1}

;

\text{CuSO}_4\cdot 5\text{H}_2\text{O} = 160 + 5(18) = 250\text{ g mol}^{-1}

Needed to establish mass fractions of solute and solvent in the crystals.

Express the mass fractions of anhydrous salt and water in

m

grams of

\text{CuSO}_4\cdot 5\text{H}_2\text{O}

crystals.

Anhydrous

\text{CuSO}_4

fraction

= \frac{160}{250}m = 0.64m

; Water fraction

= \frac{90}{250}m = 0.36m

As crystals form, they take away both anhydrous salt and water from the solution.

Set up the solubility saturation equation at

20^\circ\text{C}

\frac{110.0 - 0.64m}{200.0 - 0.36m} = \frac{20.0}{100.0} = 0.20

20^\circ\text{C}

, the remaining solution must remain saturated.

Solve for

m

110.0 - 0.64m = 0.20(200.0 - 0.36m) \implies 110.0 - 0.64m = 40.0 - 0.072m \implies 70.0 = 0.568m \implies m = 123.24\text{ g}

Isolating

m

gives the exact mass of hydrated crystals deposited.

Key Concept

Crystallization of Hydrated Salts from Saturated Solutions

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