The solubility of a salt Y (molar mass = 100\text{ g/mol}) in water is 3.5\text{ mol/dm}^3… — JAMB UTME

The solubility of a salt $Y$ (molar mass = $100\text{ g/mol}$ ) in water is $3.5\text{ mol/dm}^3$ at $70^\circ\text{C}$ and $1.5\text{ mol/dm}^3$ at $25^\circ\text{C}$ . What mass of salt $Y$ will crystallize out when $500\text{ cm}^3$ of its saturated solution is cooled from $70^\circ\text{C}$ to $25^\circ\text{C}$ ?

$100.0\text{ g}$ Answer
B
$175.0\text{ g}$
C
$200.0\text{ g}$
D
$1.0\text{ g}$

Answer

The mass of salt

Y

that crystallizes out of solution is

100.0\text{ g}

Cooling a saturated solution reduces its capacity to retain solute. The difference in concentration between

70^\circ\text{C}

(

3.5\text{ mol/dm}^3

) and

25^\circ\text{C}

(

1.5\text{ mol/dm}^3

) is

2.0\text{ mol/dm}^3

. Scaling this concentration difference for

500\text{ cm}^3

(

0.5\text{ dm}^3

) gives

1.0\text{ mole}

of salt

Y

. Converting moles to mass yields

1.0\text{ mole} \times 100\text{ g/mol} = 100.0\text{ g}

Step-by-Step Solution

Calculate the difference in solubility between

70^\circ\text{C}

and

25^\circ\text{C}

\text{mol/dm}^3

\Delta S = 3.5\text{ mol/dm}^3 - 1.5\text{ mol/dm}^3 = 2.0\text{ mol/dm}^3

Solubility decreases as temperature drops, causing solute to precipitate.

Convert solution volume from

\text{cm}^3

\text{dm}^3

and determine the moles of solute precipitated.

V = \frac{500\text{ cm}^3}{1000} = 0.5\text{ dm}^3

;

\text{Moles precipitated} = 2.0\text{ mol/dm}^3 \times 0.5\text{ dm}^3 = 1.0\text{ mole}

Solubility is given per cubic decimeter, so volume scaling is necessary.

Convert the precipitated moles into mass in grams using molar mass.

\text{Mass} = 1.0\text{ mole} \times 100\text{ g/mol} = 100.0\text{ g}

Mass equals number of moles multiplied by molar mass.

Key Concept

Crystallization calculations based on temperature-dependent solubility curves

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