The table below shows the solubility of potassium trioxonitrate(V), \text{KNO}_3, in water at two different temperatures: | Temperature (^\circ\text{C}) | Solubility (\text{g / } 100\text{ g } \text{H}_2\text{O}) | |---|---| | 60 | 110.0 | | 20 | 32.0 | What mass of \text{KNO}_3 will deposit when 42

The table below shows the solubility of potassium trioxonitrate(V), \text{KNO}_3, in water… — JAMB UTME

The table below shows the solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water at two different temperatures:

Temperature ( $^\circ\text{C}$ )	Solubility ( $\text{g / } 100\text{ g } \text{H}_2\text{O}$ )
60	110.0
20	32.0

What mass of $\text{KNO}_3$ will deposit when $420.0\text{ g}$ of a saturated solution of $\text{KNO}_3$ at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ?

A
$78.0\text{ g}$
$156.0\text{ g}$ Answer
C
$220.0\text{ g}$
D
$327.6\text{ g}$

Answer

156.0\text{ g}

60^\circ\text{C}

, every

210.0\text{ g}

of saturated solution contains

100.0\text{ g}

of water and

110.0\text{ g}

of solute. Thus,

420.0\text{ g}

of solution contains

200.0\text{ g}

of water and

220.0\text{ g}

of dissolved

\text{KNO}_3

. Upon cooling to

20^\circ\text{C}

, the solubility drops to

32.0\text{ g}

per

100\text{ g}

of water, so

200.0\text{ g}

of water can retain only

64.0\text{ g}

of solute. The excess solute that crystallizes out is

220.0\text{ g} - 64.0\text{ g} = 156.0\text{ g}

Step-by-Step Solution

Determine the composition of a saturated solution at

60^\circ\text{C}

60^\circ\text{C}

100.0\text{ g}

of water dissolves

110.0\text{ g}

\text{KNO}_3

, yielding

100.0 + 110.0 = 210.0\text{ g}

of saturated solution.

Solubility is expressed per

100\text{ g}

of water, so total solution mass is the sum of solvent and solute masses.

Calculate the mass of water and solute in

420.0\text{ g}

of saturated solution at

60^\circ\text{C}

Mass of water =

420.0 \times \frac{100.0}{210.0} = 200.0\text{ g}

. Mass of

\text{KNO}_3

420.0 \times \frac{110.0}{210.0} = 220.0\text{ g}

Scaling the ratio of components to the given solution mass.

Calculate the mass of solute remaining dissolved at

20^\circ\text{C}

20^\circ\text{C}

100.0\text{ g}

of water holds

32.0\text{ g}

\text{KNO}_3

. Therefore,

200.0\text{ g}

of water holds

2 \times 32.0 = 64.0\text{ g}

\text{KNO}_3

Solubility at the lower temperature determines the maximum solute that stays in solution.

Calculate the mass of salt crystallized (deposited)

Mass deposited =

220.0\text{ g} - 64.0\text{ g} = 156.0\text{ g}

Subtracting the solute remaining in solution from the initial mass of dissolved solute.

Key Concept

Solubility curves and fractional crystallization calculations

Estimated Time:1m 30s

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