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JAMB Chemistry 2011 Paper

Question : 9

Total: 47

From the diagram above, find the amount of solute deposited when 200 cm ${}^3$ of the solution is cooled from 55°C to 40°C.

Solution:

Concept:

The solute deposited on cooling is found from the difference in solubility at the two temperatures, then adjusted for the given volume.

Explanation:

From the solubility curve, at

55^\circ\text{C}

1000\ \text{cm}^3

of solution contains

7

moles of solute.

40^\circ\text{C}

1000\ \text{cm}^3

of solution contains

6

moles of solute.

Amount deposited from

1000\ \text{cm}^3

7 - 6 = 1

mole.

For

200\ \text{cm}^3

\text{deposit} = \frac{200}{1000} \times 1 = 0.20

mole.

Answer:

0.20

mole

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