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JAMB Chemistry 2024 Paper

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Question : 118

Total: 138

25.0g of potassium chloride were dissolved in 80g of distilled water at 30

^{0}

C. Calculate the solubility of the solute in mol dm

^{3}

. [K =39, Cl = 35.5]

3.9 moldm $^{3}$
4.6 moldm $^{3}$
4.0 moldm $^{3}$
4.2 moldm $^{3}$

Solution:

Solubilty = $\; \frac{\text{number of moles}}{\text{Volume of solvent}}$ $\times 1000$

Number of moles of KCl = $\; \frac{\text{mass}}{\text{Molar mass}}$

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol, mass = 25.0g

Number of moles of KCl = $\; \frac{25}{74.5}$ = 0.336 mole

Solubilty = $\; \frac{0.336}{80}$ $\times 1000$

= 4.2 moldm $^{3}$

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