Question

10 g of potassium chloride dissolves in 31.5 g of water at $70°\mathrm{C}$ to form a saturated solution. Calculate the solubility of potassium chloride at this temperature?

Answer 1

Solubility

• The maximum amount of solute that can be dissolved in a solvent at a specific temperature is called the solubility of that substance
• Here, it is given that 10 g of potassium chloride $\left(\mathrm{KCl}\right)$ requires 31.5 g of water $\left({\mathrm{H}}_2\mathrm{O}\right)$
• $\mathrm{solubility}\mathrm{of}\mathrm{solution}=(\mathrm{mass}\mathrm{of}\mathrm{solute}/\mathrm{mass}\mathrm{of}\mathrm{solvent})\mathrm{x}100$
• For 1 g of water, required potassium chloride is $=10/31.5=0.3174\mathrm{g}$
• For 100 g of water, required potassium chloride is $=10\mathrm{x}100/31.5=31.74\mathrm{g}$

Therefore, solubility of $\mathrm{KCl}$at $70°\mathrm{C}$ is $31.74\mathrm{g}$