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]
Solubilty = \(\frac{number of moles}{Volume of solvent}\)\(\times 1000\)
Number of moles of KCl = \(\frac{mass}{Molar mass}\)
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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