Question

Calculate the molality of KI if the density of 20% (mass/mass) aqueous solution of KI is (1)202 g mL$^{-1}$.
(Molar mass of KI is 166 g mol$^{-1}$)

• A. (1)5 mol kg$^{-1}$
• B. (1)2 mol kg$^{-1}$
• C. (1)5 mol kg$^{-1}$
• D. 0.12 mol kg$^{-1}$

Hint:

Remember to use the density of the solution to calculate the mass of the solvent when molality is involved.

Answer 1

The Correct Option is A

The mass of KI in 100 g of solution is 20 g. Thus, the mass of water in 100 g of solution is: \[ \text{Mass of water} = 100 - 20 = 80 \, \text{g} = 0.08 \, \text{kg} \] Now, using the formula for molality: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] We calculate the moles of KI: \[ \text{Moles of KI} = \frac{\text{mass of KI}}{\text{molar mass of KI}} = \frac{20 \, \text{g}}{166 \, \text{g/mol}} = 0.1205 \, \text{mol} \] Thus, molality \( m \) is: \[ m = \frac{0.1205 \, \text{mol}}{0.08 \, \text{kg}} = (1)5 \, \text{mol kg}^{-1} \] Final Answer: \[ \boxed{(1)5 \, \text{mol kg}^{-1}} \]