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 $166\, g\, mol^{-1}$)

$1.5\, mol\, kg^{-1}$

The correct answer is Option (1) → $1.5\, mol\, kg^{-1}$

A 20% (mass/mass) aqueous solution means:

• Mass of solute (KI): 20 g
• Mass of solution: 100 g
• Mass of solvent (water): $100\text{ g} - 20\text{ g} = 80\text{ g}$

2. Calculate Moles of Solute (KI):

Given the molar mass of KI is $166\text{ g mol}^{-1}$.

$\text{Moles of KI} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{20}{166} \approx 0.1205\text{ mol}$

3. Calculate Molality ($m$):

Molality is defined as the number of moles of solute per kilogram of solvent.

$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

$m = \frac{0.1205\text{ mol}}{80\text{ g} / 1000}$

$m = \frac{0.1205 \times 1000}{80} = \frac{120.5}{80}$

$m \approx 1.506\text{ mol kg}^{-1} \approx 1.5\text{ mol kg}^{-1}$