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) → $1.5\, mol\, kg^{-1}$

Step 1: Understand 20% (w/w)

$20\% \text{ (w/w)}$ means

$20\text{ g KI in } 100\text{ g solution}$

So,

• Mass of solute (KI) $= 20\text{ g}$
• Total mass of solution $= 100\text{ g}$

Step 2: Convert mass of solution $\rightarrow$ volume

We are given density:

$\text{Density} = \frac{\text{Mass}}{\text{Volume}} \text{}$

$\text{Volume} = \frac{\text{Mass}}{\text{Density}} \text{}$

$V = \frac{100}{1.202} = 83.2\text{ mL}$

Convert to litre:

$V = 0.0832 \text{ L}$

Step 3: Calculate moles of KI

Molar mass of KI:

$K = 39, I = 127$

$M = 39 + 127 = 166 \text{ g/mol}$

$\text{Moles} = \frac{20}{166} = 0.1205 \text{ mol}$

Step 4: Calculate Molarity

$M = \frac{\text{moles}}{\text{volume in L}}$

$M = \frac{0.1205}{0.0832}$

$M = 1.45 \approx 1.5 \text{ M}$