Question

What would be the molality of $20\%$ (mass/mass) aqueous solution of $KI$ ?
Molar mass of $KI$ is $166gmo{l}^{-1}$

• A. $1.51$
• B. $1.35$
• C. $1.08$
• D. $1.48$

Answer 1

The correct option is A $1.51$

$20\%$ (mass/mass) aqueous solution of $KI$ means $20g$ of $KI$ in $100g$ of solution.
Mass of solution =100g
Mass of solute=20g
Mass of solvent=mass of solution - mass of solute
$=(100-20)g=80g=0.08kg$
Molar mass of $KI$ is $166gmo{l}^{-1}$
$\text{Number of moles of solute}\left(n\right)=\frac{\text{mass of KI}}{\text{molar mass of KI}}=\frac{20}{166}=0.121mol$
So, molality of solution $m=\frac{\text{number of moles of solute}}{\text{weight of solvent (in kg)}}$
$\Rightarrow m=\frac{0.121}{0.08}=1.51mol/kg$
Hence, option (A) is the correct.