Question

The density of $20\%$ (mass/mass) aqueous KI is $1.202\,g/L$ . The molality of the solution is $x$ . Then the value of $10x$ is (nearest integer)

Answer 1

Weight of $KI$ in $100\,g$ of water $= 20\, g$
Weight of water in the solution $= 100 - 20 = 80 \,g= 0.08 \,kg$
Molar mass of $KI = 39 + 127 = 166\, g\ mol^{-1}$
Molality of the solution $\left(\right.m)=\frac{\text{Number of moles of KI}}{\text{Mass of water in kg}}$
$=\frac{(20\, g ) /\left(166\, g \,mol ^{-1}\right)}{(0 \cdot 08\, kg )}$
$= 1.506\, mol\, kg^{-1} = 1.506 \,m $
By rounding off to one place, we get the molality as $1.5$.