Question

An aqueous dilute solution containing non-volatile solute boils at $100.52^{\circ} C$. What is the molality of solution? $\left(K_{b}=0.52 \,kg \,mol ^{-1} K\right.$, boiling temperature of water $=100^{\circ} C$ )

• A. 0.1 m
• B. 0.01 m
• C. 0.001 m
• D. 1.0 m

Answer 1

Answer: (A)

$\Delta T_{b}=K_{b} \cdot m $

$T_{b}-T_{b}^{\circ}=K_{b} \cdot m$

where, $K_{b}=$ molal elevation constant

$m=$ molality

$(100.052-100.00) =(0.52)( m ) $

$0.052 =0.52 \times m$

$m =0.1$