Question

The boiling point of an aqueous solution of a non volatile solute is $100.15^{\circ }C.$ What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The values of $K_h$ and $K_f$ for water are 0.512 and $1.86K$ molality ${}^{-1}$

• A. $-0.544^{\circ }C$
• B. $-0.512^{\circ }C$
• C. $-0.272^{\circ }C$
• D. $-1.86^{\circ }C$

Answer 1

Answer: (C)

$\Delta T_b=k_b\times M...(1)$

If solution is diluted with equal volume of water, th concentration of solutions is $\frac{M}{2}$

So, $\Delta T_f=K_f\times \frac{M}{2}...(2)$

$\frac{(1)}{(2)}:\frac{\Delta T_b}{\Delta T_b}=\frac{K_b\times M}{K_f\times \frac{M}{2}}$

$=\frac{0.15}{\Delta T_f}=\frac{0.512\times 2}{1.86}$

$\Delta T_f=0.272$

$\Delta T_f=(F.P.{)}_{\text{solution }}-(F.P.{)}_{\text{solution }}$

$0.272=0^{\circ }C-(F.P.{)}_{\text{solution }}$

$(F.P.{)}_{\text{solution}}=-0.272^{\circ }C$