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 $k_b$and$k_f$ for water are $0.512$and$1.86K{\text{molality}}^{-1}$.

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

Answer 1

Answer: (C)

$-{0.272}^{\circ }C$

As we know,

$\Delta T_b=k_{b}m;0.15=0.512\times m$

$m=0.292;\Delta T_f=k_f{m}^{\prime }$

$m^{\prime }=m/2;T_1-T_0=1.86\times \frac{0.292}{2}=0.27{}^{0}C$

$T_1=0-0.27=-{0.27}^{\circ }C$

Hence, the correct option is $C$