Question

For an aqueous solution, freezing point is 0.186${}^o$C. The boiling point of the same solution is:

[$K_f={1.86}^{o}Cmo{l}^{-1}kg$ and $K_b={0.512}^{o}Cmo{l}^{-1}kg$]

• A. $0.186{}^{o}C$
• B. $100.0512{}^{o}C$
• C. $1.86{}^{o}C$
• D. $5.12{}^{o}C$

Answer 1

Answer: (B)

$100.0512{}^{o}C$

The freezing point of pure water is 0 deg C and that of the aqueous solution is 0.186 ${}^{0}C$.

The depression in the freezing point $\Delta T_f=0.186-0=0.186{}^{o}C$

$\Delta T_f=K_f\times m$

$0.186{}^{o}C=1.86{}^{o}Cm\times m$

$m=0.1m$

The aqueous solution is 1 molal.

The elevation in the boiling point

$\Delta T_b=K_b\times m$

$\Delta T_b=0.512{}^{o}Cm\times 0.1m$

$\Delta T_b=0.0512{}^{o}C$

The boiling point of pure water is 100 deg C. The boiling point of the aqueous solution is $100+0.512=100.0512{}^{o}C$

Hence, the correct option is $B$