Question

$1$ mol $NaCl$ in $1000g$ $H_{2}O$ will change the boiling point of water to

• A. $100.51$
• B. $101.04$
• C. $101.53$
• D. $101.86$
• E. $103.62$

Answer 1

The correct option is B $101.04$

Elevation in boiling point can be calculated as $\Delta T_b=i.K_b.m$, where $i$ and$K_b$ are van't hoff factor of $NaCl$ and ebullioscopic constant of water respectively.

Van't hoff factor of $NaCl=2$

Molality of $NaCl=\frac{1}{1000}\times 1000=1m$

$K_b=0.512{}^{0}C/molal$

$\Delta T_b=T_{final}-T_{initial}=i.K_b.m$

$T_{final}=100+2\times 0.512\times 1$

$\Rightarrow T_{final}=100+1.024=101.024{}^{0}C$