Question

The latent heat of vapourisation of water is $9700Cal/mole$ and if the b.p. is $100^{o}C$, ebullioscopic constant of water is.

• A. $0.513^{o}C$
• B. $1.026^{o}C$
• C. $10.26^{o}C$
• D. $1.832^{o}C$

Answer 1

Answer: (A)

$0.513^{o}C$

Answer 2

$K_{b} = \frac{M_{1}RT_{0}^{2}}{1000\Delta H_{V}} = \frac{18 \times 1.987 \times (373)^{2}}{1000 \times 9700} = 0.513^{o}C$