The latent heat of vaporisation of water is 9700 cal/mole and if the b.p. is 100°C, ebullioscopic constant of water is

Question

The latent heat of vaporisation of water is 9700 cal/mole and if the b.p. is 100°C, ebullioscopic constant of water is (A) 0.513°C (B) 1.026°C (C) 10.26°C (D) 1.832°C

Tardigrade Admin · Accepted Answer

Kb=(M1 RT02/1000 Δ HV) =(18×1.987×(373)2/1000×9700) =0.513°C

Q. The latent heat of vaporisation of water is $9700 c a l / m o l e$ and if the b.p. is $10 0^{\circ} C$ , ebullioscopic constant of water is

$0.51 3^{\circ} C$

$1.02 6^{\circ} C$

$10.2 6^{\circ} C$

$1.83 2^{\circ} C$

$K_{b} = \frac{M _{1} R T _{0}^{2}}{1000 Δ H _{V}}$
$= \frac{18 \times 1.987 \times ( 373 ) ^{2}}{1000 \times 9700}$
$= 0.51 3^{\circ} C$

Q. The latent heat of vaporisation of water is 9700cal/mole and if the b.p. is 100∘C, ebullioscopic constant of water is

0.513∘C

1.026∘C

10.26∘C

1.832∘C