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

Question

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

Tardigrade Admin · Accepted Answer

Kb=(M1 R T02/1000 Δ HV) =(18 × 1.987 ×(373)2/1000 × 9700) =0.513° C

Q. The latent heat of vaporization 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

$1.02 6^{\circ} C$

$0.51 3^{\circ} C$

$10.2 6^{\circ} C$

$1.83 2^{\circ} C$

$K_{b} = \frac{M _{1} R T _{0}^{2}}{1000Δ H _{V}}$

$= \frac{18 \times 1.987 \times ( 373 ) ^{2}}{1000 \times 9700}$

$= 0.51 3^{\circ} C$

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

1.026∘C

0.513∘C

10.26∘C

1.832∘C