10 g of non-volatile solute is dissolved in 180 g of H 2 Oresulting in lowering of vapour pressure by 0.5 %. Determine the boiling point of solution if Kb of water is 0.52 K kg mol -1

Question

10 g of non-volatile solute is dissolved in 180 g of H 2 Oresulting in lowering of vapour pressure by 0.5 %. Determine the boiling point of solution if Kb of water is 0.52 K kg mol -1 (A) 100.01° C (B) 100.15° C (C) 100.23° C (D) 100.32° C

Tardigrade Admin · Accepted Answer

p=p° x1 99.50=100 (n1/n1+n2) (n2/n1)=(1/191) Δ Tb=kb m=Kb (n2/n1) × (1000/m1) =0.52 × (1/191) × (1000/18)=0.15 Tb=10015° C

Q. $10 g$ of non-volatile solute is dissolved in $180 g$ of $H_{2}$ Oresulting in lowering of vapour pressure by $0.5%$ . Determine the boiling point of solution if $K_{b}$ of water is $0.52 K k g m o l^{- 1}$

$100.0 1^{\circ} C$

$100.1 5^{\circ} C$

$100.2 3^{\circ} C$

$100.3 2^{\circ} C$

$p = p^{\circ} x_{1}$
$99.50 = 100 \frac{n _{1}}{n _{1} + n _{2}}$
$\frac{n _{2}}{n _{1}} = \frac{1}{191}$
$Δ T_{b} = k_{b} m = K_{b} \frac{n _{2}}{n _{1}} \times \frac{1000}{m _{1}}$
$= 0.52 \times \frac{1}{191} \times \frac{1000}{18} = 0.15$
$T_{b} = 1001 5^{\circ} C$

Q. 10g of non-volatile solute is dissolved in 180g of H2​ Oresulting in lowering of vapour pressure by 0.5%. Determine the boiling point of solution if Kb​ of water is 0.52Kkgmol−1

100.01∘C

100.15∘C

100.23∘C

100.32∘C