Question

The boiling point of water at 735 torr is 99.07°c. the mass of nacl added in 100 g water to make its boiling point 100°c is (kb = 0.51 k kg mol–1)

Answer 1

You just have to use the eqn of elevation in boiling boint.

Answer 2

Answer: 10.6 grams

Explanation:

Elevation in boiling point:-

$\Delta T_b=K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

weight of solvent (water)= 100 g= 0.1 kg       (1kg=1000g)

$\Delta T_b$ = change in boiling point

$K_b$ = boiling point constant = $0.51Kkgmol^-1$

$\Delta T_b=T_b-T_b^0=(100-99.07)^0C=0.93^0C$

$0.93=0.51\times \frac{xg}{58.5g/mol\times 0.1kg}$

$x=10.6$

The amount of NaCl to be added is 10.6 grams.