A solution of 1059g sodium chloride in 125 ml of water densi

A solution of 10.59g sodium chloride in 125 ml of water (density of water is 1.00g/ml, Kb of water is 0.52°C/m, find: b. 1. 2. The molality of sodium chloride The new boiling point of water if it boils at 100.°C

Solution

a) Density of water is given as 1.00 g/mL.

The mass of 125 mL of water = (volume of water)*(density of water) = (125 mL)*(1.00 g/mL) = 125.00 g = (125.00 g)*(1 kg/1000 g) = 0.125 kg.

Mass of sodium chloride taken = 10.59 g.

Molar mass of sodium chloride = (1*22.9897 + 1*35.453) g/mol = 58.4427 g/mol.

Mole(s) of sodium chloride corresponding to 10.59 g = (10.59 g)/(58.4427 g/mol) = 0.1812 mole.

Molality of sodium chloride solution = (moles of sodium chloride)/(kilograms of water) = (0.1812 mole)/(0.125 kg) = 1.4496 mol/kg = 1.4496 m (m = molal) (ans).

b) The boiling point elevation is given as

T_b = K_b*(molality of solution) = (0.52°C/m)*(1.4496 m) = 0.753792 °C.

We define T_b = T_s – T_p where T_s = boiling point of the solution and T_p = boiling point of pure water. Given T_p = 100°C, we have

0.753792°C = T_s – (100°C)

====> T_s = (100 + 0.753792)°C = 100.753792°C 100.75°C (ans).