Question

Calculate the boiling point of a 3.5 % solution (by weight) of sodium chloride in water.

Answer 1

Answer: The boiling point of the solution is $101.02^oC$

Explanation:

We are given:

3.5 % (by weight) NaCl

This means that 3.5 g of NaCl is present in 100 g of solution

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent (water) = (100 - 3.5) g = 96.5 g

Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

The expression for the calculation of elevation in boiling point is:

$\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m$

OR

$\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Boiling point of pure solvent (water) = $100^oC$

Boiling point of solution = ?

i = Vant Hoff factor = 2 (for NaCl)

$K_b$ = Boiling point elevation constant = $0.512^oC/m$

$m_{solute}$ = Given mass of solute (NaCl) = 3.5 g

$M_{solute}$ = Molar mass of solute (NaCl) = 36.5 g/mol

$w_{solvent}$ = Mass of solvent (water) = 96.5 g

Putting values in equation 1, we get:

$\text{Boiling point of solution}-(100)=2\times 0.512\times \frac{3.5\times 1000}{36.5\times 96.5}\\\\\text{Boiling point of solution}=(1.02+100)^oC\\\\\text{Boiling point of solution}=101.02^oC$

Hence, the boiling point of the solution is $101.02^oC$