Question

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

Kb of H2O = 0.512 oC/M

Answer 1

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

Step-by-step 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* K_b* m`

OR

`\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i* K_f* \frac{m_(solute)* 1000}{M_(solute)* 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* 0.512* (3.5* 1000)/(36.5* 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`