Question

What is the boiling point of a solution made by dissolving 23.6 g of a non-ionizing solute with a molar mass of 103.5 g/mol in 188 g of water?

Answer 1

Answer: The boiling point of solution is 100.62°C

Explanation:

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

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

Boiling point of pure water = 100°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 0.512°C/m.g

$m_{solute}$ = Given mass of solute = 23.6 g

$M_{solute}$ = Molar mass of solute = 103.5 g/mol

$W_{solvent}$ = Mass of solvent (water) = 188 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=1\times 0.512^oC/m\times \frac{23.6\times 1000}{103.5g/mol\times 188}\\\\\text{Boiling point of solution}=100.62^oC$

Hence, the boiling point of solution is 100.62°C