Question

The boiling point of pure water is 373 K. Calculate the boiling point of an aqueous solution containing 18 gms of glucose (MW = 180) in 100 gms of water. Molal elevation constant of water is 0.52 K kg mol^-1.

Answer 1

 Mass of solvent (W_A) = 100 g 

Mass of solute (W_B) = 18 g 

Molecular mass of solute (M_B) = 180

Molality = \(\frac{W_B \times 1000}{M_B \times W_A}\)

\(=\frac{18\times1000}{180\times 100}\) = 1 mol. kg^-1

△ T_b = k_b x m = 0.52 x 1 = 0.52 K

Boling point of solution 

= [Boiling point of pure solvent] + [Elevation in boiling point]

= 373 + 0.52 = 373.52 K

\(\pi =\frac{n}{V}.RT\)