18 g of a non-volatile solute A is dissolved in 1 kg of water, the boiling point of water raised to 373.51 K. Given: Kb for water is 0.52 K kg mol-1. Boiling point for water is 373.15 K at 1.013 bar pressure.
The molecular weight of the solid A is
Answer & explanation
Correct answer: option 2
Weight of non-volatile solute A = 18 g
Molecular weight of solute A = M g
No of moles of solute = \(\frac{18}{M}\)
Mass of solvent = 1 kg
Molality, m = \(\frac{\text{No. of moles of solute}}{\text{Mass of solvent (in kg)}}\)
m = \(\frac{\frac{18}{M}}{1}\) = \(\frac{18}{M}\)
ΔTb = B.P. of solution - B.P of solvent = 373.51 - 373.15 = 0.36 K
ΔTb = kb x m
0.36 = 0.52 x \(\frac{18}{M}\)
M = \(\frac{0.52 × 18}{0.36}\) = 26.0 g mol-1