Question

1.0 x10^-3Kg of urea when dissolved in 0.0985 Kg of a solvent, decreases freezing point of the solvent by 0.211 k. 1.6x10 Kg of another non-electrolyte solute when dissolved in 0.086 Kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)

Answer 1

Mass of solute (A), i.e. urea = 1.0 × 10^-3 kg

Molar mass of solute (A) = 60

Mass of solvent in which solute A is dissolved = 0.0985 kg

Mass of solute (B) = 1.6 × 10^-3 kg

Molar mass of solute (B) = ?

Mass of solvent in which solute B is dissolved = 0.086 kg

${\displaystyle \frac{\Delta T_{f_A}}{\Delta T_{f_B}}=\frac{m_A}{m_B}}$

${\displaystyle \frac{0.211}{0.34}=\frac{\frac{\text{Mass of solute (A)}}{\text{Molecular mass of solute(A)×Kg of solvent}}}{\frac{\text{Mass of solute(B)}}{\text{Molecular mass of solute(B)×Kg of solvent}}}}$

${\displaystyle \frac{0.211}{0.34}=\frac{\frac{1\times {10}^{-3}}{60\times 0.0985}}{\frac{1.6\times {10}^{-3}}{\text{Molecular mass of solute(B)}\times 0.086}}}$

${\displaystyle \frac{0.211}{0.34}=\frac{1\times {10}^{-3}\times \text{Molecular mass of solute(B)}\times 0.086}{60\times 0.0985\times 1.6\times {10}^{-3}}}$

${\displaystyle \text{Molecular mass of solute(B)}=\frac{0.211\times 60\times 0.0985\times 1.6\times {10}^{-3}}{0.34\times 1\times {10}^{-3}\times 0.086}}$

Molar mass of another solute = 68.24