Question

A solution prepared from 0.3 g of an unknown nonvolatile solute and 30 g CCI, has a boiling point of 0.392 °C higher than that pure CCl. What is the molecular weight of the solute?

Answer 1

Answer:-

(We know that molal boiling point elevation constant of ccl₄ is 5.02 ^oC/m)

Solute = 0.3g

Solvent= 30 g

$\Delta t_b=0.392 \ ^oC \\ \Delta t_b=k_b×m \\$

$=k_b×{w_{solute}\over m_{solute}×({w_{solvent}\over1000})kg}$

$0.392=5.02×{0.3×1000\over M_b×30 }$

$\boxed{M_b=128.06 g} \ answer$