A solution of 0.5 g of a solute (molar mass =150 g mol- 1 ) in 50 g of a solvent yields a boiling point elevation of 0.40K . Another solution of 0.60g of an unknown solute in the same mass of solvent exhibits a boiling point elevation of 0.8 K . The molar mass of unknown solute is :

Question

A solution of 0.5 g of a solute (molar mass =150 g mol- 1 ) in 50 g of a solvent yields a boiling point elevation of 0.40K . Another solution of 0.60g of an unknown solute in the same mass of solvent exhibits a boiling point elevation of 0.8 K . The molar mass of unknown solute is : (A) 60 g mol- 1 (B) 90 g mol- 1 (C) 120 g mol- 1 (D) 180 g mol- 1

Tardigrade Admin · Accepted Answer

$60 g m o l^{- 1}$

$90 g m o l^{- 1}$

$120 g m o l^{- 1}$

$180 g m o l^{- 1}$

Q. A solution of 0.5g of a solute (molar mass =150gmol−1 ) in 50g of a solvent yields a boiling point elevation of 0.40K . Another solution of 0.60g of an unknown solute in the same mass of solvent exhibits a boiling point elevation of 0.8K . The molar mass of unknown solute is :

60gmol−1

90gmol−1

120gmol−1

180gmol−1

$60 g m o l^{- 1}$

$90 g m o l^{- 1}$

$120 g m o l^{- 1}$

$180 g m o l^{- 1}$