Home
Class 12
CHEMISTRY
When 1.5 g of a non-volatile solute was ...

When 1.5 g of a non-volatile solute was dissoved in 90 g of benzene, the boiling point of benzene was raised from `353.23^(-1)` K to 353.93 K. Calculate the molar mas of solute (`K_(b)` for benzene=2.52 K kg `mol^(-1)`).

Text Solution

Verified by Experts

The correct Answer is:
60.0 g `mol^(-1)`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xx W_(A))`
`W_(B)=1.5g , W_(A)=90.0 g=0.09" kg", DeltaT_(b)=0.7" K"`
`K_(b)=2.52" k kg mol"^(-1)`
`M_(B)=((2.52" K kg mol"^(-1))xx(1.5" g"))/((0.7" K")xx(0.09" kg"))=60.0" g mol"^(-1)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

The boiling a point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_(b) for benzene is 2.53 K kg mol^(-1) .

The boiling point of benzene is 353 .23 K. When 1.80 gram of non- volatile solute was dissolved in 90 gram of benzene , the boiling point is raised to 354.11 K. Calculate the molar mass of solute . [K_(b) for benzene = 2.53 K kg mol^(-1)]