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)`.

Answer

57.5g `"mol"^(-1)`

Explanation

Weight of solvent `(W_(1)) = 90 g`
Weight of solute `(W_(2)) = 1.80 g`
Mol. wt. of solute `(M_(2))=? DeltaT_(b)=354.11 -353.23=0.88K`
`K_(b)=2.53K`
Now `DeltaT_(b)=K_(b) xx m or m=(DeltaT_(b))/(K_(b))=(0.88)/(2.53)=0.3478"mol kg"^(-1)`
`m=("Weight of solue" xx1000)/("Gram mol. wt. of solute" xx "Weight of solvent")`
Hence, `0.3478=(1.8 xx 1000)/(M_(2)xx90) or M_(2)=(1.8xx1000)/(90xx0.3478)`
`=57.5"g mol"^(-1)`