A solution containing `0.5216g` of naphthalene (mol.wt.`=128.16`) in 50mL of `"CC"l_(4)` shows boiling point elevation of `0.402^(@)` while a solution of `0.6216g` of an unknown solute in the same weight of solvent gave a boiling point elevation of `0.647^(@)` . Find the molecular mass of the unknown solute.

Explanation

In this case, the first data is used to find the value of `K_(b)`. This value of `K_(b)` is used in the second data to find the molecular mass.
Srep I. To find `K_(b)` for `C Cl_(4)` from data on napthalene solution
`w_(2)=0.5126 g, w_(1)=50.00 g, M_(2)=128 g mol^(-1), Delta T_(b)=0.402^(@)C=0.402 K^(**)`
`K_(b)=(Delta T_(b).w_(1)M_(2))/(1000 w_(2))=(0.402 K xx 50 g xx 128 g mol^(-1))/(1000 g kg^(-1)xx0.5126 g)=5.02" K kg mol"^(-1)`
Steo II. To find the mol. mass of unknown solute.
The given data are : `w_(2)=0.6216 g, w_(1)=50.00 g, Delta T_(b)=0.647^(@)C=0.647 K`
`K_(b)=5.02 "K kg mol"^(-1)` (calculated above)
`M_(2)=(1000 K_(b)w_(2))/(w_(1)Delta T_(b))=(1000 g kg^(-1)xx5.02"K kg mol"^(-1)xx 0.6216 g)/(50g xx0.647 K)=96.46 g mol^(-1)`