A solution containing 0.513 g of naphthalene (molar mass = 128) in 50 g of `C Cl_(4)` gives a boiling point elevation of `0.402^(@)C` while a solution of 0.625 g of an unknown solute gives a boiling point elevation of `0.650^(@)C` . Find the molar mass of the unknown solute .

Answer

`96.54 g mol^(-1)`

Explanation

Calculation of `k_(b)` :
`k_(b) = (Delta T_(b) xx w_(A) xx M_(B))/(1000 xx w_(B))`
` = (0.402 x 50 xx 128)/(1000 x 0.513) = 5.02 K m^(-1)`
`M_(B) = (5.02 xx 0.625 xx 1000)/(0.650 xx 50) =96.54`