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The boiling point of CHCl(3) was raised ...

The boiling point of `CHCl_(3)` was raised by `0.323^(@)C` when `0.37 g` of naphthalene was dissolved in `35 g CHCl_(3)`. Calculate the molecular weight of naphthalene. `(k'_(b) "for" CHCl_(3)= 3.9Kmol^(-1) kg)`.

Text Solution

Verified by Experts

Given that, `w= 0.37 g, W= 35 g`.
`k_(b)= 3.9 mol^(-1) kg, DeltaT= 0.323^(@)C`
`:. DeltaT=(1000k_(b)w)/(mW)`
`:. 0. 323=(100xx3.9xx0.37)/(mxx35)`
`m= 127.6`
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