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# The Freezing point of ether was lowered by $0.60^{\large\circ}$C on dissolving 2.0gm of phenol in 100g of ether.Calculate the molar mass of phenol and comment on the result.Given $K_f$(ether)=$5.12Kkg mol^{-1}$

$\begin{array}{1 1}170.7gmol^{-1}\\270.7gmol^{-1}\\370.7gmol^{-1}\\470.7gmol^{-1}\end{array}$

Let $M$ be the molar mass of phenol in ether.
We will have ,molality of solution =$\large\frac{2.0}{(100/1000)}$
Now using the expression $-\Delta T_f=K_fm$
We get,
$0.60K=(5.12Kkg mol^{-1})(\large\frac{20}{M}\frac{g}{kg})$
Which gives $M=\large\frac{5.12\times 20}{0.60}$$g mol^{-1}$
$\Rightarrow 170.7gmol^{-1}$
Since the calculated molar mass is about twice of its actual molar mass $(94gmol^{-1})$,it may be concluded that the phenol in ether is almost present in dimerized form