If molality of ethyl alcohol in aqueous solution is 4.45 then the mole fraction of ethyl alcohol in the solution is

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Solution

Given,

Molality of the ethyl alcohol is $4.45$

The mole of the solute is:

$1 k g$ of $H_{2} O = \frac{1}{18} \times \frac{m o l e o f H_{2} O}{g H_{2} O}$

$= 55.556 m o l$

So, mole fraction is: mole of solute is divided by the total mole of the solution

$= \frac{4.45}{55.56 + 4.45} = 0.074$

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Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0. 25 hence the percentage of ethyl alcohol by weight is:

5.2

C H_{3} O H

(C H_{3} C H_{2} O H, m o l a r m a s s = 46.0 g m o l^{- 1})

C H_{3} C H_{2} O H

^{- 1}

= 0.789 m L^{- 1}

= 1.00 g m L^{- 1}

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