If mole fraction of urea in 11.11 molal aqueous solution?

0.1

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0.17

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0.2

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0.37

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Solution

The correct option is D $0.17$
Molality= $\frac{n o . o f m o l e s o f u r e a}{w t . o f s o l v e n t (H_{2} O) i n k g s}$

11.11= $\frac{n_{u r e a}}{w t . o f H_{2} O k g s}$

n $_{u r e a}$ =(11.11) x wt. of $H_{2}$ O in kgs.

Mole fraction of urea= $\frac{n o . o f m o l e s o f u r e a}{n o . o f m o l e s o f u r e a + n o . o f m o l e s o f w a t e r}$

$n_{u r e a}$ (11.11) x wt. of $H_{2}$ O in kgs

Let the wt. of $H_{2} O$ in $= x k g s$ = $1000 x g r a m s$

Moles of $H_{2} O$ = $\frac{1000}{18}$ $x = 55.55 x$

n $_{u r e a}$ =(11.11) x $x$ = 11.11 $x$

Molefraction of urea = $\frac{11.11 x}{11.11 x + 55.55 x}$ = $\frac{11.11 x}{66.66 x}$ = $\frac{11.11}{66.66} = \frac{1}{6}$
$= 0.17$
Molefraction of urea $= 0.17$

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