JEE Advance - Chemistry (1993 - No. 17)
What weight of the non-volatile solute, urea(NH2 - CO - NH2) needs to be dissolved in 100g of water, in order to decrease the vapour pressure of water by 25%? What will be the molality of the solution?
100 g, 16.67 m
75 g, 12.5 m
111.1 g, 18.52 m
50 g, 8.33 m
150 g, 25 m
Explanation
Given, mass of water (W1) = 100 g.
Let, the vapour pressure of water = p0
As given, vapour pressure of urea solution (p) $$ = ({p^0} - 0.25{p^0}) = 0.75{p^0}$$
According to Raoults' law, $${{{p^0} - p} \over {{p^0}}} = {{{W_2}/{M_2}} \over {{W_1}/{M_1} + {W_2}/{M_2}}}$$
or, $${{{p^0} - 0.75{p^0}} \over {{p^0}}} = {{{W_2}/60} \over {100/18 + {W_2}/60}}$$ [molar mass of urea = 60 g mol$$-$$1 & that of water = 18 g mol$$-$$1]
or, $${1 \over 4} = {{{W_2}/60} \over {{{1000 + 3{W_2}} \over {180}}}}$$ or, $${W_2} = 111.1$$
$$\therefore$$ 111.1 g of urea needs to be dissolved in the solution.
No. of moles of dissolved urea $$ = {{111.1} \over {60}} = 1.85$$ mol
Molality (m) of the solution $$ = {{1.85} \over {100}} \times 1000 = 18.5$$ mol kg$$-$$1
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