The mass of urea required to make 2.5 kg of | vedclass

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(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given: Molality $(m)$ = $0.25 \ mol/kg$,Mass of solution = $2.5

Vedclass · Accepted Answer

$37$

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(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given: Molality $(m)$ = $0.25 \ mol/kg$,Mass of solution = $2.5 \ kg$.Let the mass of urea be $w \ g$. Molar mass of urea $(NH_2CONH_2)$ = $60 \ g/mol$.Moles of urea $(n)$ = $\frac{w}{60}$.Mass of solvent = Mass of solution - Mass of solute = $(2500 - w) \ g = \frac{2500 - w}{1000} \ kg$.Using the formula: $m = \frac{n}{Mass \ of \ solvent \ (kg)}$.$0.25 = \frac{w/60}{(2500 - w)/1000}$.$0.25 = \frac{w}{60} 	imes \frac{1000}{2500 - w}$.$0.25 = \frac{1000w}{60(2500 - w)}$.$15(2500 - w) = 1000w$.$37500 - 15w = 1000w$.$1015w = 37500$.$w \approx 36.94 \ g \approx 37 \ g$.

The mass of urea required to make $2.5 \ kg$ of $0.25 \ molal$ aqueous solution is (in $g$)

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