Calculate the mass of urea $N H_{2} C O N H_{2}$ required in making 2.5 kg of 0.25 molal aqueous solution.

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Solution

0.25 molal aqueous solution of urea means that 0.25 mole of urea are present in 1 kg of water
Moles of urea = $0.25 m o l e$
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea ( $N H_{2} C O N H_{2}$ ) = 60 g mol
1. $0.25 m o l e o f u r e a = 0.25 m o l \times 60 g m o l 1 = 15 g$
$T o t a l m a s s o f s o l u t i o n = 1000 + 15 g = 1015 g = 1.015 k g$

Thus; 1.015 kg of solution contain urea = $15 g \times 2.5 k g$ of solution will require urea
$\frac{15 g}{1.015 k g} \times 2.5 k g$ = 37 g

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