Calculate the mass of urea $(NH_2CONH_2)$ required in making $2.5 \, kg$ of $0.25 \, \text{molal}$ aqueous solution.
(D) Molar mass of urea $(NH_2CONH_2) = 2(14 + 2 \times 1) + 12 + 16 + 14 + 2 = 60 \, g \, mol^{-1}$.
$0.25 \, \text{molal}$ aqueous solution means $0.25 \, mol$ of urea is dissolved in $1000 \, g$ of water.
Mass of urea $= 0.25 \, mol \times 60 \, g \, mol^{-1} = 15 \, g$.
Total mass of solution $= \text{Mass of solute} + \text{Mass of solvent} = 15 \, g + 1000 \, g = 1015 \, g$.
In $1015 \, g$ of solution,urea present is $15 \, g$.
Therefore,in $2500 \, g$ $(2.5 \, kg)$ of solution,mass of urea $= \frac{15 \times 2500}{1015} \, g \approx 36.95 \, g \approx 37 \, g$.
$0.25 \, \text{molal}$ aqueous solution means $0.25 \, mol$ of urea is dissolved in $1000 \, g$ of water.
Mass of urea $= 0.25 \, mol \times 60 \, g \, mol^{-1} = 15 \, g$.
Total mass of solution $= \text{Mass of solute} + \text{Mass of solvent} = 15 \, g + 1000 \, g = 1015 \, g$.
In $1015 \, g$ of solution,urea present is $15 \, g$.
Therefore,in $2500 \, g$ $(2.5 \, kg)$ of solution,mass of urea $= \frac{15 \times 2500}{1015} \, g \approx 36.95 \, g \approx 37 \, g$.
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