The molality (mol/kg) of 27 g of urea dissolved in 500 g of water is:

The correct answer is Option (2) → 0.90

Molality (m) is the number of moles of solute dissolved in 1 kg of solvent:

    \( \displaystyle m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)

Because it uses the mass of the solvent, not the volume of the solution, molality does not change with temperature.

Urea is \( \ce{CO(NH2)2} \), with molar mass 12 + 16 + 2(14 + 2) = 60 g mol⁻¹.

    \( \displaystyle \text{Moles of urea}=\frac{27}{60}=0.45\ \mathrm{mol} \)

    \( \displaystyle \text{Mass of water}=500\ \mathrm{g}=0.500\ \mathrm{kg} \)

    \( \displaystyle m=\frac{0.45}{0.500}=0.90\ \mathrm{mol\,kg^{-1}} \)

So the molality of the solution is 0.90 mol kg⁻¹. (0.45 is only the number of moles of urea; it must still be divided by the mass of water in kg.)