Question

The molality of a urea solution in which 0.0200 g of urea $\left(N{H}_{2}CON{H}_2\right)$ is added to $0.400d{m}^3$ of water at STP is

• A. $0.555molk{g}^{-1}$
• B. $0.555\times {10}^{-4}molk{g}^{-1}$
• C. $8.33\times {10}^{-4}molk{g}^{-1}$
• D. $33.3molk{g}^{-1}$

Answer 1

The correct option is C $8.33\times {10}^{-4}molk{g}^{-1}$

$1d{m}^3=1000c{m}^3=1L=1kg$
$0.400d{m}^{3}of{H}_{2}OatSTP=0.400kgof{H}_{2}O$ (density $=1.0gc{m}^{-3}$)
Moles of urea $=\frac{0.0200}{60}mol$
$\therefore \text{Molality}=\frac{\text{Moles of solute (urea)}}{\text{kg of solvent}}$
$\text{Molality}=\frac{0.020/60}{0.400kg}mol$
$\text{Molality}=8.33\times {10}^{-4}molk{g}^{-1}$
Hence option (c) is correct.