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Q.
The molality of a urea solution in which 0.0200 g of urea $(NH_{2}CONH_{2})$ is added to $0.400\, dm^{3}$ of water at STP is
Solutions
Solution:
$1\,dm^{3}=1000\,cm^{3}=1L=1kg$
$0.400\,dm^{3}$ of $H_{2}O$ at stp
$=0.400\,Kg$ of $H_{2}O$
(density = 1.0 g cm )
Moles of urea $=\frac{0.0200}{60}$
$\therefore $ Molality $=\frac{\text{Moles of solute (urea) }}{\text{ kg of solvent}}$
$=\frac{0.020/60}{0.400}$
$=8.33 \times 10^{-4}\,mol\,kg^{-1}$