Question

The molality of a urea solution in which $0.0100 \,g$ of urea, $[(NH_2)_2CO]$ is added to $0.3000\, dm^3$ of water at STP is

• A. $5.55 \times 10^{-4}\, M$
• B. $33.3\, M$
• C. $3.33 \times 10^{-2}\, M$
• D. $0.555\, M$

Answer 1

Answer: (A)

Molality $= \frac{\text{moles of the solute}}{\text{mass of the solvent in kg}}$
Moles of urea $(n_{urea}) = \frac{0.0100\,g}{6.\,g\,mol^{-1}}$
Mass of $1000 \,mL$ of solution = volume $\times$ density
$ = 300\,mL \times \frac{1\,g}{mL}$ $[ \because 1\,dm^3 = 1000\,mL]$
$ = 300\,g$
Mass of solvent $= 300 \,g - 0.0100 \,g$
$= 299.9 \,g = 0.2999\, kg$
Molality $= \frac{0.0100}{60\ times 0.2999} $
$= 5.55 \times 10^{-4}\, mol\, kg^{-1}$