Question

The molality of a urea solution in which $0.0100g$ of urea is added to $0.3000d{m}^3$ of water at $STP$ is:

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

Answer 1

Answer: (B)

$0.555\times {10}^{-4}m$

Weight of urea $=0.0100$ $g$

Molecular mass of $\left[\right(N{H}_2{)}_{2}CO]=60$ $g/mol$

$\therefore$ No. of moles $=\frac{0.0100}{60.06}=1.67\times {10}^{-4}$ $mol$

Now, Mass $=$ Volume $\times$ Density

Density of water $={10}^3$ $g/d{m}^3$ and Volume $=0.3000$ $d{m}^3$

$\therefore$ Mass of water $=0.3000\times {10}^3=300$ $g=0.3$ $kg$

$Molality=\frac{Molesofsolute}{Massofsolvent\left(kg\right)}=\frac{1.67\times {10}^{-4}}{0.3}=5.57\times {10}^{-4}$ $m$