Question

Urea solution has a molarity of $4.0{\text{mol L}}^{-1}$ and molality $4.0{\text{mol kg}}^{-1}{H}_{2}O$. Thus, the density $\left({\text{g mL}}^{-1}\right)$ of the solution is (Report the answer upto two decimals)

Answer 1

$1000\text{mL}$ solution = 4.0 moles urea = $240\text{g}$ urea
Since, $\frac{m}{V}=$ density $\left(d\right)$
$\therefore$ Mass $=V\times d$
$1000\times d\text{g}$ solution $=240\text{g}$ urea
Solvent $\left(H_{2}O\right)=(1000d-240)\text{g}=\left(\frac{1000d-240}{1000}\right)\text{kg}$
Molality $=\frac{4\text{mol}\times 1000}{(1000d-240)}{\text{mol kg}}^{-1}$
Given, molality $=4{\text{mol kg}}^{-1}$
$\therefore \frac{4000}{(1000d-240)}=4$
$=\frac{4000}{4}=(1000d-240)$
$1000=1000d-240$
$1000d=1240$
$\therefore d=1.24{\text{g mL}}^{-1}$
Note: We can calculate density by the following relation,
Molality $=\frac{1000\times \text{molarity(M)}}{[1000d-M{m}_1(\text{molar mass}\left)\right]}$