Question

$100\text{g}$ $C_6{H}_{12}{O}_6(aq.)$ is dissolved in water. The solution has vapour pressure is equal to $40\text{torr}$ at a certain temperature. Vapour pressure of $H_{2}O$ (l) is $40.18\text{torr}$ at same temperature. If this solution is cooled to
$-0.93{}^{\circ }C,$ what mass of ice will be separated out? $(K_f=1.86kgmo{l}^{-1})$
(Give your answer upto two decimal point)

Answer 1

$\text{Molality of solution}=\frac{P^0-P}{P}\times \frac{1000}{M_w}$
$\text{Where}{P}^0=\text{Vapour pressure of pure solvent}=40.18\text{atm}$
$P=\text{Vapour pressure of solution}$
$M_w=\text{Molar mass of solvent}$
$\therefore \text{Molality (m)}=\frac{40.18-40}{40}\times \frac{1000}{18}$
$\Rightarrow \text{Molality}=0.25\text{m}$
$\text{Now, Molality}=\frac{n_{\text{glucose}}}{W_{\text{water (1kg)}}}$
$0.25=\frac{\frac{W}{180}}{1}$
$\Rightarrow W=0.25\times 180=45\text{g}$
$\text{Now,}$
$\text{1000 g water contains 45 g glucose}$
$\text{So, 100 g solution contains}$
$=\frac{45}{1045}\times 100=4.306\text{g glucose}$
$\Rightarrow \text{weight of water = 100 g – 4.31 g = 95.69 g}$
$\text{Now, we know,}$
$\Delta T_f=K\times m$
$\Rightarrow 0-(-0.93)=1.86\times m$
$\Rightarrow m=\frac{0.93}{1.86}=0.5m$
$\text{Now, m}=\frac{n_{\text{glucose}}}{W_{\text{water}}\left(kg\right)}$
$0.5=\frac{W_{\text{glucose}}/180}{1}$
$\Rightarrow W_g=0.5\times 180=90g$
$\text{1000 g of water contains 90 g glucose}$
$\Rightarrow \text{90 g glucose is present in 1000 g water}$
$\therefore \text{4.31 g glucose will be present in}=\frac{1000}{90}\times 4.31$
$=47.88g$
$\therefore \text{Mass of ice}=95.7-47.88g=47.82g$