Question

A non-volatile solute ${}^{\prime }{X}^{\prime }$ completely dimerises in water if the temperature is below $-3.72{}^{\circ }C$ and the solute completely dissociates as $X\to Y+Z$, if the temparature is above ${100.26}^{\circ }C$. In between these temperature (including both temperature), ${}^{\prime }{X}^{\prime }$ is neither dissociated nor associated. One mole of ${}^{\prime }{X}^{\prime }$ dissolved in $1\text{kg}$ water.
Given : $K_b\text{of water}=0.52{\text{K. kg mol}}^{-1}$
$K_f\text{of water}=1.86{\text{K.kg mol}}^{-1}$
Which of the following statement (s) is/are true regarding the solution ?

• A. The freezing point of the solution is $-{1.86}^{\circ }C$
• B. The boiling point of the solution is ${101.04}^{\circ }C$
• C. When the solution is cooled to $-{7.44}^{\circ }C,75\%$ of water present initially will separate as ice
• D. When the solution is heated to $102.086^{\circ }C,50$% of water present initially will escape out as vapour

Answer 1

Answer: (A), (B), (D)

When the solution is heated to $102.086^{\circ }C,50$% of water present initially will escape out as vapour

(A) $\Delta T_f=\text{Depression in freezing point}=T_f^0-T_f$
$T_f=\text{Freezing point of the solution}$
$T_f^0=\text{freezing point of pure solvent}$
We know,
$\Delta T_f=i{K}_{f}m$
$\text{Molality (m)}=\frac{1\text{mole}}{1Kg}=1$
$i=1,\text{van't Hoff factor}$
$\Delta T_f=1\times 1.86\times 1$
$\Delta T_f={1.86}^{\circ }C$
$T_f=T_F^o-\Delta T_f=0-1.86=-{1.86}^{\circ }C$

(B)
$\Delta T_b=\text{elevation of boiling point}=T_b-T_b^{\circ }$
$T_b^{\circ }\text{and}{T}_b\text{are boiling point of pure solvent and solutions respectively.}$
$\Delta T_b=i{K}_{b}m$
$i=\text{van't Hoff factor}=2$
Since, $X\to Y+Z$
$\text{molality}\left(m\right)=1$
$\therefore \Delta T_b=2\times 0.52\times 1$
$\Rightarrow \Delta T_b={1.40}^{\circ }C$
$T_b=\Delta T_b+T_b^{\circ }=1.04+100={101.04}^{\circ }C$
$(T_b^{\circ }=\text{boiling point of water}={100}^{\circ }C)$

(C) $\text{Here},\Delta T_f={7.44}^{\circ }C$
$\therefore \Delta T_f=i\times K_f\times m$
$\left(\text{So, given temparature}\right(-{7.44}^{\circ }C)<(-{3.72}^{\circ }C)$
So here dimerization will take place and van't Hoff factor would be $\left(i\right)=1/2$
$\therefore 7.44=\frac{1}{2}\times 1.86\times \frac{1}{w}\times 1000$
Here, $w=\text{mass of solvent}$
$\therefore w=\frac{1.86\times 1000}{7.44\times 2}$
$\text{Mass of solvent (w)}=125g$
$\text{Ice separate out}=1000g-125g$
$=875\text{g}$
So, when the solution is cooled to $-{7.44}^{\circ }C,87.5\%$ of water present initially will separate as ice
(D)
$\Delta T_b=T_b-T_b^{\circ }$
$\Delta T_b=102.08-100=2.08$
$\Delta T_b=2\times 0.52\times \frac{1}{w}\times 1000$
Here, $w=\text{mass of solvent}$
$2.08=1.04\times \frac{1}{w}\times 1000$
$\Rightarrow w=500g$
So, when the solution is heated to $102.086^{\circ }C,50$% of water present initially will escape out as vapour