Liquid Solution

Magnesium nitrate separates into ions in solution as shown below:

$ \mathrm{Mg}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{NO}_3^{-} $

This means that one unit of magnesium nitrate produces a total of 3 particles after complete dissociation (1 magnesium ion and 2 nitrate ions).

If the salt is only 40% dissociated, it means $\alpha = 0.4$.

The Van’t Hoff factor, $ i $, which shows how many times the number of effective particles increases due to dissociation, is given by:

$ i = 1 + \alpha (n - 1) $

Substituting the values, we get:

$ i = 1 + 0.4(3 - 1) = 1 + 0.8 = 1.8 $

Elevation of boiling point depends on the vant' Hoff factor (i).

Urea is non-electrolyte, $i=1$

$\mathrm{CaCl}_2$ dissociates into 3 ions, $i=3$

So, effective particles $=0.05 \times 3=0.15 \mathrm{~mol}$ for $\mathrm{CaCl}_2$ is more than 0.1 mol for urea is equal amount of solvent.

Hence, $\mathrm{CaCl}_2$ solution will show greater elevation.

$ \alpha=0.5 $ (50\% dissociated)

The reaction involves is

$ \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightleftharpoons 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4} $

$\begin{array}{1} \text { Initial moles } & 1 & 0 & 0 \\ \text { Moles after dissociation } & 1-\alpha & 4 \alpha & \alpha \end{array}$

Total moles at equilibrium $ =1-\alpha+4 \alpha+\alpha $

Total $ =1+4 \alpha $

$ \therefore $ van't Hoff factor $ (i)=\frac{1+4 \alpha}{1}=1+4 \times 0.5=3 $

Osmotic pressure, $ \pi=i C R T $

$ \begin{aligned} C & =0.1 \mathrm{M}=10^{2} \mathrm{~mol} \mathrm{~m}^{-3} \\ \pi & =3 \times 10^{2} \times 8.314 \times 300 \\ & =7.483 \times 10^{5} \mathrm{Nm}^{-2} \end{aligned} $

or $ \quad \pi=7.48 \mathrm{~atm} $

$ 58.5 \mathrm{~g} \mathrm{NaCl}=\frac{58.5}{58.5} \mathrm{~mol} $.

$ \begin{aligned} & =1 \mathrm{~mol} \\ 180 \mathrm{~g} \text { of glucose } & =\frac{180}{180}=1 \mathrm{~mol} \end{aligned} $

Thus, we have 1 M NaCl and 1 M glucose solution.

But NaCl dissociates in the solution and glucose does not dissociates.

Thus particle concentration is more in NaCl .

Hence, NaCl solution will have higher elevation in boiling point.

According to Raoult's law, relative lowering of vapour pressure is equal to mole fraction of solute,

$\therefore$ ${{p - {p_s}} \over p} = {n \over {n + N}} = {w \over m}{{w/m} \over {{w \over m} + {W \over M}}} \Rightarrow {w \over m} < < < {W \over M}$

$\therefore$ ${{p^\circ - {p_s}} \over {p^\circ }} = {{w/m} \over {{W \over M}}} = {w \over m} \times {M \over W}$

$0.0125 = {{w \times 18} \over {m \times W}} \Rightarrow {w \over {mW}} = {{0.0125} \over {18}}$

$ = {{Weight\,of\,solute\,(w) \times 1000} \over {Molar\,mass\,of\,solute\,(m) \times Weight\,of\,{H_2}O(W)}}$

Now, molality $m = {{0.0125} \over {18}} \times 1000 = 0.69 \approx 0.7$

$\therefore$ ${K_b} = {{RT_b^2{M_1}} \over {\Delta {H_{vap}}}} = {{8.314 \times {{(373)}^2} \times 0.0180} \over {40585}}$ = 0.513 K kg/mol

Now, molecular weight of $A{l_2}{(S{O_4})_3} = 342$

$\mathop {A{l_2}{{(S{O_4})}_3}}\limits_{1 - \alpha } $ $\rightleftharpoons$ $\mathop {2A{l^{3 + }}}\limits_{2\alpha } $ + $\mathop {3SO_2^{2 - }}\limits_{2\alpha } $

i = 1 $-$ $\alpha$

$\Delta$T_b = i $\times$ K_b . m

$ = 5 \times 0.513 \times {{5.6} \over {342 \times 1}} = 0.042$

Boiling point of solution = 100 + 0.042 = 100.042$^\circ$C