Question

For a dilute solution containing \[2.5g\] of non-volatile non-electrolyte solute in \[100g\] of water, the elevation in boiling point at \[1atm\] pressure is \[2^\circ C\]. Assuming concentration solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is:
$[{K_b} = 0.76KKgmo{l^{ - 1}}]$
A. $724$
B. $740$
C. $736$
D. $718$

Answer 1

Hint: We must remember that the boiling point of the solvent above a solution is greater than the boiling point of the pure solvent whether the solution contains a non-volatile solute or a volatile solute. Experimentally, the change in boiling point of the solvent above a solution from that of the pure solvent is directly proportional to the mole concentration of the solute.

Complete step by step answer:
As we know that, if a solute is non-volatile, then the solute does not evaporate at ambient temperature and pressure and non-electrolyte solute means the solute does not conduct electricity.
The results that in dilute ideal solutions, the extent of boiling-point elevation is directly proportional to the molal concentration (amount of substance per mass) of the answer consistent with the equation:
$\Delta {T_b} = {K_b} \times m$------\[\left( 1 \right)\]
$\Delta {T_b} = $Boiling point elevation, defined as ${T_{b(solution)}} - {T_{b(pure \;solvent)}}$
${K_b}$= ebullioscopic constant
$m = Molality = \dfrac{{{\text{Weight of solute}}}}{{{\text{Molecular weight of solute}}}} \times \dfrac{{1000}}{{{\text{Weight of solvent}}}}$
Given, weight of non-volatile non-electrolyte solute $ = 2.5g$, weight of the solvent $ = 100g$ and the elevation boiling point $\Delta T = 2^\circ C$.
${K_b} = 0.76KKgmo{l^{ - 1}}$
Apply these value in \[\;\left( 1 \right)\] equation,
$2 = 0.76 \times \dfrac{{2.5}}{{{\text{Molecular weight of solute}}}} \times \dfrac{{1000}}{{1000}}$
Molecular weight of solute = $9.5g$
The relative lowering of vapour pressure from the Roult’s law,
$\dfrac{{{p^0} - {p_s}}}{{{p^0}}} = \dfrac{{{n_2}}}{{{n_1} + {n_2}}} = \dfrac{{{n_2}}}{{{n_1}}}$ $[\because {n_1} > > {n_2}]$
Where, $p = $ vapour pressure of the solution, \[{p^0} - {p_s}\] = lowering of vapor pressure, $\dfrac{{{p^0} - {p_s}}}{{{p^0}}}$ = relative lowering of vapour pressure and \[\dfrac{{{n_2}}}{{{n_1} + {n_2}}}\] = mole fraction of the solute in the solution.
We know that, vapour pressure of the solvent at \[1atm\] = $760mmHg$
Rewrite the above equation with respect to \[\;\left( 1 \right)\] equation,
$\dfrac{{{p^0} - p}}{{{p^0}}} = \dfrac{{\left[ {\dfrac{{{\text{Weight of solute}}}}{{{\text{Molecular weight of solute}}}}} \right]}}{{\left[ {\dfrac{{{\text{Weight of solvent}}}}{{{\text{Molecular weight of solvent}}}}} \right]}}$
Now we can substitute the given values we get,
$\dfrac{{760 - p}}{{760}} = \dfrac{{\left[ {\dfrac{{2.5}}{{9.5}}} \right]}}{{\left[ {\dfrac{{100}}{{18}}} \right]}}$[Molecular weight of solvent, water $ = 18g$]
\[760 - p = \dfrac{{\left[ {\dfrac{{2.5}}{{9.5}}} \right]}}{{\left[ {\dfrac{{100}}{{18}}} \right]}} \times 760\]
$p = 760 - 36$
On simplification we get,
$p = 724$
The vapour pressure of the solution is $724$$mmHg$.

Hence option A is correct.

Note: $\Delta {T_b} = {K_b} \times m$
We know that this equation is the assumption of the non-volatility of the solute together with Clausius-Clapeyron relation & Raoul’s law. Surrounding pressure is depended by the boiling point of the liquid.
${K_b}$ unit is \[Kkgmo{l^{ - 1}}\].
The elevation in boiling point of the solution which may be theoretically produced when one mole of the solute is dissolved in $1$kg of the solvent, it is defined by a molal boiling point elevation constant, when the molality of the solvent is $1$. Then ${K_b} = \Delta {T_b}$.