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For dilute solutions containing 2.5g of a non-volatile non electrolyte solute in 100g of water,the elevation in boiling point at 1atm pressure is $2^{\large\circ}C$.Assuming concentration of solute is much lower than the concentration of solvent,the vapour pressure (mm of Hg) of the solution is (take $K_b=0.76Kkgmol^{-1})$

$\begin{array}{1 1}(a)\;740\\(b)\;718\\(c)\;724\\(d)\;736\end{array}$

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Answer: 724 torr
Elevation of Boiling Point $\Delta\; T_b = i\;K_b\; m$, where $i$ is the van't Hoff factor, $K_b$ is the Boiling Point Elevation Constant and $m$ is the molality.
$\Rightarrow$ Molality $m = \large\frac{\Delta T_b}{K_b}$$ = \large\frac{2}{0.76} $$ = 2.63 \; mol \; Kg^{-1}$
We know that the molality of pure water is $55.6 \; mol \; Kg^{-1}$
From Raoult's law we know that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
Adding of a solute lowers the vapour pressure of the solvent and the relative lowering of vapour pressure $= \large\frac{P_0 - P}{P_0}$ which is equal to $\large\frac{n}{n+N}$ where $n$ is the number of moles in solute and N is the number of moles of solvent.
This relationship was deduced by Raoult’s and is only applicable to dilute solution BUT if solution is VERY DILUTE then $n \lt \lt N$ such that $N+n \approx N$, we get:
In our case $ \large\frac{P_0 - P}{P_0}$ $= \large\frac{n}{N}$$\rightarrow \large\frac{760-P}{760} $$= \large\frac{2.63}{55.56}$ (Note: For water, the vapor pressure reaches the standard sea level atmospheric pressure of 760 mmHg at 100$^{\circ}$C)
$\Rightarrow P = 724\; torr$
answered Feb 13, 2014 by sreemathi.v
edited Jul 17, 2014 by balaji.thirumalai

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