Question

Estimate the lowering of vapour pressure due to the solute (glucose) in a 1.0 M aqueous solution at ${100}^{o}C$.

• A. $10$ torr
• B. $18$ torr
• C. $13.45$ torr
• D. $24$ torr

Answer 1

The correct option is C $13.45$ torr

First, we have to calculate the mole fraction of gas.

Let us assume that mole fraction of gas is $x$ before as we know, the relation between mole fraction and

Molality $\approx$ Molarity $=\frac{x\times 1000}{(1-x)M}$

where M is the molecular weight of a solvent for aqueous solution, the solvent is water.

The molecular weight of water, $M=18g/mol$

Now,

$1=\frac{x\times 1000}{(1-x)\times 18}$

$\Rightarrow 18(1-x)=1000x$

$\Rightarrow 18=(1000+18)x$

$\therefore x=\frac{18}{1018}$

Now use formula

Relative lowering of $VP$ = mole fraction gas

$\frac{\Delta p}{p_0}=x$

here $p_0$ is initial pressure, at STP

$p_0=760mmHg$

So,

$\Delta p$$=\frac{760\times 18}{1018}=13.44$ torr approx

Hence $13.45$torr