For a dilute solution containing 2.5 grams of non volatile non electrolyte solute in hundred grams of water, the elevation in the boiling point at 1 atmosphere pressure is to degree Celsius asuming concentration of solute is much lower than concentration of solvent, the vapour pressure of the solution is?

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Solution

According to

Elevation of Boiling Point $Δ T_{b} = i K_{b} m$

Where,

i= Vant’s Hoff factor,

Kb= Boiling point Elevation Constant

m= molarity
So,

$M o l a r i t y = \frac{Δ T_{b}}{K_{b}}$

$= \frac{2}{0.76} = 2.63 m o l K g^{- 1}$

Now,

We know that the molarity of pure water $= 55.6 m o l K g^{- 1}$

Also given that

Atmospheric pressure of $760 m m H g a t 100^{0} C$

So,

$\frac{P_{0} - P}{P_{0}} = \frac{n}{N}$

$\frac{760 - P}{760} = \frac{2.63}{55.56}$

$P = 724 t o r r$

Hence,

The vapour pressure of solution $P = 724 t o r r$
This is the required solution.

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. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is:

[

K_{b} = 0.76 K K g m o l^{- 1}

]

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in $100 g$ of water, the elevation in boiling point at $1 a t m$ pressure is $2 ℃$ . Assuming that the concentration of solute is much lower than the concentration of solvent, the vapour pressure $(in mm of Hg)$ of the solution is: