Question

The vapour pressure of a pure solvent at 500C is 400mm. If 0.2 mole of a non volatile solute is added to 1 mole of the solvent the vapour pressure of that solution at 500C is nearly​

Answer 1

Answer:

Applying Raoult's law in the following form:

PsP0−Ps=WmwM=W/Mw/m

= No. of moles of solute per mole of benzene

or Nn=167(268−167)=0.6047=0.605

Alternative method: We know that, Ps= Mole fraction of solvent ×P0

or 167 = Mole fraction of solvent ×268

So, Mole fraction of solvent =268167=0.623

Mole fraction of solute =1−0.623=0.377

Nn=Mole fraction of solventMole fraction of solute=0.6230.377=0.605

Answer 2

The vapor pressure of that solution at 500C is 480.2 mm.

Given:

The vapor pressure of a pure solvent at 500C is 400mm. 0.2 mole of a nonvolatile solute is added to 1 mole of the solvent.

To Find:

The vapor pressure of that solution is at 500C.

Solution:

To find the vapor pressure of that solution at 500C we will follow the following steps:

Vapor pressure is the pressure of vapors that occurs when the liquid evaporates at its boiling point and covers the surface of the liquid.

Also,

Relative lowering of vapor pressure is directly proportional to the mole fraction of solute.

The formula of relative lowering of vapor pressure =

$\frac{p0 - ps}{p0} = x2$

po = vapor pressure of the solvent

ps = vapor pressure of solution = 400 mm

x2 is the mole fraction of solute.

$x2 = \frac{0.2}{1 + 0.2} = \frac{0.2}{1.2} = 0.167$

In putting values we get,

$1 - \frac{400}{p0} = 0.167$

$\frac{400}{p0 } = 1 - 0.167 = 0.833$

$p0 = \frac{400}{0.833} = 480.2 mm$

Henceforth, the vapor pressure of that solution at 500C is 480.2 mm.

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