Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Solution:

Step-1 What is given in problem? 

Molar mass of solute = 40g/mol

Mass of solvent = 114 g

Vapour Pressure of solution is  = 80% of pure liquid

Step-2 What to find?

We have to find mass of non volatile solute

Let mass of non volatile solute = x gram

We know the partial pressure of solution is 80% of pure liquid

If vapour pressure of pure is = P^o

80 % of pure pressure P^o = 80×P^o/100 = 0.8P^o

The formula of decrease in partial pressure = pressure of pure liquid × molar fraction of solute 

P =P^o × X_solute                                                                           …….(1)

Here we have to find mole fraction of solute

      ……(2)

So we need to calculate no of moles 

 

We have mass of solute = x gram

And mass of solvent = 114g

Molar mass of solute= 40 g/mol

Molar mass of solvent (octaneC8H18) = 114g/mol

Number of moles of solute = x/40 = 0.025x

Number of moles of solvent = 114/114= 1 moles

Plug the value in equation (2)

Mole fraction of solvent = 1/(1+0.025x)

Now plug the value in equation first we get

0.8P^o=P^o×1/(1+0.025x)

Cross multiply we get

(1+0.025x))0.8P^o= P^o

Divide by 0.8 P^o we get

1+0.025x = 1.25

Subtract 1 both side we get

0.025x = 0.25

Now divide by 0.025 we get

X = 10g