The pressure imposed by a gas on a liquid (or solid) in a closed system at a certain temperature is known as vapour pressure. When liquid molecules try to exit the surface of the liquid through evaporation, they are subjected to upward pressure. Simply put, it is the pressure exerted by the vapour existing above the liquid surface. Raoult’s law describes how adding a solute lowers the vapour pressure. Now let’s get into the meat of the matter.
Let us suppose that the non-volatile solute’s necessary mass is ‘x’ g.
The solute’s molar mass is 40 g/mol.
Therefore, the number of moles of solute will be x/40 = 0.025x moles.
Now, mass of octane = 114 g and molar mass of octane = 114 g/mol.
Therefore, the number of moles of octane = 1 mole.
Now, according to Raoult’s law, the relative lowering of vapour pressure is proportional to the mole fraction of the solute.
So, mole fraction of solute is 0.025x/1+0.025x .
Now, the vapour pressure is to be lowered by 20%.
So, we can write that 0.025x/1+0.025x=20/100 Now, solving for x,
we will get- 100 × 0.025x = 20(1 + 0.025x)
0r,2.5x = 20 + 0.5x
or, 2x=20 or, x=10
We can see from the above calculation that the mass of a non-volatile solute that should be dissolved in 114g of octane to lower its vapour pressure by 20% is 10g.