State Raoult's law. Why is the vapour pressure of a solvent lowered by the addition of nonvolatile solute to it? A solution containing 18 g of non-volatile solute in 200 g of water freezes at 272.07 K. Calculate the molecular mass of the solute. (Given `K_f` = 1.86 K/m).

Explanation

Raoult.s law may be stated in two different .ways:
(i) Solution containing a non-volatile solute. According to the law, the vapour pressure of the solution at a given temperature is equal to the product of the vapour pressure of the pure solvent and its mole fraction.
`p = p^@ X_A`
where p = vapour pressure of solution
`p^@ =` vapour pressure of pure solvent
`X_A =` mole fraction of pure solvent.
(ii) Solution containing both yolatile solute and solvent. According to Raoult.s law,
the vapour pressure of each component in the solution is equal to the product of the vapour pressure of the pure solvent and its mole fraction.
`P_A= P_A^@ X_A, P_(B)^(@) X_B`
When a non-volatile solute is added to a solvent (generally water), the vapour pressure of the solution gets lowered. Actually, evaporation is a surface phenomenon i.e. it takes place from the surface of the container. The non-volatile solute particles occupy certain surface area and as a result, the tendency of the solvent molecules to change to vapours decreases. In other words, the vapour pressure of the solution falls.
Numerical
Mass of the solute (WB) = 18 g
Mass of water (WA) = 200 g
`Delta T_f =273 -272 .07 =0.93 K`
`K_f =1.86 K//m`
`M_b=?`
We know that,
`M_B =(K_f xx W_Bxx 1000 )/(Delta T_f xx W_A) `
`=(1.86 xx 18 xx 1000 )/( 0.93 xx 200 )=180.`