JEE Main 2018 — Solutions Question with Solution

Relative lowering of vapour pressure,

   $$\frac{\Delta P}{P}$$   =   $$\frac{n_2}{n_1}$$

n₂  =  Number of moles of solute

n₁   =   Number of moles of solvent.

Given that,

Here is 5 molal solution, means 5 moles of solute are dissolved in 1 kg or 1000 g of solvent.

$$\therefore$$ Number of moles of solute = 5

Number of moles of solvent X  =  $$\frac{1000}{M_X}$$

Number of moles of solvent Y = $$\frac{1000}{My}$$

$$\therefore$$ $${\left(\frac{\Delta P}{P}\right)}_x$$   =   $$\frac{5}{\frac{1000}{M_x}}$$ = $$\frac{5M_x}{1000}$$

$${\left(\frac{\Delta P}{P}\right)}_y$$   =   $$\frac{5}{\frac{1000}{M_y}}$$   =   $$\frac{5M_y}{1000}$$

Accoding to the question.

$$\frac{5M_x}{1000}$$   =   m $$\times$$ $$\frac{5M_y}{1000}$$

$$\Rightarrow$$ M_x  =   m $$\times$$ M_y

$$\Rightarrow$$ $$\frac{3}{4}$$ M_y = m $$\times$$ M_y   [as given,   M_x = $$\frac{3}{4}$$ M_y]

$$\Rightarrow$$ m   =   $$\frac{3}{4}$$