According to Raoult, the relative lowering of vapour pressure of a solvent over a solution containing a non-volatile solute is equal to:
Athe mole fraction of the solvent
Bthe mole fraction of the solute
Cthe molality of the solution
Dthe molarity of the solution
Answer & Solution
Correct answer: B. the mole fraction of the solute
1. From $p_1 = x_1 p_1^0$, the lowering is $\Delta p_1 = p_1^0(1 - x_1) = x_2 p_1^0$.
2. Dividing by $p_1^0$: $(p_1^0 - p_1)/p_1^0 = x_2$.
3. The left side is the relative lowering of vapour pressure and equals $x_2$, the mole fraction of the solute.
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.15_
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