0.10 mol of CaCl2 is dissolved in 100 g of water (molar mass 18). The salt is 80% dissociated (alpha = 0.80) into Ca^2+ and Cl-. Treating each ion as an independent particle, so that i = 1 + 2alpha and (p^0-p)/p^0 = (i n)/(n_1 + i n), the relative lowering of vapour pressure is

JEE Advanced levelSingle correct optionDifficulty 3 of 3so050

$0.10 mol$ of $CaCl X_{2}$ is dissolved in $100 g$ of water (molar mass $18$ ). The salt is $80%$ dissociated ( $α = 0.80$ ) into $Ca X^{2 +}$ and $Cl X^{-}$ . Treating each ion as an independent particle, so that $i = 1 + 2 α$ and $(p^{0} - p) / p^{0} = \frac{in}{n _{1} + in}$ , the relative lowering of vapour pressure is

i = 1 + 2 (0.80) = 2.60

, so particles

= 2.60 \times 0.10 = 0.26 mol

. Water

= 100/18 = 5.556 mol

. Relative lowering

= 0.26/ (5.556 + 0.26) = 0.0447

. Using

i = 3

(complete dissociation) gives 0.0512.