12.2 g of benzoic acid (molar mass 122) is dissolved in 156 g of benzene (molar mass 78), in which it is 80% associated into dimers (alpha = 0.80). The acid is non-volatile. Using i = 1 - alpha/2 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 3so051

$12.2 g$ of benzoic acid (molar mass $122$ ) is dissolved in $156 g$ of benzene (molar mass $78$ ), in which it is $80%$ associated into dimers ( $α = 0.80$ ). The acid is non-volatile. Using $i = 1 - α /2$ and $(p^{0} - p) / p^{0} = \frac{in}{n _{1} + in}$ , the relative lowering of vapour pressure is

n = 12.2/122 = 0.100 mol

;

n_{1} = 156/78 = 2.00 mol

. For dimerisation

i = 1 - α /2 = 0.60

, so particles

= 0.060 mol

. Relative lowering

= 0.060/ (2.00 + 0.060) = 0.0291

. Ignoring association gives 0.0476.