The van der Waals constant b of a gas is 0.0400 L mol^-1. Treating molecules as hard spheres with b = 4N_A((4)/(3)pi r^3) and N_A = 6.022 x 10^23 mol^-1, the molecular radius r in pm (nearest integer) is

JEE Main levelNumerical answerDifficulty 2 of 3rg017

The van der Waals constant $b$ of a gas is $0.0400 L mo l^{- 1}$ . Treating molecules as hard spheres with $b = 4 N_{A} (\frac{4}{3} π r^{3})$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , the molecular radius $r$ in pm (nearest integer) is

Numerical answer type, answer in pm, tolerance 2.

Volume of one molecule

= \frac{b}{4 N _{A}} = \frac{0.0400 \times 1 0 ^{- 3} m ^{3}}{4 \times 6.022 \times 1 0 ^{23}} = 1.661 e - 29 m^{3}

. Then

r = (\frac{3 V}{4 π})^{1/3} = 1.583 e - 10 m = 158.3 pm

, i.e.

158 pm