Question

The van der Waals' constant ${}^{\prime }{b}^{\prime }$ of a gas is $4\pi \times {10}^{-4}L/mol$. How near can the centres of the two molecules approach each other? [Use : $N_A=6\times {10}^{23}$]

• A. ${10}^{-7}m$
• B. ${10}^{-10}m$
• C. $5\times {10}^{-11}m$
• D. $5\times {10}^{-9}m$

Answer 1

The correct option is B ${10}^{-10}m$

$b=4\times \frac{4}{3}\times \pi r^3\times N_A$
$4\times \pi \times {10}^{-4}\times 1000=4\times \frac{4}{3}\times \pi \times r^3\times 6\times {10}^{23}$
$r=5\times {10}^{-9}cm$
Distance of closest approach$=2r={10}^{-8}cm$ or ${10}^{-10}m$.