Question

The van der Waal's constant 'b' 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

Answer: (B)

$\text{b}=4\times \frac{4}{3}\times \pi {\text{r}}^3\times {\text{N}}_{\text{A}}$ ;

We know that, $1L=1000c{m}^3$

$4\times \pi \times 10^{-4}\times 1000=4\times \frac{4}{3}\times \pi \times {\text{r}}^3\times 6\times 10^{23}$

$\text{r}=5\times 10^{-9}cm$

Distance of closest approach $=2r$

$=10^{-8}cm$ or $10^{-10}m$