Question

The van der Waal's constant 'b' of a gas is $4 \pi \times 1 0^{- 4} L/mol$ . How near can the centres of the two molecules approach each other ? [Use : $N_{A}=6\times 10^{23}$ ]

• A. $1 0^{- 7} m$
• B. $1 0^{- 10} m$
• C. $5 \times 1 0^{- 1 1} m$
• D. $5 \times 1 0^{- 9} m$

Answer 1

Answer: (B)

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

We know that, $1L=1000 \, cm^{3}$

$4 \times \pi \times 1 0^{- 4} \times 1 0 0 0 = 4 \times \frac{4}{3} \times \pi \times \text{r}^{3} \times 6 \times 1 0^{2 3}$

$\text{r} = 5 \times 1 0^{- 9} cm$

Distance of closest approach $=2r$

$= 1 0^{- 8} cm$ or $1 0^{- 10} m$