Question

The van der Waals constant ${}^{\prime }{b}^{\prime }$ for water vapour is 0.03 litre $mo{l}^{-1}$. The radius of the water vapour molecule is:

• A. 1.448 $cm$
• B. 1.448 $\stackrel{o}{A}$
• C. 1.448 $pm$
• D. 1.448 $nm$

Answer 1

The correct option is B 1.448 $\stackrel{o}{A}$

$b=0.03\frac{litre}{mol}=0.03\times \frac{{10}^{-3}{m}^3}{6.023\times {10}^{23}}=4.981\times {10}^{-29}{m}^3$

$Volume=\left(\frac{4}{3}\pi r^3\right)\times 4\times N$ water vapor molecule $⟹N=1$

$4.981\times {10}^{-29}{m}^3=\left(\frac{4}{3}\pi r^3\right)\times 4\times 1r^3=\frac{1.189}{4}\times {10}^{-29}r=1.448\times {10}^{70}mr=1.448°A$

Radius of water vapour molecule is $1.448°A.$