Question

The Van der Waals constant ${}^{\prime }{b}^{\prime }$ for $N_2$ and $H_2$ has the values $0.04litmo{l}^{-1}$ and $0.025litmo{l}^{-1}$. The density of solid $N_2$ is $1g.c{m}^{-3}$. Assuming the molecules in the solids to be close packed with the same percentage void, the density of solid $H_2$ would be: (in $g.c{m}^{-3}$)

• A. 0.114
• B. 0.682
• C. 1.466
• D. 0.071

Answer 1

Answer: (A)

0.114

The expression for the magnitude of the van der Waals constant, b, the volume V, molecular weight M and density d for nitrogen and hydrogen is as given below:

$\frac{b_{N_2}}{b_{H_2}}=\frac{V_{N_2}}{V_{H_2}}=\frac{(M/d{)}_{N_2}}{(M/d{)}_{H_2}}$

Substitute values in the above expression:

$d_{H_2}=\frac{0.040}{0.025}\times \frac{2}{28}=0.114$

Thus the density of hydrogen is $0.114$ $gc{m}^{-3}$.