The van der Waal's constant 'b' for `N_2` and `H_2` has the values 0.04 lit `"mol"^(-1)` and 0.025 lit `"mol"^(-1)`.The density of solid `N_2` is 1 g `cm^(-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 `cm^(-3)`)

Options

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

Answer

A

Explanation

`b_(N_2)/b_(H_2)=V_(N_2)/V_(H_2)=((M//d)_(N_2))/((M//d)_(H_2))implies d_(H_2)=0.040/0.025xx2/28=0.114`