A gaseous compound X contained 44.4 %C, 51.9 % N and 3.7% H . Under like conditions 30 cm^(3) of X diffused through a pinhole in 25 sec and the same volume of H_(2) diffused in 4.81 sec . The molecular formula ofX is

A gaseous compound X contained 44.4 %C, 51.9 % N and 3.7% H . Under li

1.	A gaseous compound X contained 44.4 %C, 51.9 % N and 3.7% H . Under like conditions 30 cm^(3) of X diffused through a pinhole in 25 sec and the same volume of H_(2) diffused in 4.81 sec . The molecular formula ofX is
Answer» `C_(2) H_(2) N` `C_(2) H_4 N_2` `C_2 H_2 N_2` `C_4 H_2 N_2` Solution :RATIO of number of atoms in the molecule = Ratio of gram atoms of elements in he COMPOUND `C : N : H= (44.4)/(12) : (51.9)/(14) : (3.7)/(1)` `= 3.7 : 3.7 : 3.7 = 1 : 1 :1 ` `therefore ` empirical FORMULA of `X = CNH` Ratio of RATE of diffusion of X and that of `H_(2)` is given by `(r_(x))/(r_(H_(2))) = (30 xx 4.81 )/(25 xx 30) = sqrt((2)/(M))` On solving , M = 54 `therefore` molecular formula of x is `C_(2) H_(2) N_(2)`