question_answer

Beryllium gives a compound \[\text{X}\] with the following percentage composition: Be Molecular weight of \[-61%,\,N-37.8%,\,Cl-48%,\,H-8.1%\] \[\text{X}\] is \[\text{148 g mo}{{\text{l}}^{-}}^{\text{1}}\] and that of Be is \[\text{9 g mo}{{\text{l}}^{\text{-1}}}.\] The molecular formula of the compound is

A) \[\text{Be}{{\text{N}}_{\text{4}}}\text{C}{{\text{l}}_{\text{2}}}{{\text{H}}_{\text{12}}}\]

B) \[~\text{Be}{{\text{N}}_{\text{2}}}\text{Cl}{{\text{H}}_{6}}\]

C) \[\text{Be}{{\text{N}}_{\text{4}}}\text{C}{{\text{l}}_{\text{2}}}{{\text{H}}_{6}}\]

D) \[\text{Be}{{\text{N}}_{\text{4}}}\text{Cl}{{\text{H}}_{\text{8}}}\]