Question

The active mass of `64 g` of `HI` in a `2-L` flask would be
A. (A) `22:3:7`
B. (B) `0.5:3:7`
C. (C) `1:3:1`
D. (D) `1:3:0.5`

Answer 1

Correct Answer - D
Molar conc.`=(22)/(44)=(1)/(2)`
Moles of `H_(2)=(3)/(2) rArr "Moles of" N_(2)=(7)/(28)=(1)/(4)`
Ratio of active massess`=1//2:3//2:1//4 "or"1:3:0.5`.