A) \[3/2\]
B) \[1/2\]
C) 2
D) 1
E) \[5/2\]
Correct Answer: D
Solution :
\[N{{H}_{2}}COON{{H}_{4}}\] ⇌ \[2N{{H}_{3}}+C{{O}_{2}}\] \[\alpha =\frac{D-d}{(n-1)\,d}\] where \[D\] is the density (initial) \[D=\frac{mol.\,wt}{2}=\frac{78}{2}=39\] \[n=\]no. of product = 3 \[d=\] final density \[\alpha =\frac{39-13}{(3-1)\,13}=1\], so \[\alpha =1\]You need to login to perform this action.
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