A) 1
B) 4
C) 3
D) 8
Correct Answer: B
Solution :
\[\frac{{{r}_{{{H}_{2}}}}}{{{r}_{{{C}_{n}}}}_{{{H}_{2n-2}}}}=\sqrt{\frac{{{M}_{{{C}_{n}}{{H}_{2n-2}}}}}{{{M}_{{{H}_{2}}}}}}\] \[=\sqrt{\frac{{{M}_{{{C}_{n}}{{H}_{2n-2}}}}}{2}}\] \[\because \] \[\sqrt{\frac{{{M}_{{{C}_{n}}{{H}_{2n}}-2}}}{2}}=3\sqrt{3}=\sqrt{27}\] \[\Rightarrow \] \[{{M}_{{{C}_{n}}{{H}_{2n-2}}}}=27\times 2=54\] Hence, \[12n+(2n-2)\times 1=54\] \[\Rightarrow \] \[14n=56\Rightarrow n=4\] Thus, hydrocarbon is \[{{C}_{4}}{{H}_{6}}.\]
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