A) \[{{C}_{6}}{{H}_{7}}\]
B) \[{{C}_{6}}{{H}_{8}}\]
C) \[{{C}_{5}}{{H}_{6}}\]
D) None of these
Correct Answer: A
Solution :
\[C=10.5\,\,gm=\frac{10.5}{12}mol=0.87\,\,mol\] \[H=1\,\,gm=\frac{1}{1}=1\,\,mol\] \[\therefore \,{{({{C}_{0.87}}{{H}_{1}})}_{7}}={{C}_{6.09}}{{H}_{7}}\approx {{C}_{6}}{{H}_{7}}\] \[PV=nRT\]; \[PV=\frac{w}{m}RT\] \[1\times 1=\frac{2.4}{m}\times 0.082\times 400\] \[m=2.4\times 0.082\times 400\]\[=78.42\approx 79\].
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