A) \[{{C}_{6}}{{H}_{7}}\]
B) \[{{C}_{6}}{{H}_{6}}\]
C) \[{{C}_{5}}{{H}_{6}}\]
D) \[{{C}_{6}}{{H}_{6}}\]
Correct Answer: A
Solution :
| [a]\[C=10.5g=\frac{10.5}{12}mol=0.87mol\] |
| \[H=1g=\frac{1}{1}mol=1mol\] |
| \[\therefore \] \[{{({{C}_{Q.87}})}_{7}}={{C}_{6.09}}{{H}_{7}}={{C}_{6}}{{H}_{7}}\] |
| \[pV=nRT=\frac{w}{m}RT\] |
| \[1\times 1=\frac{24}{m}\times 0.082\times 400\] |
| \[m=79\] |
| Hence, the hydrocarbon is \[{{C}_{6}}{{H}_{7}}\]. |
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