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A definite amout of gaseous hydrocarbon ...

A definite amout of gaseous hydrocarbon was burnt with just sufficient of `O_(2)`. The volume of all reactants was `600ml`, after the explosion the volume of the products `[CO_(2)(g)` and `H_(2)O(g)` was found to be `700ml` under the similar conditions the molecular formula of the compound is .

A

`C_(3)H_(8)`

B

`C_(3)H_(6)`

C

`C_(3)H_(4)`

D

`C_(4)H_(10)`

Text Solution

Verified by Experts

The correct Answer is:
A
Mass `%` of `O=((60/(98)xx4xx16+(40)/(80)xx3xx16)xx100)/(100) = 63.18%`
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