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A definite amount of gaseous hydrocarbon having `("carbon atoms less than" 5)` was burnt with sufficient amount of `O_(2)`. The volume of all reactants was `600mL`. After the explosion the volume of the product `[CO_(2)(g)` and `H_(2)O (g)]` was found to be `700 mL` 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)` |
Answer» Correct Answer - A `{:(,C_(x)H_(y),+(x+(y)/(4))O_(2)rarr,xCO_(2)+,(y)/(2)H_(2)O),(,a,(x+(y)/(4))a,ax,(ay)/(2)),(,a, +(x+(y)/(4))a = 600,,),(,,ax+a(y)/(2)=700,,):}` `6x+3y=7+7x+7y//4` `7+x=5y//4` `x lt 5` put the value if `x=3 " " 10=5y//4 " " y=8` Ans is `C_(3)H_(8)` |
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