InterviewSolution
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InterviewSolution
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Fifteen milliliters of gaseous hydrocarbo (A) was required for complete combustion 357 ml of air (21 % oxygen by volume) and gaseous products occupied 327 ml (all volumes being measured at STP). The molecular formula of the hydrocarbon (A) is:A. `C_(2)H_(6)`B. `C_(2)H_(4)`C. `C_(3)H_(6)`D. `C_(3)H_(8)` |
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Answer» Correct Answer - D Let the formula of hydrocarbon A is CxHy. `C_(x)H_(y)(g) +(x+(y)/(4))O_(2) (g) rarr xCO_(2)(g) + (y)/(2)H_(2)O(I)` 1 ml `(x + (y)/(4))ml` x ml ---- 15 ml `(x + (y)/(4))ml` 15x ml ---- Volulme of `O_(2) =(357xx21)/(100) =75 ml` , volume of `N_(2)` `=357 - 75 =282 ml` Volume of `CO_(2) =327 - 282 =45 ml` `:. 15x=45, x=3` `:. ({:(15(x+y/4)=75),(x+y/4=5):})` `3 + (y)/(4) =5` , On solving, we get y = 8. Formula of (A) `= C_(3)H_(8)` |
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