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10 ml of gaseous hydrocarbon on combution gives 20ml of `CO_(2)` and 30ml of `H_(2)O(g)`. The hydrocarbon is :-A. `C_(4)H_(5)`B. `C_(2)H_(6)`C. `C_(4)H_(8)`D. `C_(4)H_(10)` |
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Answer» Correct Answer - B for unknown hydrocarbon ` C_(X)H_(Y)+(X+(Y)/(4))O_(2)toCO_(2)+(Y)/(2)H_(2)O` `{:(1mL,,,xmL,(y)/(2)mL),(10mL,,,10xmL,10(y)/(2)mL):}` 10ml = 20 and` 10 (Y)/(2) = 30` Hence formula is `C_(2)H_(6)` X =2 Y = 6 |
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