0.58 g of hydrocarbon on combustion gave 0.9
g of water. The percentage of carbon is about:

75.8

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82.7

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27.85

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68.8

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Solution

The correct option is D 82.7
Mole of water formed $= 0.9 / 18 = 0.05$ ,
so mole of hydrogen $0.05 * 2 = 0.1$
mass of hydrogen obtained $= 0.1 * 1 = 0.1$ gm
So mass of carbon in hydrocarbon is $0.58 - 0.1 = 0.48$ gm
% of carbon in hydrocarbon $= (0.48 / 0.58) * 100 = 82.75$ %

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0.58 g of hydrocarbon on combustion gave 0.9g of water. The percentage of carbon is about:

The correct option is D 82.7Mole of water formed =0.9/18=0.05,so mole of hydrogen 0.05∗2=0.1mass of hydrogen obtained =0.1∗1=0.1gmSo mass of carbon in hydrocarbon is 0.58−0.1=0.48 gm% of carbon in hydrocarbon =(0.48/0.58)∗100=82.75 %

0.58 g of hydrocarbon on combustion gave 0.9
g of water. The percentage of carbon is about:

The correct option is D 82.7
Mole of water formed $= 0.9 / 18 = 0.05$ ,
so mole of hydrogen $0.05 * 2 = 0.1$
mass of hydrogen obtained $= 0.1 * 1 = 0.1$ gm
So mass of carbon in hydrocarbon is $0.58 - 0.1 = 0.48$ gm
% of carbon in hydrocarbon $= (0.48 / 0.58) * 100 = 82.75$ %