what would be the molecular formula of the alkane , 5.6 litres of which was found to weigh 11 g at normal temperature and pressure ??
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the correct answer is propane C3H8 .
If we divide the given volume by 22.4 it will give the number of moles that is 0.25.
Then divide given mass by 0.25 the mola mass of the alkane will be obtained that is 44 gram.
I HOPE IT WILL YOU.
If we divide the given volume by 22.4 it will give the number of moles that is 0.25.
Then divide given mass by 0.25 the mola mass of the alkane will be obtained that is 44 gram.
I HOPE IT WILL YOU.
sneha9531:
please mark it as brainliest answer
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