number of carbon atoms present in 180 mg of glucose C6H12O6 is
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6 atoms of carbon will be present in 180mg of glucose
I think so
I think so
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General molecular weight of glucose is 180 g
In 180 g→ 6 NA carbon atoms are present
In 180*10^-3 g →(6*180*10^-3 /180)NA
Therefore there are 6*10^-3 NA carbon atoms present in 180 ng of glucose
NA = 6.023*10^23
sriramamallela:
NA is avagadro number
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