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Q)

Calculate the percentage composition of element in methanol $(CH_2OH)$

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A)
Solution :
Molecular mass of methanol $=1 \times 12.01 +4 \times 1.008 + 1 \times 16.0$
$\qquad= 32.042\;gm$
Percentage composition of carbon =$\large\frac{12.01}{32.042}$$\times 100=37.48\%$
Percent composition of Oxygen $=\large\frac{16 \times 100}{32.04}$$=49.93\%$
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