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69 mg of an a...

Question

69 mg of an alcohol $(R - O H)$ was treated with $C H_{3} M g B r$ and the gas evolved was measured to be 33.6 ml at STP. The molar mass (g/mol) of alcohol is .

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Solution

$R - O H 1 m o l + C H_{3} M g B r \to C H_{4} 1 m o l + M g (B r) (O R)$

1 mole at STP = 22.4 L
Let , x be the molar mass.
$\frac{0.069}{x} = \frac{33.6}{22400}$
$\Rightarrow x = 46$

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