1.12 ml of a gas is produced at STP by the action of 4.12 mg of alcohol, with methyl magnesium iodide. The molecular mass of alcohol is

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41.2

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82.4

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156

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Solution

The correct option is C 82.4
$R O H + C H_{3} M g I \to C H_{4} + M g <_{I}^{O R}$
1 mol 1 mol = 22400 cc
1.12 mL is obtained from 4.12 mg
$∴$ 22400 mL will be obtained from
$\frac{4.12}{1.12} \times 22400 m g = 84.2 g$

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