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

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

(A) 16.0

(B) 41.2

(C) 82.4

(D) 156.0

Sol: ROH (1 mol) + CH3MgBr  → CH4(1 mol) + Mg(OH)Br

As 1.12 ml of gas is obtained by 4.12 mg of alcohol

22400 ml of CH4 at STP  $\large = \frac{4.12}{1.12} \times 22400 mg $ 

$\large = \frac{4.12}{1.12} \times 22400 \times \frac{1}{1000} \;gm $ 

= 82.4 gm (Mol. Wt.)

Ans: (C)