Question

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

• A. 16
• B. 41.2
• C. 82.4
• D. 156

Answer 1

The correct option is C 82.4

$ROH+C{H}_{3}MgI\to C{H}_4+Mg{<}_I^{OR}$
1 mol 1 mol = 22400 cc
1.12 mL is obtained from 4.12 mg
$\therefore$ 22400 mL will be obtained from
$\frac{4.12}{1.12}\times 22400mg=84.2g$