Question

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

• A. $16.0$
• B. $41.2$
• C. $82.4$
• D. $156.0$

Answer 1

The correct option is C $82.4$

$ROH+C{H}_{3}MgI⟶C{H}_4+MgROI$

$22400mLofC{H}_{4}producedatSTP$

$\text{= 1mol of ROH}$

$1.12mLofC{H}_4\text{produced at STP}=\frac{1.12}{22400}mol$

$\text{Now molecular mass of ROH}=\frac{4.12\times {10}^{-}3}{5\times {10}^{-5}}$

$\therefore \text{mol mass of ROH=82.4.}$