Question

$$0.037$$g of an alcohol, $$R-OH$$ was added to $$CH_3MgBr$$ and the gas evolved measured $$11.2$$ mL at STP. The molecular mass of $$R-OH$$ will be?

Answer 1

Correct option is C. $$74$$

$$\underset{1 mol}{R-OH}+CH_3MgBr\rightarrow \underset{\underset{\underset{22400mL at STP}{or}}{1 mol}}{CH_4}+Mg(Br)OR$$

$$11.2$$mL $$CH_4$$ at STP is formed by $$0.037$$g ROH.

$$\therefore 22400$$mL $$CH_4$$ at STP will be formed by

$$\dfrac{0.037}{11.2}\times 22400$$g ROH

$$=74$$g ROH

$$\therefore$$ Molar mass of ROH$$=74$$.

Option C is correct.