Question

A sample of 1.79 mg of a compound of molarmass $90g{mol}^{-1}$ when treated with ${CH}_{3}Mgl$ releases 1.34 ml of a gas at STP. The number of active hydrogen in the molecule is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

Moles of compound= $\frac{1.79}{90}\times {10}^{-3}=1.98\times {10}^{-5}\approx 2\times {10}^{-5}$

Moles of gas= $\frac{1.34}{22400}\approx 6\times {10}^{-5}$

$2$ moles of compound gives $6$ moles of gas $\left(C{H}_4\right)$ or $1$ mol of compound gives $3$ moles of $C{H}_3$

$\therefore 1$ molecule of the compound contains $3$ active hydrogen .