A 100mL solution of CH _ 3 CH _ 2 MgBr on treatment with methanol...

In the above reaction the Ethyl group of grignard reagent acts as a strong base.

As per the stoichiometric equation the number of moles of methanol is equal to the number of moles of ethane.

Volume of methanol=Volume of ethane=$2 . 24ml$

Hence, number of moles of ethane

$n = \dfrac{2 . 24 \times {10}^{- 3}}{22 . 4} = {10}^{- 4}$

Weight of the gas= Number of moles$\times$ Molecular mass

$W = n \times M$

$= {10}^{- 4} \times 30 = 3mg$