`0.46 g` of a compound active hydrogen (molar mass `=92 mol^(-1)`) on treatment with excess of `CH_(3)MgI` releases `336 ml` of a gas at `STP` (`1 atm` pressure). The number of active `H` atoms per molecule of compound is

Options

A. `3`
B. `4`
C. `2`
D. `1`

Answer

A

Explanation

`CH_(3)MgI+underset(1mol)(HZ)rarr underset(1 mol)(CH_(4)+ZMgX`
According to the above equation, 1 mol of `CH_(4)` gas in equivalent to `1 mol` of active `H` atoms. Thus, the moles of `CH_(4)` gas are equal to the moles of active `H` atoms
`n_("compound")=("Mass of compound")/("Molar mass of compound")=(0.46g)/(92 "g "mol^(-1))`
`0.005`
`n_(CH_(4))=(336 "mol")/(22400mL mol^(-1))= 0.015`
Thus, the number of moles of active `H` atoms per mole of compound is
`(n_(CH_(4)))/(n_("comp"))=(0.015)/(0.005)=3`
This implies that are three active `H` atoms per molecule of compound.