Correct Answer - B
Reduction of `H_2O` takes place at cathode since reduction potential of `H_2Ogt` reduction potential of `K^(o+)` ions.
Molecular mass of B `(Me--=-H)=40gm`
1 mol of propyne gas (i.e., `40gm`) = 1 mol of `C_2H_6(g)`
`:. 4gm` of propyne
`implies(1)/(40)xx4=0.1` mol of `C_2H_6(g)`
`implies0.1xx22.4=2.24` litres
So the answer is (b).