`0.037gm` of an alcohol was added to `CH_(3)Mgl` and the gas evolved measured `11.2ml` at `STP`. What is the molecular weight of the alcohol ?
Identify `(A)` to `(E)`.

Explanation

`11.2 ml` of `CH_(4)`(gas) is obtained by `0.037 gm`of `ROH`
`22,400 ml` of `CH_(4)` is obtained by `implies(0.037xx22400)/( 11.2)=74gm`
Molecular mass of `ROH=74 gm`
Since ROH on oxidation gives an aci with the same number of `C` atoms, thus `ROH` is a `1^(@)` alcohol.
`C_(N)H_(2n+1). CH_(2)OH=12n+2n+1+12+2+16+1=74`.
`n=3`.
The molecular formula of alcohol `(A)=(C_(3)H_(7)CH_(2)OH)`.

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