1.12 ml of a gas is produced at STP by the action of 4.12 mg of alcohole, with methyl magnesium iodide. The molecular mass of alcohol is

Options

A. `16.0`
B. `41.2`
C. `82.4`
D. `156.0`

Answer

C

Explanation

`underset(1mol)(ROH)+CH_(3)MgBrtounderset("at S.P.T.")underset(22400mL)(CH_(4)) uarr+BrMgOR`
22400 mL of `CH_(4)` at S.T.P. = 1 mol of ROH
1.12 mL of `CH_(4)` at S.T.P. `=(1molxx1.12mL)/(22400mL)`
`=5xx10^(-5)`mol
`5xx10^(-6)`mol of ROH `=4.12 mg`
`=4.12xx10^(-3)g`
1 mol of ROH `=(4.12xx10^(-3))/(5xx10^(-5))`
`=0.824xx10^(2)=82.4g`
`:.` Mol. mass of ROH = 82.4