KCET 2014 Chemistry Solved Paper

At STP 112cm3 of methane liberated, 22400cm3=1mol at ST
1cm3=‌

1

22400

mol
112cm3=‌

1

22400

×112mol
= 0.005 mol of methane liberated
1 mol of monohydric alcohol react with 1 mol of methyl magnesium iodide to produces 1 mol of methane, according to the reaction,
Monohydric alcohol+CH3Mgl

Either

––––––––>

CH4
⇒ 0.005 mol of monohydric alcohol produces 0.005 mol of methane
Now, molar mass of monohydric alcohol
=‌

0.44

0.005

=88g
Monohydric alcohol from the given option having 88 g molar mass is (CH3)3C−CH2OH
When react with PCC, aldehyde is formed which give positive silver mirror test.