Question

0.44 g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP, 112 cc of methane. With PCC the same alcohol forms a carbonyl compound that answers silver mirror test. The monohydric alcohol is...

​

Answer 1

Firstly you need to understand that silver mirror test is only given by aldehydes in order to give an acid hence the answer should be a primary alcohol which on oxidation with PCC(pyridium chromyl chloride) will give you an aldehyde

112/22400ml=0.005 moles of the methyl compound

now we can find the molecular weight of the compound by the formula

moles=weight/mol weight

mol.wt=0.44/0.005=88g

b)5(C)+(O)+12(H)=12×5+16+12×1

=88g

I hope this helps

PLEASE MARK BRANLIEST IF HELPS