Question

A sample of 4.5mg of unknown alcohol is added to CH3MgBr when 1.68ml of CH4 at STP is obtained . The unknown alcohol is?

Answer 1

Dear student!

At room temperature, we know, for gases 1 mole = 22.4 L

So, for 1.68 ml methane ,no.of moles of methane = 1.68/22400 = 75 x10^-6 mol

Since with Grignard reagent, 1 mole of alcohol produces 1 mol of methane

So, here 4.5 mg alcohol (= 0.0045gm) will also have 7.5x10^-7 moles

So, molar mass of alcohol = given mass/ no. of moles =0.0045 g / 75x10^-6 =60 gm/mol

Since, molar mass is 60 which corresponds to the Propanol (C₃H₇OH )thus the alcohol will be propanol.