Question

The treatment of ch3oh with ch3mgi releases 1.04ml of a gas at stp. The mass of ch3oh added is

Answer 1

Ch3 + Ch3Mgl=Ch4+Mg(OH)4l

Molar mass =16 g

22400 ml is occupied by  16 g of gas

so 1.04 ml is occupied by  16/22400*1.04=7.43*10-4 g

One mole of  Ch3OH is required to produce one moles of  Ch4

Equal mole of gas occupy equal volume at STP

No of moles= 7.43*10-4=7.43*10-4/16=4.64*10-5 mol

Mass=4.64*10-5 mol*32=1.48*10-3 g

Hope it helps!!