In a Victor Meyer's determination, 0.23 g of a volatile substance displaced air which measured 112 mL at S.T.P. Calculate the vapour density and molecular weight of the substance (1 litre of `H_2` at S.T.P. weighs 0.09 g).

Explanation

Mass of compound `=017gm`
Volume of air collected `=34.2ml`
Temperature `=15^@C`
Atmospheric pressure `=750mm`
Vapour pressure of water at `15^@C=13mm`
Calculate he vaour density and the molcular mass of the compound:
Given `V_1=34.2ml`
`V_2=?`
`P_1=(750-13)=737`mm
`P_2=760mm`
`T_1=(15+273)=288K`
`T_2=273K`
By gas equation we get
`V_2=(737xx34.2)/(288)xx(273)/(760)`
`=31.43767ml`
vapour density `=(W)/(V_2xx0.00009)`
`=(0.17)/(31.4376xx0.00009)=60.08`
Molecualar mass `=2xx`vapour density
`=2xx60.8=120.16`