A duma bulb full of air weighs 22.567 gm at `20^@C` and 755 mm pressure. Full of vapours of a substance at `120^@C` and the same pressure. It weighs 22.8617 gm. The capacity of the bulb is 200 ml. Find out the molecular mass of the substance. [density of air `=0.00129(gm)/(ml)`]

Explanation

Given
`{:(V_(1)="Volume of bulb" =200mL,V_(2)=?),(T_(1)=(20-273)=293K,T_(2)=273K),(P_(1)=755mm,P_(2)=760mm):}}NTP "conditions"`
So, `V_(2)=` Volume of buld at NTP
`= (200 xx 755)/(293) xx (273)/(760) = 185.122 mL`
Mass of air `=V_(2) xx 0.00129 = 185.122 xx 0.00129`
`= 0.2388g`
Mass of empty bulb `= (22.567 - 0.2388)`
`= 22.3282g`
Mass of vapours `= (22.8617 - 22.3282)`
`= 0.5335g`
Let the volume of vapours at NTP be V.
`V = (200 xx 755)/(393) xx (273)/(760) = 138mL`
Mo. mass of the substance `= ("Mass of vapours")/("Vol. of vapours at NTP") xx 22400`
`= (0.5335)/(138) xx 22400 = 86.59`