A Duma's bulb full of air weighs 22.567 g at `20^@C` and 755 mm pressure. Full of vapours of a substance at `120^@C` and the same pressure, it weighs 22.8617 g. The capacity of the bulb is 200 mL. Find out the molecular mass of the substance. [Density of air=0.00129 g/mL) : 86.69, 80.64, 78.92, 96.75

Options

A. 86.59
B. 80.64
C. 78.92
D. 96.75

Answer

A

Explanation

Given `V_1` = volume of bul = 200 mL `V_2=?`
`T_1 = (20+273) = 293 K " " T_2 =273 K`
`P_1 = 755 mm " " P_2 = 760 mm`
So, `V_2` = Volume of bulb at NTP `= (200xx755)/(293)xx(273)/(260)=185.122mL`
Mass of air `=V_2 xx 0.00129 = 185.122 xx0.00129 = 0.2388 g`
Mass of empty bulb ` = (22.567 - 0.2388) = 22.3282 `
Mass of vapours = `(22.8617- 22.3282) = 0.5335 g`
Let the volume of vapours at NTP be V , `V = (200xx755)/(393) xx(273)/760 = 138 mL`
Molecular mass of the substance = `("mass of vapours")/("volume of vapours at NTP") xx 22400`
`= (0.5335 )/138 xx22400 = 86.59 `