Question

At 0.483 atm and 52 degress C, the density of gas "X" is 2.35 g/L. A volume of 45.0 mL of gas "X" effuses through an apparatus in 3.60 seconds. The rate of effusion of gas "Y" through the same apparatus and under the same conditions is 18.4 mL/sec. What is the molar mass of gas "Y"?

Answer 1

Apply graham law:

Rate 1/ Rate 2 = sqrt(MW2/MW1)

Get MW of X via

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure; V = total volume of gas

n = moles of gas

T = absolute Temperature; R = ideal gas constant

D = mass / volume and Molar volume = n/V

MW = molar weight of sample

Then,

PV = nRT; get molar volume

n/V = P/(RT)

get mass -> n = mass/MW

mass/(MW*V) = P/(RT)

D = mass/V

D/MW = P/(RT)

D = P*MW/(RT)

2.35  = (0.483)(MW)/(0.082*(52+273))

MW = 2.35 * (0.082*(52+273))/0.483

MW of X = 129.66 G/mol

Rate 1 = 45/3.6 = 12.5 mL/s

Rate 2 = 18.4 mL/s

substitute

Rate 1/ Rate 2 = sqrt(MW2/MW1)

12.5 / 18.4 = sqrt(MW2/129.66 )

MW of Y = 129.66 *(12.5 / 18.4 )^2

MW of Y = 59.83 g/mol