Question

A gaseous mixture contain oxygen and another unknown gas in molar ratio of 4:1 diffuses through

Answer 1

Diffusion is the transfer of moles of gas in a specific duraton of time and can be shown as:

d = n /t

where, n is moles of gas

t is time in sec

d1 /d2 = (M2 /M1)1/2 ----(1)

M1 is molar mass of oxygen gas = 32 g /mol

M2 is molar mass of unknown gas

d1 = n /t = 4 /245

d2= 1 /220

Putting the value in (1)

(4*220) /(1*245) = (M2 /32)1/2

3.59 = (M2 /32)1/2

12.89 = M2 /32

M2 = 412.63 g /mol

Hence, molar mass of unknown gas is 412.63 g /mol

Answer 2

Answer:

70.4

Explanation:

let no. of moles of oxygen =4 and no. of moles of unknown gas = 1 and let molecular mass of unknown gas be m

let's first calculate the average molecular mass of the mixture

avg molecular mass= x1m1+x2m2

where x1 and x2 are the mole fraction and m1 and m2 are the molecular mass

so average molecular mass of the mixture = 4/5 x 32 + 1/5 x m

= (128+m)/5

r1/r2=(m2/m1)1

since the same volume of oxygen is used

therefore r1/r2=v/245/v/220 = 220/245 = 44/49

44/49 = ((32 x 5)/(128+m))½

1936/2401=160/(128+m)

247808+1936m= 384160

therefore m = 70.4

so final ans :- 70.4

Thank you