Given,
Volume,
V=500cm3=5×102×10−6m3 =5×10−4m3 Pressure,
p=400kPa=4×105Pa Temperature, T = 300 K
Total mass, m = 0.76 g
We know that,
Molar mass of hydrogen,
M1= 2 g
Molar mass of oxygen,
M2 = 32 g
Let
n1 and
n2 be the number of moles of hydrogen and oxygen, respectively.
Total number of moles of mixture,
n=n1+n2 By using ideal gas equation, pV = nRT
Substituting the values, we get
4×105×5×10−4=n×8.314×300 n= 0.08 mol
Then,
n1+n2=0.08 Let mass of hydrogen in mixture =
m1 and mass of oxygen in oxygen =
m2 n1M1+n2M2=m [∵n=⇒m=nM] Substituting the values, we get
2n1+32n2=0.76 n1+16n2=0.38 ...(ii)
Solving Eqs. (i) and (ii), we get
n1=0.06 and
n2=0.02 ∴ Ratio of masses,
==0.02×320.06×2= i.e.
m2:m1=16:3