JEE Main 31-Aug-2021 Shift 2 Solved Paper

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

M

⇒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,

m2

m1

=

n2M2

n1M1

=0.02×320.06×2=

16

3

i.e. m2:m1=16:3