Question

600 ml of a mixture of ozone and oxygen gas weighs 1g at NTP. calculate the volume of ozone in the mixture?

Answer 1

Use PV = nRT to calculate the number of moles of gas:
At NTP,
P = 1 atm
​T = 298 K
R = 0.0821 L atm K^-1 mol^-1
​Putting the values to get number of moles of gaseous mixture at NTP :
n = PV/ RT
= 1 atm x 0.6 L / 0.0821 L atm K^-1 mol^-1​ x 298 K
​= 0 0.0245 mol.
Now determine the weight of individual gases as follows :
Moles of oxygen = x g/32 g
Moles of ozone = 1-x g /32 (note : total weight of gases is x gram).
Therefore total number of moles :
x/32 + (1-x)/48 = 0.0245
multiply each term by 32 to get:
x + 2/3 - (2/3 x) = 0.784
1/3 x = 0.118
x = 0.355 g of O₂
So, ozone would be = 1 g -0.355 g = 0.645 g

Now Calculate volume of ozone at NTP:
For ozone n = 0.645 g/48 g mol^-1
Keeping the value of P, n R and T in PV = nRT we get
V = nRT/P
= 0.645 g/48 g mol^-1 x 0.0821 L atm K^-1 mol^-1 x 298 K /1 atm
= 0.329 L
= 329 m L