Problem on weight fraction

Question

A gas contains 350 ppm of H₂S in CO₂ at 72°F and 1.53 atm pressure. If the gas is liquified, what is the weight fraction H₂S?

Answer 1

Now we consider 1liter of CO₂ at 72deg F {[K] ≡ (72+ 459.67) × 5/9}=295.3722 K 1.53 atm

So No moles n= PV/RT =1.53atm*1liter/(0.08205liter-atm/K-mole*295.3722) =0.063131mole

So wt of CO = 0.063131mole*28g/mole =1.76767g

Now 350 ppm H₂S would have a volume 350 microlitre ( µL ) per litre, = 350 ppm =350E-6 litre

So No of moles H₂S = n= PV/RT =1.53atm*(350E-6)liter/(0.08205liter-atm/K-mole*295.3722K)
   = 2.20959E-05 mole

The MW H₂S = 34.116g/mole

So wt of H₂S = 0.000753823 g

So total weight =1.768423823g

So %wt H₂S = (0.0007538236/1.768423823g)*100%=0.04262%