Explanation
Given: Mass of $$N_{2}$$ = 56g
Mass of $$CO_{2}$$ = 44g
Mass of $$CH_{4}$$ = 16g
Solution: Moles of $$N_{2}$$ = $$\dfrac{56}{28}$$ = 2
Moles of $$CO_{2}$$ = $$\dfrac{44}{44}$$ = 1
Moles of $$CH_{4}$$ = $$\dfrac{16}{16}$$ = 1
Now, as we know, partial pressure of a gaseous component = mole fraction of the component * total pressure
Thus, partial pressure of $$CH_{4}$$ = ($$\dfrac{1}{2 + 1 +1}$$ * 720) mm Hg
= 180 mm Hg
Thus, the correct option is (B).
What is a weight of $$CO_2$$ in a $$10L$$ cylinder at $$5atm$$ and $$27^0C$$?
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