O2 and SO2 gases are filled in ratio of 1 - 3 by moles in a closed container of 3 L at temperature of 27-C- If the partial pressure of O2 is 0-54 atm- then the concentration of SO2 in mol L-1 would be R-0-082 L atm K -1mol-1

We have P_O_2 = χ_O_2× P_total ∴ 0.54 = n_O_2n_O_2 + n_SO_2× P_total ⇒ 0.54 = n_O_2n_O_2 + 3n_O_2× P_total ⇒ nbsp; 0.54 = 14× P_total ⇒ nbsp; P_total = 0.54 ...

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Byju's Answer

Standard XIII

Chemistry

Dalton's Law of Partial Pressures

O2 and SO2 ga...

Question

$O_{2}$ and $S O_{2}$ gases are filled in ratio of $1 : 3$ by moles in a closed container of $3 L$ at temperature of $27^{\circ} C$ . If the partial pressure of $O_{2}$ is $0.54 atm$ , then the concentration of $S O_{2} ({in mol L}^{- 1})$ would be $(R = 0.082 {L atm K}^{- 1} m o l^{- 1})$

0.66

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0.066

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6.6

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66

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Solution

The correct option is B $0.066$
We have
$P_{O_{2}} = χ_{O_{2}} \times P_{t o t a l} ∴ 0.54 = \frac{n_{O_{2}}}{n_{O_{2}} + n_{S O_{2}}} \times P_{t o t a l}$
$\Rightarrow 0.54 = \frac{n_{O_{2}}}{n_{O_{2}} + 3 n_{O_{2}}} \times P_{t o t a l} \Rightarrow 0.54 = \frac{1}{4} \times P_{t o t a l} \Rightarrow P_{t o t a l} = 0.54 \times 4 = 2.16 atm ∵ P_{O_{2}} + P_{S O_{2}} = 2.16 atm \Rightarrow P_{S O_{2}} = 2.16 - P_{O_{2}} = 2.16 - 0.54 = 1.62 atm$
$∴ Concentration of S O_{2} = \frac{1.62}{0.082 \times 300} = 0.066$

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Similar questions

O_{2}

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O2 and SO2 gases are filled in ratio of 1:3 by moles in a closed container of 3 L at temperature of 27∘C. If the partial pressure of O2 is 0.54 atm, then the concentration of SO2 (in mol L−1) would be (R=0.082 L atm K−1mol−1)

The correct option is B 0.066We have PO2=χO2×Ptotal∴0.54=nO2nO2+nSO2×Ptotal ⇒0.54=nO2nO2+3nO2×Ptotal⇒0.54=14×Ptotal⇒Ptotal=0.54×4=2.16 atm∵PO2+PSO2=2.16 atm⇒PSO2=2.16−PO2=2.16−0.54=1.62 atm ∴ Concentration of SO2=1.620.082×300=0.066

$O_{2}$ and $S O_{2}$ gases are filled in ratio of $1 : 3$ by moles in a closed container of $3 L$ at temperature of $27^{\circ} C$ . If the partial pressure of $O_{2}$ is $0.54 atm$ , then the concentration of $S O_{2} ({in mol L}^{- 1})$ would be $(R = 0.082 {L atm K}^{- 1} m o l^{- 1})$