O 2 and SO 2 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 O 2 is 0-54 atm- then the concentration of SO 2 in mol -L -1 would be R -0-082 L atm K -1 mol -1A- 6-6B- 66C- 0-66D- 0-066

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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Pressure vs Velocity

O 2 and SO 2 ...

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 $0.54 a t m$ , then the concentration of $S O_{2} (i n m o l L^{- 1})$ would be $(R = 0.082 L a t m K^{- 1} m o l^{- 1})$

0.66

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.066

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6.6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

66

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

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}

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})

S O_{2}

at STP contained in a flask was replaced by

O_{2}

under identical conditions of pressure, temperature and volume. Then the weight of

O_{2}

will be ______ of

S O_{2}

Q. A mixture of

S O_{3}, S O_{2}

and

O_{2}

gases are maintained at equilibrium in 10 litre flask at a temperature at which

K_{c}

for the reaction,

2 S O_{2} (g) + O_{2} (g) \Leftrightarrow 2 S O_{3} (g)

is 100

m o l^{- 1}

litre. At equilibrium,

a) If no of mole of

S O_{3}

and

S O_{2}

in the flask are same, how many moles of

O_{2}

are present?
b) If no. of mole of

S O_{3}

in the flask are twice the no of mole of

S O_{2}

, how many moles of

O_{2}

are present?

Q. In a closed container and at constant temperature

0.3

mole of

S O_{2}

and

0.2

mole of

O_{2} g a s

750

torr are kept with a catalyst. If at equilibrium

0.2

mole of

S O_{3}

is formed the partial pressure of

S O_{2}

t o r r

Q. In the equilibrium,

2 {S O}_{2} + O_{2} ⇌ 2 S O_{3}

the partial pressure of

{S O}_{2}, O_{2}

and

S O_{3}

are 0.662,0.10 and 0.331 respectively.What should be the partial pressure of oxygen so that the equilibrium concentration of

{S O}_{2}

and

S O_{3}

are equal?

Join BYJU'S Learning Program

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=XO2×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 $0.54 a t m$ , then the concentration of $S O_{2} (i n m o l L^{- 1})$ would be $(R = 0.082 L a t m K^{- 1} m o l^{- 1})$