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Question

At 340 K and 1 atm pressure, N2O4 is 66% dissociated into NO2. What volume does 10 g N2O4 occupy under these conditions?

A
5.023 L
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B
10.4 L
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C
15.4 L
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D
0.05 L
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Solution

The correct option is A 5.023 L
The dissociation reaction is as follows:

N2O42NO2
Initial number of moles 1 0
Number of moles at equilibrium 1a 2a
Here, a is the degree of dissociation which is equal to 0.66.

Total number of moles at equilibrium =1a+2a=1+a=1+0.66=1.66

1 mole of N2O4 present initially corresponds to 1.66 moles at equilibrium.

1092 moles of N2O4 present initially corresponds to 1.66×1092=0.18 moles at equilibrium.

The ideal gas equation is PV=nRT.
Substituting values in the above expression, we get
1×V=0.18×0.0821×340
V=5.02 L

Hence, the volume occupied by 10 g of N2O4 will be 5.02 L.

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