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When NO2 is cooled to room temperature, some of it reacts to form a dimer N2O4 through the reaction,
2NO2(g)N2O4(g)
In the reaction, 15.2 g of NO2 is placed in a 10.0 L flask at high temperature and the flask is cooled to 25C. The total pressure is measured to be 0.500 atm. Thus, the partial pressure of NO2 is found to be:

A
0.31 atm
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B
0.19 atm
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C
0.10 atm
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D
0.40 atm
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Solution

The correct option is B 0.19 atm
Initial moles of NO2=15.246=0.33
Molecular weight of NO2=14+2×16=46
2NO2(g)N2O4(g)
After the dimer formation, number of N atom remains unchanged.
Let, N-atom in NO2=nNO2
N-atom in N2O4=2nN2O4 ...(i)
Also by Dalton's law
PNO2+PN2O4=0.5 atm
RTVnNO2+RTVnN2O4=0.5
nNO2+nN2O4=0.5VRT
nNO2+nN2O4=0.5×100.0821×298
nNO2+nN2O4=0.204 ...(ii)
Solving Equations (i) and (ii), we get
nNO2=0.078 mol nN2O4=0.126 mol
Total moles =0.078+0.126=0.204
Mole fraction of NO2=nNO2nNO2+nN2O4=0.0780.204=0.38
PNO2=χNO2×Ptotal=0.38×0.5=0.191 atm

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