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Q.

20% of N2O4 molecules are dissociated in a sample of gas at 27ºC and 760 torr. Mixture has the density at equilibrium equal to

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a

1.48 g/L

b

2.25 g/L

c

1.84 g/L

d

3.12 g/L

answer is D.

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Detailed Solution

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Degree of decomposition, α =M-mmn-1

M = MW of undissociated species

m = MW of equilibrium mixture (observed molecular weight)

n = number of particles produced by the dissociation of one molecule

N2O4(g) 2NO2(g)

n = 2

α = 0.2

T = (27 + 273)K

P = 760 torr (or) 1 atm

\large {M_{{N_2}{O_4}}} = 92
\large 0.2 = \frac{{92 - m}}{{m\left( {2 - 1} \right)}}
\large 0.2 = \frac{{92 - m}}{m}
\large m = \frac{{92}}{{1.2}} = 76.66

Observed molecular weight = 76.66

for an ideal gas, density (d)

\large = \frac{{Pm}}{{RT}}
\large = \frac{{1\left( {76.66} \right)}}{{\left( {0.0821 \times 300} \right)}}
\large = 3.11g/\ell

 

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