Mole ConceptHard

Question

A quantity of 2.3 g of a mixture of NO2 and N2O4 has a pressure of 0.82 atm, at temperature TK in a container of volume V litres such that the ratio, T: V is 300: 1 in magnitude. What is the degree of dissociation of N2O4?

Options

A.0.17
B.0.33
C.0.67
D.0.70

Solution

$M = \frac{WRT}{PV} = \frac{2.3 \times 0.082}{0.82} \times \frac{300}{1} = 69$

Now, $\alpha = \frac{M_{o} - M}{(n - 1).M} = \frac{92 - 69}{(2 - 1) \times 69} = 0.33$

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