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(C) The relationship between the degree of dissociation $\alpha$,theoretical vapour density $D$,and observed vapour density $d$ is given by the formul

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(C) The relationship between the degree of dissociation $\alpha$,theoretical vapour density $D$,and observed vapour density $d$ is given by the formula: $\alpha = \frac{D - d}{(n - 1)d}$.For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the number of moles of product formed from one mole of reactant is $n = 2$.Substituting $n = 2$ into the formula: $\alpha = \frac{D - d}{(2 - 1)d} = \frac{D - d}{d}$.Rearranging the equation: $\alpha = \frac{D}{d} - 1$,which gives $\frac{D}{d} = \alpha + 1$.At point $A$ on the graph,the degree of dissociation $\alpha = 0$ (this represents the initial state before any dissociation has occurred).Therefore,at point $A$,$\frac{D}{d} = 0 + 1 = 1$.

Similar Questions

Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:
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