Answer
The molar concentration of $N_2O_4$ is equal to $1.3 \space M$
Work Step by Step
1. Write the equilibrium constant expression:
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ N_2O_4 ]}{[ NO_2 ]^{ 2 }}$$
2. Solve for the missing concentration:
$$K_c \times [ NO_2 ]^{ 2 } = [N_2O_4]$$
3. Evaluate the expression:
$$ [N_2O_4] = { 5.0 \times ( 0.50 )^{ 2 }{}}$$
$$[N_2O_4] = 1.3 \space M$$
