Initially 3 moles of '$\mathrm{A}$' was taken in a $\text{1}\text{L}$ container. The approx. moles of A left in the container when the following equilibrium established is $\mathrm{x}\times 10^{-6}$ .The value of ' $\mathrm{x}$ ' is.
$3\mathrm{A}\rightleftharpoons \mathrm{B};{\mathrm{K}}_{\mathrm{C}}=8\times 10^{15}$

Solution

Now,
$\frac{1-\mathrm{x}}{(3\mathrm{x}{)}^3}=8\times 10^{15}$
$3\mathrm{x}=5\times 10^{-6}$

Hence, the correct answer is 5.