For $\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C}$, the equilibrium concentrations of A and B at a temperature are 15 mole/lit. When volume is doubled the reaction has equilibrium concentration of A as 10 mol/lit. The concentration of C in original equilibrium is

Volume of vessel is doubled, concentrations become halved.

But according to Lechatelier's principle increase in volume favours reaction involving increase in numberr of moles. Thus reaction proceeds in backward direction at new equilibrium is attained.

At new equilibrium, [A] = 10M

At new equilibrium, [C] decreases whereas [A] & [B] increases

Due o doubling of volume [A] must be 7.5 M but it is given that [A] = 10 M

Increase in concentration of 'A' is 2.5 M and increase in concentration of 'B' must be 2.5 M

Decrease in concentration of 'C' must be 2.5 M

At new equilibrium

Temperature is constant, K_C remains constant.

Comparing (1) & (2)

[C]_eq in original equilibrium = x = 45 M