See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

] Y [] X [ ] Z [ K c  a ] Z [ 2 1 ] Y [ 2 1 ] X [    (let)  [Z] = 2a, [Y] = 2a, [X] = a  104 = a 2 a a 2  a = 10–4  [Z] = 2a = 2 × 10–4 mol L1.