When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is _______.

Reduction at cathode: 2H+(aq) + 2e⁻ → H₂(g)
At N.T.P, 22.4 L (or 22400 mL) of H₂ = 1 mole of H₂
112 mL of H₂ = 112/22400 × 1 = 0.005 mole of H₂
Moles of H₂ produced = I(A) × t(s) / [96500 (C/mole⁻) × mole ratio]
0.005 mol = I(A) × 965 s / [96500 (C/mole⁻) × 1 mol H₂ / 2 mol e⁻]
I = 1 A