When an electric current is passes 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:

At normal temperature and pressure (N.T.P.), one mole of any ideal gas occupies a molar volume of $$22.4\ \text{L}=22400\ \text{mL}.$$

We have collected only $$112\ \text{mL}$$ of hydrogen. The number of moles of hydrogen obtained is therefore

$$n(\text{H}_2)=\frac{\text{volume of H}_2}{\text{molar volume}}=\frac{112\ \text{mL}}{22400\ \text{mL mol}^{-1}}=0.005\ \text{mol}.$$

During electrolysis of acidified water, hydrogen forms at the cathode according to the half-reaction

$$2\text{H}^+(aq)+2e^- \longrightarrow \text{H}_2(g).$$

This equation shows that $$2$$ moles of electrons are needed to produce $$1$$ mole of $$\text{H}_2.$$ Hence, to produce $$0.005\ \text{mol}$$ of $$\text{H}_2,$$ the moles of electrons required are

$$n(e^-)=2\times n(\text{H}_2)=2\times0.005=0.010\ \text{mol}.$$

The charge carried by one mole of electrons is the Faraday constant $$F=96500\ \text{C mol}^{-1}.$$ Therefore, the total charge that has passed through the solution is

$$Q=n(e^-)\,F=0.010\ \text{mol}\times96500\ \text{C mol}^{-1}=965\ \text{C}.$$

The problem states that this charge was delivered in $$t=965\ \text{s}.$$ Electric current is defined by the relation $$I=\dfrac{Q}{t}.$$ Substituting the values gives

$$I=\frac{965\ \text{C}}{965\ \text{s}}=1.0\ \text{A}.$$

Hence, the correct answer is Option D.