Describe the process of electrolysis of dilute sulphuric acid and determine the formula of metal oxide.

$$
Q = I \times t
$$
$$
Q = 4 \text{ A} \times (10 \times 60) \text{ s}
$$
$$
Q = 4 \text{ A} \times 600 \text{ s}
$$
$$
Q = 2400 \text{ C}
$$

$$
\text{Moles of X} = \frac{\text{Mass of X}}{\text{Relative atomic mass of X}}
$$
$$
\text{Moles of X} = \frac{1.78 \text{ g}}{118 \text{ g/mol}}
$$
$$
\text{Moles of X} \approx 0.01508 \text{ mol}
$$

Using Faraday's constant ($F = 96485 \text{ C/mol}$):
$$
\text{Moles of electrons} = \frac{Q}{F}
$$
$$
\text{Moles of electrons} = \frac{2400 \text{ C}}{96485 \text{ C/mol}}
$$
$$
\text{Moles of electrons} \approx 0.02487 \text{ mol}
$$

The deposition reaction is $X^{z+}(aq) + z e^- \to X(s)$.
$$
z = \frac{\text{Moles of electrons}}{\text{Moles of X}}
$$
$$
z = \frac{0.02487 \text{ mol}}{0.01508 \text{ mol}}
$$
$$
z \approx 1.65
$$
Since the charge must be an integer, this value is approximately 2 (due to rounding or slight experimental variation). Therefore, the charge on the metal ion is $X^{2+}$.

If the metal ion is $X^{2+}$ and the oxide ion is $O^{2-}$, the formula of the metal oxide is $XO$.
$$
\text{Formula of the metal oxide} = \boxed{XO}
$$