The weight percent of chlorine in the compound is given as 55.0%. This implies that the weight percent of the metal is 45.0%.

Now, we need to find the molar mass of the compound. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L. We know that 100 mL (or 0.1 L) of the metal chloride vapor weighs 0.57 g. Therefore, the molar mass of the compound (MClx) is:

$$M_{MCl_x} = \frac{0.57 \, \text{g}}{0.1 \, \text{L}} \times 22.4 \, \text{L/mol} = 127.68 \, \text{g/mol}$$

Given the weight percent of the metal and chlorine, we can find the molar mass of the metal (MM) using the equation:

$$MM = M_{MCl_x} \times \frac{45.0}{55.0} = 104.04 \, \text{g/mol}$$

Subtracting the atomic mass of chlorine from the molar mass of the metal gives the molar mass of the metal:

$$MM - x \cdot 35.5 = 104.04 \, \text{g/mol}$$

Where 'x' is the number of chlorine atoms in the compound.

Solving this equation for 'x' gives:

$$x = \frac{104.04 - MM}{35.5}$$

We find that 'x' is approximately 2. Therefore, the molecular formula of the compound is MCl₂.

So, the correct answer is MCl₂.