A metal chloride contains $$55.0\%$$ of chlorine by weight. $$100$$ mL vapours of the metal chloride at STP weigh $$0.57$$ g. The molecular formula of the metal chloride is
(Given: Atomic mass of chlorine is $$35.5$$ u)

A metal chloride has 55% chlorine by weight, and 100 mL of its vapour at STP weighs 0.57 g. The molecular formula can be determined by finding the molar mass from the STP data.

At STP (Standard Temperature and Pressure), one mole of an ideal gas occupies 22400 mL, so the molar mass $$M$$ is calculated as $$M = \frac{\text{mass} \times 22400}{\text{volume}} = \frac{0.57 \times 22400}{100} = \frac{12768}{100} = 127.68\;\text{g/mol}$$.

Since 55% of the compound’s mass is chlorine, the mass of Cl in one mole is $$55\% \times 127.68 = 70.22$$ g. The number of Cl atoms per molecule is then $$\frac{70.22}{35.5} \approx 1.98 \approx 2$$.

The remaining mass corresponds to the metal: $$127.68 - 2(35.5) = 127.68 - 71 = 56.68$$ g/mol. This value is close to the atomic mass of Mn (54.94) or Fe (55.85), indicating that the metal is M and the formula is $$MCl_2$$.

Therefore, the molecular formula is $$MCl_2$$, corresponding to Option 3.