Question

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 35.5u)

• A. MCL
• B. MCL₂
• C. MCL₃
• D. MCL₄

Answer 1

The Correct Option is B

To determine the molecular formula of the metal chloride, we need to follow these steps:

1. Calculate the Molar Mass of the Metal Chloride:
   • The chlorine content by weight is 55.0%. Hence, the metal content is 45.0%. 
   • Let the atomic mass of the metal be \(M\).
   • The formula weight of the compound according to percentages can be represented as:
   • \(\frac{M}{M + X \cdot 35.5} = \frac{45}{100}\)
2. Determine the Molar Volume of the Gas at STP:
   • At STP, 22.4 liters of gas is equivalent to 1 mole.
   • Given that 100 mL (or 0.1 L) of vapors weigh 0.57 g, we find the molecular weight using proportionality:
   • \(22.4 \, \text{L} \, \longrightarrow \, 1 \, \text{mole}\)
   • \(0.1 \, \text{L} \, \longrightarrow \, \text{Molar Mass (in grams) = } \frac{22.4}{0.1} \times 0.57 = 127.68 \, \text{g/mole}\)
3. Solve for the Metal Chloride Formula:
   • Assume the metal chloride formula is \(\text{MCl}_n\).
   • We know the molecular weight of \(Cl\) is 35.5 u. Hence, the equation becomes:
   • \(M + n \cdot 35.5 = 127.68\)
   • From the percentage calculation:
   • \(\frac{M}{M + n \cdot 35.5} = \frac{45}{100}\)
   • Solving this, \(M = 0.45 \times 127.68 \, \text{= 57.456} \, \text{u}\)
   • Checking possible integer values of \(n\) such that \(57.456 \, \text{+} \, n \cdot 35.5 \approx 127.68\), we find that for \(n = 2\):
   • \(57.456 + 2 \times 35.5 = 128.456\) (close to 127.68).

Hence, the correct molecular formula of the metal chloride is MCl₂.