The chloride of a metal contains 37.321% chlorine. 1 litre of metal chloride vapour at STP weighs 16.96gm. The formula of metal chloride will be-:

a)MCl b)MCl2 c)MCl3 d)None of the above

To determine the formula of the metal chloride based on the given information, we need to analyze the composition of the compound and the weight of the vapor. Let's break it down step by step.

Understanding the Composition

The problem states that the chloride of a metal contains 37.321% chlorine. This percentage tells us how much of the compound's mass is due to chlorine. We can use this information to find the molar mass of the metal chloride.

Calculating the Molar Mass of Metal Chloride

First, let's denote the molar mass of the metal as \( M \) and the molar mass of chlorine (Cl) as approximately 35.453 g/mol. The formula for the metal chloride can be represented as \( MCl_x \), where \( x \) is the number of chlorine atoms in the formula.

Since we know the mass percentage of chlorine in the compound, we can set up the following equation:

• Mass percentage of Cl = (mass of Cl in the formula / molar mass of metal chloride) × 100
• Given that mass percentage of Cl = 37.321%

From this, we can express the mass of chlorine in terms of \( M \) and \( x \):

Mass of Cl = \( 35.453x \) g

Molar mass of metal chloride = \( M + 35.453x \) g

Setting Up the Equation

Now, substituting these into the mass percentage equation gives:

37.321 = (35.453x / (M + 35.453x)) × 100

Rearranging this equation allows us to solve for \( M \) in terms of \( x \).

Using the Weight of the Vapor

Next, we know that 1 liter of metal chloride vapor at standard temperature and pressure (STP) weighs 16.96 g. At STP, 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the molar mass of the metal chloride using the weight of the vapor:

Molar mass = (Weight of vapor / Volume at STP) × 22.4

Molar mass = (16.96 g / 1 L) × 22.4 L = 380.544 g/mol

Finding the Metal's Molar Mass

Now we have two equations: one from the mass percentage of chlorine and one from the vapor weight. We can substitute the molar mass of the metal chloride into our earlier equation:

380.544 = \( M + 35.453x \)

Solving for Different Values of x

Let's test the possible values of \( x \) based on the options provided:

• If \( x = 1 \): \( M + 35.453 = 380.544 \) → \( M = 380.544 - 35.453 = 345.091 \) g/mol
• If \( x = 2 \): \( M + 2(35.453) = 380.544 \) → \( M = 380.544 - 70.906 = 309.638 \) g/mol
• If \( x = 3 \): \( M + 3(35.453) = 380.544 \) → \( M = 380.544 - 106.359 = 274.185 \) g/mol

Analyzing the Results

Now, we need to consider the typical molar masses of metals. The calculated values for \( M \) should correspond to known metals. The value for \( x = 2 \) gives us a reasonable molar mass for a metal, suggesting that the formula of the metal chloride is likely \( MCl_2 \).

Thus, the answer to the question is:

b) MCl2