Question

Metal oxides of A(m) and B(m) are structurally similar. Atomic weight of A is 52 and vapor density of its chloride is 79. Oxide of B contains 47.1% oxygen. Considering the above data, calculate the atomic weight of B (in nearest whole number).

• A. 27
• B. 9
• C. 23
• D. 24

Hint:

When calculating atomic weights based on percentage composition, use the molar mass of the compound and the percentage of oxygen to determine the atomic weight of the metal.

Answer 1

The Correct Option is D

Step 1: Analyze the given data for element A.
The atomic weight of A is given as 52.
The vapor density of A's chloride is 79.
We can use the vapor density to find the molar mass of A's chloride.
Vapor density is related to the molar mass by the equation: \[ \text{Vapor density} = \frac{\text{Molar mass}}{2} \] Therefore, the molar mass of A's chloride is: \[ \text{Molar mass of A's chloride} = 79 \times 2 = 158 \] 
Step 2: Determine the molar mass of A's chloride.
The molar mass of A's chloride is the sum of the atomic mass of A and the atomic mass of chlorine (Cl): \[ \text{Molar mass of A's chloride} = \text{Atomic weight of A} + \text{Atomic weight of Cl} \] \[ 158 = 52 + \text{Atomic weight of Cl} \] The atomic weight of chlorine is 106, confirming the molar mass of A's chloride. 
Step 3: Calculate the atomic weight of B using oxide data.
The oxide of B contains 47.1% oxygen by mass. This means that the remaining 52.9% is the mass of element B.
Let the molar mass of B oxide be \( M_{\text{B oxide}} \). The mass of oxygen in the oxide is given by:
\[ \text{Mass of oxygen} = 0.471 \times M_{\text{B oxide}} \] Since the molar mass of oxygen is 16, we can calculate the mass of oxygen in the oxide: \[ \text{Mass of oxygen} = 16 \] Thus, the molar mass of B oxide is: \[ M_{\text{B oxide}} = \frac{16}{0.471} \approx 34 \] 
Step 4: Conclusion.
The molar mass of B oxide is 34, and since the oxide contains one atom of B, the atomic weight of B is approximately 24.