Question

A metal exists as an oxide with formula M_0.96​O. Metal M can exist as M⁺² and M⁺³ in its oxide M0.96​O. The percentage of M⁺³ in the oxide is nearly

• A. 9.6%
• B. 8.3%
• C. 4.6%
• D. 5%

Answer 1

The Correct Option is B

The given oxide formula is \( M_{0.96}O \). Let the amount of metal M present as \( M^{+2} \) be \( x \) and the amount of metal M present as \( M^{+3} \) be \( y \). The total number of metal atoms in the oxide is \( 0.96 \), so: \[ x + y = 0.96. \] The charge balance equation for the oxide is: \[ 2x + 3y = 2 \quad \text{(since the oxide is neutral, and oxygen has a charge of -2)}. \] Now, solve the system of equations: 1. \( x + y = 0.96 \) 2. \( 2x + 3y = 2 \) From equation (1), solve for \( x \): \[ x = 0.96 - y. \] Substitute this into equation (2): \[ 2(0.96 - y) + 3y = 2, \] \[ 1.92 - 2y + 3y = 2, \] \[ 1.92 + y = 2, \] \[ y = 2 - 1.92 = 0.08. \] So, the percentage of \( M^{+3} \) in the oxide is: \[ \frac{y}{0.96} \times 100 = \frac{0.08}{0.96} \times 100 \approx 8.3\%. \]

Thus, the correct answer is (B): \( 8.3\% \).

Answer 2

Let the number of moles of metal M in the oxide be \( x \). The number of moles of oxygen is \( 0.96 \). Let the amount of \( \text{M}^{2+} \) be \( y \) and the amount of \( \text{M}^{3+} \) be \( z \). The total charge from the metal atoms must balance the charge from the oxygen atoms. Oxygen atoms carry a charge of \( -2 \times 0.96 = -1.92 \). Therefore, the total positive charge must be \( +1.92 \). This gives the equation: \[ 2y + 3z = 1.92 \] Also, the total amount of metal M is \( y + z = 1 \) because there is 1 mole of metal. Now, solve these two equations: 1. \( 2y + 3z = 1.92 \) 2. \( y + z = 1 \) From the second equation, \( y = 1 - z \). Substituting this into the first equation: \[ 2(1 - z) + 3z = 1.92 \] \[ 2 - 2z + 3z = 1.92 \] \[ 2 + z = 1.92 \] \[ z = 1.92 - 2 = -0.08 \] So, \( z = 0.08 \). Therefore, the percentage of \( \text{M}^{3+} \) in the oxide is: \[ \frac{0.08}{1} \times 100 = 8.3\% \] Thus, the correct answer is 8.3%.