Let's break down the problem step by step: ### Step 1: Identify X₂ - We have a 0.105-g sample of X₂ containing \( 8.92 \times 10^{20} \) molecules. - Find the molar mass of X₂: Number of moles in 0.105 g of X₂: \[ \text{Moles of } X_2 = \frac{8.92 \times 10^{20}}{6.022 \times 10^{23}} = 1.482 \times 10^{-3} \text{ moles} \] \[ \text{Molar mass of } X_2 = \frac{0.105 \text{ g}}{1.482 \times 10^{-3} \text{ moles}} = 70.89 \text{ g/mol} \] Since X₂ is diatomic, the molar mass of X is about half of 70.89 g/mol: \[ \text{Molar mass of X} = \frac{70.89}{2} = 35.45 \text{ g/mol} \] This matches the atomic mass of **chlorine (Cl)**. ### Step 2: Find M in MX₃ Given: MX₃ consists of 54.47% X (Cl) by mass. Let the molar mass of M be \( M \). \[ \frac{3 \times 35.45}{M + 3 \times 35.45} = 0.5447 \] \[ 3 \times 35.45 = 106.35 \] \[ \frac{106.35}{M + 106.35} = 0.5447 \] \[ 106.35 = 0.5447(M + 106.35) \] \[ 106.35 = 0.5447M + 57.95 \] \[ 106.35 - 57.95 = 0.5447M \] \[ 48.4 = 0.5447M \] \[ M = \frac{48.4}{0.5447} \approx 88.9 \] This is approximately the atomic mass of **Yttrium (Y)**. ### Step 3: Name the compound Since Yttrium forms a +3 ion and the anion is chloride, the compound is **Yttrium(III) chloride**. ### Step 4: Match with the Options - Option (b): M:Y, X:Cl; Yttrium(III) chloride **Correct Answer:** > **b. M:Y, X:Cl; Yttrium(III) chloride**