How many anions are in 0.500 g of MgBr₂? A)1)64 × 10²¹ anions B)3)27 × 10²¹ anions C)2)22 × 10²⁶ anions D)4)43 × 10²⁶ anions

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To determine the number of anions in 0.500 g of MgBr₂, we need to first determine the formula weight of the compound. The atomic weights of Mg, Br, S, and O are approximately 24, 80, 32, and 16 g/mol, respectively. Since the compound contains one magnesium ion (Mg2+) and three anions (Br-, S2-, and O2-), we can calculate the formula weight as follows: MgBr₂: (1 x 24 g/mol) + (1 x 80 g/mol) + (1 x 32 g/mol) + (1 x 16 g/mol) + (2 x 16 g/mol) + (3 x 32 g/mol) = 296 g/mol Now we can use the formula weight to convert the mass of the compound to moles: 0.500 g / 296 g/mol = 0.001689 moles Since there are three anions per formula unit of the compound, we can calculate the number of anions as: 0.001689 moles x 3 = 0.00507 moles of anions Finally, we can convert moles of anions to the number of anions using Avogadro's number: 0.00507 moles x 6.022 x 10²³ ions/mol = 3.06 x 10²¹ ions, which is approximately 27 x 10²⁰ or 27 x 10¹⁰³ anions. Therefore, the correct answer is B, which is 27 x 10¹⁰³ or 27 x 10²¹.

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