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Question
compound 1 has a composition of 46.7 mass % of element a and 53.3 mass % of element b. a and b also form a second binary compound 2. if two compositions of the two compounds are consistent with the law of multiple proportions, which of the following compositions could be that of compound 2? options: 33.3 mass % a, 66.7 mass % b; 23.4 mass % a, 76.6 mass % b; 30.4 mass % a, 69.6 mass % b; 73.3 mass % a, 26.7 mass % b; 52.3 mass % a, 47.7 mass % b
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To solve this, we use the law of multiple proportions. First, find the ratio of masses of B that combine with a fixed mass of A (or vice versa) for Compound I and the candidate compounds.
For Compound I: Mass of A = 46.7 mass % (so mass of B = 100 - 46.7 = 53.3 mass %). Let's take 100 g of Compound I: mass A = 46.7 g, mass B = 53.3 g. The ratio of B to A is \( \frac{53.3}{46.7} \approx 1.141 \).
Now check each option:
- Option 1: 33.1 mass % A, 66.9 mass % B
Mass A = 33.1 g, Mass B = 66.9 g. Ratio \( \frac{66.9}{33.1} \approx 2.021 \). Ratio of this to Compound I’s ratio: \( \frac{2.021}{1.141} \approx 1.77 \) (not a small whole number).
- Option 2: 23.4 mass % A, 76.6 mass % B
Mass A = 23.4 g, Mass B = 76.6 g. Ratio \( \frac{76.6}{23.4} \approx 3.273 \). Ratio to Compound I’s ratio: \( \frac{3.273}{1.141} \approx 2.87 \) (not a small whole number).
- Option 3: 30.4 mass % A, 69.6 mass % B
Mass A = 30.4 g, Mass B = 69.6 g. Ratio \( \frac{69.6}{30.4} \approx 2.289 \). Ratio to Compound I’s ratio: \( \frac{2.289}{1.141} \approx 2 \) (whole number, good).
- Option 4: 73.3 mass % A, 26.7 mass % B
Mass A = 73.3 g, Mass B = 26.7 g. Ratio \( \frac{26.7}{73.3} \approx 0.364 \). Ratio to Compound I’s ratio: \( \frac{0.364}{1.141} \approx 0.319 \) (not a small whole number).
- Option 5: 52.2 mass % A, 47.8 mass % B
Mass A = 52.2 g, Mass B = 47.8 g. Ratio \( \frac{47.8}{52.2} \approx 0.916 \). Ratio to Compound I’s ratio: \( \frac{0.916}{1.141} \approx 0.803 \) (not a small whole number).
The correct answer is the option with 30.4 mass % A and 69.6 mass % B (the third option in the list).