two 250 ml samples of water are drawn from a deep well bored into a large underground salt (nacl) deposit. sample #1 is from the top of the well, and is initially at 42 °c. sample #2 is from a depth of 150 m, and is initially at 8 °c. both samples are allowed to come to room temperature (20 °c) and 1 atm pressure. an nacl precipitate is seen to form in sample #1. predicted observation (choose one) a bigger mass of nacl precipitate will form in sample #2. a smaller mass of nacl precipitate will form in sample #2. the same mass of nacl precipitate will form in sample #2. no precipitate will form in sample #2. i need more information to predict whether and how much precipitate will form in sample #2.

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# Brief Explanations:
To determine the precipitation in Sample #2, we analyze the solubility of NaCl. Solubility of NaCl in water increases with temperature (though the change is small) and is relatively insensitive to pressure for aqueous solutions at moderate pressures. Sample #1 is from the top (42°C), Sample #2 from 150 m depth (8°C initially, then brought to 20°C and 1 atm). When Sample #1 (higher initial temp, then cooled to 20°C) forms precipitate, Sample #2, when brought to the same final conditions (20°C, 1 atm) as Sample #1, will have similar solubility conditions for NaCl. However, wait—actually, the key is that the final conditions (20°C, 1 atm) are the same for both after equilibration. But Sample #1 was at 42°C initially, so when it's cooled to 20°C, NaCl solubility decreases slightly, causing precipitation. Sample #2 was at 8°C initially, then heated to 20°C. The solubility of NaCl at 20°C is known, and when heating from 8°C to 20°C, solubility increases (since solubility of NaCl generally increases with temperature). Wait, maybe I got the initial temps reversed? Wait, Sample #1: top of well, initial temp 42°C. Sample #2: depth 150 m, initial temp 8°C. Both are brought to room temp (20°C) and 1 atm. So Sample #1 is cooled from 42°C to 20°C: solubility of NaCl decreases (since solubility of NaCl in water increases with temperature, though slightly). So at 42°C, it could hold more NaCl, at 20°C, less, so precipitate forms. Sample #2 is heated from 8°C to 20°C: solubility of NaCl increases (since 20°C is warmer than 8°C). So at 8°C, it had some amount of dissolved NaCl, and at 20°C, the solubility is higher, so it can hold more, so no precipitate should form? Wait, no—wait, the problem says "both samples are allowed to come to room temperature (20°C) and 1 atm pressure". So we need to consider the initial concentration. But the problem is about predicting the precipitation. Wait, maybe the key is that the solubility of NaCl is relatively constant over small temperature changes, but also, the pressure at 150 m depth: pressure at depth $ P = P_0 +
ho gh $, where $ P_0 = 1 $ atm, $
ho = 1000 $ kg/m³, $ g = 9.8 $ m/s², $ h = 150 $ m. So $ P = 1 + \frac{1000 \times 9.8 \times 150}{101325} \approx 1 + 14.5 = 15.5 $ atm initially for Sample #2. But when brought to 1 atm, the pressure decreases. However, the solubility of NaCl in water is not strongly dependent on pressure (unlike gases). So the main factors are temperature and concentration. But the problem states that Sample #1 (top, 42°C initial) forms precipitate at 20°C, 1 atm. Sample #2 (depth, 8°C initial, 15.5 atm initial) is brought to 20°C, 1 atm. When we reduce pressure from 15.5 atm to 1 atm, for NaCl (a solid solute), pressure has little effect on solubility. The temperature change for Sample #2 is from 8°C to 20°C (increase), so solubility increases (since NaCl solubility increases with T). For Sample #1, temperature change is from 42°C to 20°C (decrease), solubility decreases, so precipitate forms. For Sample #2, heating from 8°C to 20°C: solubility increases, so if it was saturated or nearly saturated at 8°C and 15.5 atm, at 20°C and 1 atm (lower pressure, but pressure doesn't affect NaCl solubility much), the solubility is higher, so it can hold more NaCl, so no precipitate should form. Wait, but maybe the initial concentration: if both samples have the same amount of dissolved NaCl initially (but at different T and P), but the problem doesn't state that. Wait, the problem is about two 250 mL samples of water drawn from the well. So maybe the water is from the same source, so same initial concentration of NaCl, but different T and P initially. Then, when brought to 20°C and 1 atm: Sample #1 (cooled from 42°C to 20°C): solubility at 42°C is higher than at 20°C, so if it was saturated at 42°C, at 20°C it's supersaturated, so precipitate. Sample #2 (heated from 8°C to 20°C): solubility at 8°C is lower than at 20°C, so if it was saturated at 8°C, at 20°C it's unsaturated, so no precipitate. But the problem says "An NaCl precipitate is seen to form in Sample #1". So Sample #1 was at 42°C, then cooled to 20°C, became supersaturated, precipitated. Sample #2 was at 8°C (lower T, so lower solubility) initially, then heated to 20°C (higher T, higher solubility). So if the initial concentration of NaCl in both samples was the same (since they're from the same well), then at 8°C, Sample #2 could hold less NaCl, but when heated to 20°C, it can hold more, so no precipitate. Alternatively, maybe the pressure effect: at 150 m depth, the pressure is higher, which for some solutes (but not NaCl) could increase solubility, but NaCl's solubility is pressure-insensitive. So the main factor is temperature. Since Sample #2 is heated to 20°C (same as Sample #1's final temp) from a lower temp, its solubility increases, so it can dissolve more NaCl, so no precipitate forms. Therefore, the correct prediction is "No precipitate will form in Sample #2."
# Answer:
No precipitate will form in Sample #2.

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