Module 7 · Air, Water, and SolutionsPractice · 空气、水与溶液

A low result means too little water flowed in. Too little phosphorus leaves some oxygen unused; a leak lets outside air in; and opening the clamp while the jar is still warm means the gas is still expanded, so less water enters. D and E make the result too high: lowering the phosphorus slowly lets heated air escape from the jar, and an open clamp lets gas escape through the tube during burning.

The phosphorus uses up the oxygen in the jar, and the product, $\ce{P2O5}$, is a solid, so the gas lost equals the volume of oxygen. After cooling, the same volume of water flows in: 250 mL × 1/5 = 50 mL.

$\ce{MnO2}$ is a catalyst, so its mass doesn’t change. But oxygen keeps escaping, so the total mass of the solid keeps falling and $\ce{MnO2}$’s share of it rises. Once all the potassium chlorate has decomposed, the solid’s mass stops changing and so does the percentage. For example, with 24.5 g of potassium chlorate and 5 g of manganese dioxide, it starts at about 16.9% and ends at about 25.1%, when 14.9 g of KCl is left.

Iron burns fiercely in oxygen, throwing off sparks and forming black $\ce{Fe3O4}$. The molten droplets that fall are very hot and would crack the glass if they landed on it directly. Burning iron makes no gas, so the water isn’t there to absorb one; it’s when sulfur is burned that water in the jar absorbs the poisonous $\ce{SO2}$.

$\ce{2H2O(l) ->[\text{electrolysis}]}$ $\ce{2H2(g) +}$ $\ce{O2(g)}$. Oxygen forms at the positive electrode and hydrogen at the negative one, in a volume ratio of about 2 : 1 hydrogen to oxygen (the ratio of the coefficients). So 10 mL × 2 = 20 mL of hydrogen collects at the negative electrode.

Adding solute or evaporating solvent at constant temperature works for any solution. Changing the temperature depends on how the solubility changes: calcium hydroxide becomes less soluble as it gets warmer (0.185 g at 0 °C but only 0.077 g at 100 °C, per 100 g of water), so you raise the temperature, not lower it. Adding water makes the solution even less saturated.

At 60 °C, 100 g of water dissolves 110 g, making 210 g of saturated solution. So 105 g of solution holds 105 × 100/210 = 50 g of water and 55 g of potassium nitrate. At 20 °C, 50 g of water can dissolve only 50 × 31.6/100 = 15.8 g, so 55 − 15.8 = 39.2 g crystallizes. Note that the problem gives the mass of solution, not of water.

Potassium nitrate’s solubility changes a lot with temperature (13.3 g at 0 °C, 246 g at 100 °C), while sodium chloride’s barely changes (35.7 g → 39.8 g). Make a hot saturated solution and cool it: most of the potassium nitrate crystallizes while the small amount of sodium chloride stays dissolved, and filtering gives fairly pure potassium nitrate. Evaporating to dryness would leave both solids together, and since both dissolve in water, filtering alone can’t separate them.

The mass of solute doesn’t change on dilution: 50 g × 20% = 10 g. The diluted solution must weigh 10 g ÷ 5% = 200 g, so the water added is 200 g − 50 g = 150 g.

First check how much dissolves: 50 g of water dissolves at most 50 × 34.0/100 = 17 g, so not all 20 g dissolves. The solution is saturated and holds only 17 g of solute. Mass percent = 17 ÷ (17 + 50) × 100% ≈ 25.4%. Using 20 ÷ 70 gives about 28.6%, which is wrong: undissolved solid isn’t part of the solution.