How much water would be needed to prepare 500 mL of a saturated Potassium Iodide (KI) solution containing 100 g KI per 100mL of solution? (density of saturated solution = 1.7 g/mL, density of water = 1.00 g/mL)

Question

BlackTom AI · Accepted Answer

A saturated KI solution specified as 100 g KI per 100 mL solution means mass composition and solution density must be combined to find how much water is present in 500 mL.

1) Relevant knowledge points and concepts
- Concentration given as g solute per mL of solution (100 g KI / 100 mL solution). This scales linearly with volume.
- Solution density (ρ = mass/volume) allows conversion between solution volume and total solution mass.
- Mass of water = total mass of solution − mass of solute. With water density 1.00 g/mL, mass of water in grams equals its volume in mL.

2) Step-by-step reasoning and calculations
- Scale solute amount to 500 mL solution:
  100 g KI per 100 mL ⇒ for 500 mL: (500 mL ÷ 100 mL) × 100 g = 5 × 100 g = 500 g KI.
- Compute total mass of 500 mL of saturated solution using its density:
  density = 1.7 g/mL ⇒ mass_total = 1.7 g/mL × 500 mL = 850 g (total solution mass).
- Find mass of water in that solution:
  mass_water = mass_total − mass_KI = 850 g − 500 g = 350 g.
- Convert mass of water to volume (density of water = 1.00 g/mL):
  volume_water = mass_water / (1.00 g/mL) = 350 mL.

3) Intermediate steps shown above (scaling, multiplication, subtraction, division).

4) Why the correct answer is correct
The calculation accounts for both the solute mass required for the specified concentration and the increased density of the saturated solution. Subtracting solute mass from total solution mass yields the actual mass (and thus volume) of water present. That gives 350 mL of water.

5) Why other options are incorrect (brief)
- 150 mL and 206 mL are too small; they ignore either the proper solute scaling or the solution density.
- 294 mL is close but results from an incorrect subtraction or density application.
- 500 mL would imply the solution is pure water (no solute), which contradicts the 100 g KI/100 mL specification.

6) Conclusion
Therefore, the required volume of water is 350 mL, matching the provided correct answer (d. 350 mL).

How much water would be needed to prepare 500 mL of a saturated Potassium Iodide (KI) solution containing 100 g KI per 100mL of solution? (density of saturated solution = 1.7 g/mL, density of water = 1.00 g/mL)单项选择题

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