Question textYou are given Rhodamine B at 1 mg/mL in absolute ethanol. You have mixed 0.5 mL of the 1mg/mL Rhodamine B in absolute ethanol with 0.5 mL of the DI water. The new stock solution concentration is Answer 1 Question 2[input] µg/mL in Answer 2 Question 2[select: , absolute ethanol, 50:50 (v/v) Ethanol:Water].多项填空题
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Question

Question textYou are given Rhodamine B at 1 mg/mL in absolute ethanol. You have mixed 0.5 mL of the 1mg/mL Rhodamine B in absolute ethanol with 0.5 mL of the DI water. The new stock solution concentration is Answer 1 Question 2[input] µg/mL in Answer 2 Question 2[select: , absolute ethanol, 50:50 (v/v) Ethanol:Water].多项填空题

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BlackTom AI · Accepted Answer

Equal-volume dilution reduces concentration by half, so the 1 mg/mL stock becomes 0.5 mg/mL (500 µg/mL) when mixed 1:1 with water; the solvent composition becomes 50:50 (v/v) ethanol:water.

1) Relevant concepts
- Concentration units: 1 mg/mL = 1000 µg/mL.  
- Conservation of solute mass: adding solvent changes volume but not the amount (mass) of dissolved solute.  
- Volume fractions: mixing 0.5 mL ethanol with 0.5 mL water yields 0.5/1.0 = 50% ethanol by volume.

2) Step-by-step reasoning
- Start with 1 mg/mL Rhodamine B in absolute ethanol.  
- Take 0.5 mL of that solution and add 0.5 mL DI water. Total volume = 0.5 + 0.5 = 1.0 mL.  
- The mass (amount) of Rhodamine B in the 0.5 mL taken remains constant when diluted.

3) Calculations (all intermediate steps)
- Initial concentration: 1 mg/mL = 1000 µg/mL.  
- Mass of dye in 0.5 mL = concentration × volume = 1000 µg/mL × 0.5 mL = 500 µg.  
- New concentration after dilution = mass / total volume = 500 µg / 1.0 mL = 500 µg/mL.

4) Why this is correct
- Because only solvent was added (no additional dye), the dye amount is fixed at 500 µg and is spread into a total volume of 1.0 mL, giving 500 µg/mL. The 1:1 volume mix halves the original concentration.

5) Why other options are incorrect
- "Absolute ethanol" would be incorrect for the solvent description because DI water was added; the final mixture is an equal-volume mixture of ethanol and water, i.e., 50:50 (v/v). Any concentration other than 500 µg/mL would violate mass conservation or miscalculate the volume change.