You are required to make a 500mL solution of sodium chloride containing 24 mg/L chloride ions in water. Assume that your balance has a sensitivity of 0.05 g, and you can accurately weigh 1 g of sodium chloride (table salt) powder (MW of NaCl = 58, MW of Na+ = 23, MW of Cl- = 35). Use this 1 g of sodium chloride powder to prepare a stock solution, such that 5 mL of the stock solution can be diluted to 500 mL water to produce a final solution containing 24 mg/L chloride. What is the required volume for the stock solution?

Question

BlackTom AI · Accepted Answer

Mass of chloride required in 500 mL = 24 mg/L × 0.5 L = 12.0 mg Cl−.

1) Relevant concepts
- Concentration in mg/L multiplied by volume (L) gives mass of solute (mg).
- The chloride fraction in NaCl = MW(Cl−)/MW(NaCl) = 35/58; to get mass of NaCl required to supply a given mass of Cl− multiply by 58/35.
- Dilution: mass of solute in an aliquot of a stock = (mass of solute initially dissolved) × (aliquot volume / stock volume).

2) Step-by-step reasoning
- Find how much Cl− is needed in the final 500 mL.
- Convert that needed Cl− mass to the equivalent mass of NaCl.
- Use the fact 1.000 g NaCl is dissolved to make a stock of unknown volume V. Five mL of this stock must contain the needed NaCl mass. Solve for V.

3) Calculations (all intermediate steps)
- Cl− required: 24 mg/L × 0.500 L = 12.0 mg Cl−.
- Convert to NaCl mass: mass_NaCl = 12.0 mg × (58 / 35) = 12.0 × 1.6571429 = 19.8857 mg NaCl = 0.0198857 g.
- Let V be stock volume in mL. Mass of NaCl in 5 mL aliquot = 1.000 g × (5 mL / V).
  Set this equal to required mass: 1.000 × (5 / V) = 0.0198857 g.
- Solve for V: V = 5 / 0.0198857 = 251.34 mL.

Rounded appropriately, V ≈ 251.4 mL.

4) Why this is correct
Every step preserves mass balance: the 5 mL aliquot must contain exactly the NaCl necessary to provide 12 mg Cl− in the final 500 mL. Converting Cl− mass to NaCl mass using molar mass ratios ensures the correct amount of salt is accounted for.

5) Common mistakes (why other answers would be wrong)
- Using 24 mg instead of 24 mg/L × 0.5 L (i.e., not accounting for the 0.5 L final volume).
- Forgetting to convert Cl− mass to NaCl mass via the 35/58 ratio.
- Mixing units (mg vs g) or using the wrong dilution factor (e.g., using 500 mL instead of 5 mL aliquot).
Any of these errors changes the computed V and yields an incorrect result.

6) Conclusion
The required stock volume is 251.4 mL, which matches the provided answer.

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