How much Sodium Fluoride (NaF; MW = 42, F- ; AW = 19) is required to prepare 100 mL of a solution such that 30 mL of this solution diluted to 1 litre with drinking water, will produce a final concentration of F- of 1 part per million? (Note: The drinking water already contains 0.5 ppm F- ) Assume the NaF available is 100% pure.

Question

BlackTom AI · Accepted Answer

Extra fluoride required = target 1.0 ppm − existing 0.5 ppm = 0.5 ppm, which is 0.5 mg F− per 1 L of water.

Relevant concepts:
- ppm for aqueous solutions ≈ mg/L.
- Dilution: mass of solute delivered = concentration × volume delivered.
- Stoichiometry: convert mass of F− needed to mass of NaF using mass fraction F = AW(F)/MW(NaF) = 19/42.

Step-by-step calculation:
1) Mass of F− needed in the final 1 L: 0.5 mg (because 0.5 mg/L × 1 L = 0.5 mg).
2) This 0.5 mg of F− must come from 30 mL (0.03 L) of the prepared stock. Therefore the stock must contain 0.5 mg F− in 0.03 L, so its concentration = 0.5 mg / 0.03 L = 16.6667 mg/L (or 16.6667 ppm F−).
3) We will prepare 100 mL (0.1 L) of that stock, so total mass of F− to add to 0.1 L = concentration × volume = 16.6667 mg/L × 0.1 L = 1.66667 mg F−.
4) Convert F− mass to NaF mass. Fraction of F in NaF = 19/42, so NaF mass = (mass F−) × (MW NaF / AW F) = 1.66667 mg × (42/19) = 1.66667 × 2.210526 = 3.68421 mg ≈ 3.68 mg NaF.

Why this answer is correct:
- It accounts for the existing 0.5 ppm in the drinking water and calculates only the additional fluoride needed.
- It uses the correct dilution (only 30 mL of stock contributes to 1 L final) and stoichiometric conversion from F− to NaF.

Why the other choices are incorrect:
- a (0.83 mg): about half of the F− mass (1.67 mg) and ignores conversion to NaF; not consistent with volume/dilution.
- b (1.67 mg): equals the mg of F− required in 100 mL, not the mass of NaF; fails to convert F− → NaF.
- d (16.66 mg): equals the stock concentration in mg/L, not the mass needed in 100 mL; incorrect scaling.
- e (18.4 mg): far larger and inconsistent with stoichiometry and dilution; likely from a wrong MW ratio or wrong volume.

Conclusion: The required mass of NaF to prepare 100 mL of the stock solution is 3.68 mg, matching answer c.

类似问题

How much Isopropanol 90 % v/v is needed to prepare a 500 mL solution of Isopropanol 40% v/v in water?

How many mL of Glycerin would be needed to prepare 500 g of an ointment containing 5% w/w Glycerin ? Note: Density of Glycerin = 1.25g per mL

You are required to prepare 200 g of Sodium Benzoate 5% w/w in sorbolene cream from existing supplies of Sodium Benzoate 10% w/w (in sorbolene cream) and sorbolene cream. How much Sodium Benzoate 10% w/w cream is required?

How much Isopropanol 90 % v/v is needed to prepare a 500 mL solution of Isopropanol 40% v/v in water?

How many mL of a 1:60 stock solution of Ephedrine Sulphate would be needed to make 30 mL of a product to treat nasal congestion containing 0.25 % Ephedrine Sulphate?

What volumes of Ethanol 95% and distilled water should be mixed to produce 1000mL of Ethanol 70%? Part B) What volume of distilled water is required?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单