1. Approximately how many mL of 5% BaCl2. 2H2O solution would be required to pre

Question

1. Approximately how many mL of 5% BaCl2. 2H2O solution would be required to precipitate all the sulfate if we assume that your sample are pur sodium sulfate? assume that the density of the barium chloride soultion is 1.00g/mL and that you should be using 50% excess precipitant?

2.If the sample were pure potassium sulfate would you require a smaller or larger volume of barium chloride solution than the amount calculated in 1 above? why?

these 2 questions are related, so I post together. thanks for helping!!!

Explanation / Answer

Answer: According to the given informations , molar mass of 244.34 g

Hecne the 5 % of 244.34 g is = 12.217 gram

and we have given the density = 1.00 g / ml = 1000 g / L

Hence the required volume is = mass / density = 12.217 / 1000 = 0.012217 L

means = 12.217 ml

This is all about the given question .

b] If the Potassium sulphate is used we required Low volume of barium chloride solution.

Thank you :)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.