I have already calculated the volume of ethyl alcohol need to dissolve 0.300 g of sulfanilamide at 78 degrees Celsius. I got 1.429 mL ethyl alcohol. My question is: how do I calculate how many grams of sulfanilamide will remain dissolved in the mother liquor (ethyl alcohol) after the mixture has cooled to O.O degrees Celsius? Given that the solubility of sulfanilamide at 78 degrees Celsius is 210 mg/mL, and at 0.0 degrees Celsius is 14 mg/mL. This is for Pre-Lab calculations. Can anyone show me the dimensional analysis involved? Thank you!

asked by Sasha

9 years ago

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1 answer

For the first one, which you have done correctly, it is
210 mg/mL x ? mL = 300 mg
? mL ethanol - 1.429 mL needed.

When it cools to zero C, the solubility is
14 mg/mL x 1.429 mL = ? mg dissolved.

answered by DrBob222

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Question

I have already calculated the volume of ethyl alcohol need to dissolve 0.300 g of sulfanilamide at 78 degrees Celsius. I got 1.429 mL ethyl alcohol. My question is: how do I calculate how many grams of sulfanilamide will remain dissolved in the mother liquor (ethyl alcohol) after the mixture has cooled to O.O degrees Celsius? Given that the solubility of sulfanilamide at 78 degrees Celsius is 210 mg/mL, and at 0.0 degrees Celsius is 14 mg/mL. This is for Pre-Lab calculations. Can anyone show me the dimensional analysis involved? Thank you!

1 answer

For the first one, which you have done correctly, it is
210 mg/mL x ? mL = 300 mg
? mL ethanol - 1.429 mL needed.

When it cools to zero C, the solubility is
14 mg/mL x 1.429 mL = ? mg dissolved.

Ask a New Question or answer this question.

DrBob222 · Answer

For the first one, which you have done correctly, it is 210 mg/mL x ? mL = 300 mg ? mL ethanol - 1.429 mL needed. When it cools to zero C, the solubility is 14 mg/mL x 1.429 mL = ? mg dissolved.