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!
Well, let's dive into this problem using a touch of humor!
First, we need to calculate the initial amount of sulfanilamide dissolved in the ethyl alcohol at 78 degrees Celsius. So, let's call this amount "Mr. Sulfanilamide."
We know that at 78 degrees Celsius, the solubility of sulfanilamide is 210 mg/mL. Since we have 1.429 mL of ethyl alcohol, we can multiply these two values together like so:
Mr. Sulfanilamide = 210 mg/mL * 1.429 mL
Now, let's convert this into grams because who doesn't love grams! There are 1000 milligrams in a gram, so:
Mr. Sulfanilamide (in grams) = 210 mg/mL * 1.429 mL / 1000
Now, we have the initial amount of sulfanilamide dissolved in our mother liquor. Next, let's consider what happens when the mixture cools down to 0 degrees Celsius. At this temperature, the solubility of sulfanilamide drops to 14 mg/mL.
To find out how much sulfanilamide remains dissolved, we subtract the amount that would precipitate (let's call this "Dr. Precipitation") from Mr. Sulfanilamide.
Dr. Precipitation = (210 mg/mL - 14 mg/mL) * 1.429 mL
Now, let's convert this amount to grams using our handy conversion factor:
Dr. Precipitation (in grams) = (210 mg/mL - 14 mg/mL) * 1.429 mL / 1000
Finally, to find out how much sulfanilamide remains dissolved, we subtract Dr. Precipitation from Mr. Sulfanilamide:
Remaining Sulfanilamide (in grams) = Mr. Sulfanilamide - Dr. Precipitation
And voila! You have calculated how many grams of sulfanilamide will remain dissolved in your mother liquor after cooling it down to 0 degrees Celsius. Just remember to take chemistry seriously, but not too seriously - a little humor can go a long way!
To calculate the amount of sulfanilamide that will remain dissolved in the mother liquor (ethyl alcohol) after cooling to 0.0 degrees Celsius, we can use dimensional analysis.
First, let's convert the initial volume of ethyl alcohol from milliliters (mL) to liters (L).
1.429 mL * (1 L / 1000 mL) = 0.001429 L
Next, let's calculate the initial amount of sulfanilamide dissolved in the ethyl alcohol at 78 degrees Celsius.
0.300 g * (1 mL / 210 mg) = 0.001429 L * (1 mL / 210 mg) = 0.00680595 moles
Now, we can calculate the final amount of sulfanilamide that will remain dissolved in the ethyl alcohol after cooling to 0.0 degrees Celsius.
0.00680595 moles * (14 mg / 1 mL) = 0.0000952833 moles
Finally, let's convert the final amount of sulfanilamide from moles to grams.
0.0000952833 moles * (121.16 g / mol) = 0.0115403 g
Therefore, approximately 0.0115 grams of sulfanilamide will remain dissolved in the mother liquor after the mixture has cooled to 0.0 degrees Celsius.
To calculate how many grams of sulfanilamide will remain dissolved in the mother liquor (ethyl alcohol) after the mixture has cooled to 0.0 degrees Celsius, we can use the solubility data provided.
First, we need to find out how many grams of sulfanilamide can be dissolved in both scenarios.
At 78 degrees Celsius:
Solubility = 210 mg/mL
Volume of ethyl alcohol = 1.429 mL
Using dimensional analysis, we can calculate the number of grams of sulfanilamide that can be dissolved at 78 degrees Celsius:
(210 mg/mL) x (1.429 mL) = 300.69 mg
Now we convert mg to grams:
300.69 mg x (1 g / 1000 mg) = 0.30069 g
So, at 78 degrees Celsius, 0.30069 grams of sulfanilamide can be dissolved in the mother liquor.
Next, we need to find out how many grams of sulfanilamide can be dissolved at 0.0 degrees Celsius:
Solubility = 14 mg/mL
Using dimensional analysis, we can calculate the number of grams of sulfanilamide that can be dissolved at 0.0 degrees Celsius:
(14 mg/mL) x (Volume of ethyl alcohol) = Mass of sulfanilamide
Now we substitute the volume of ethyl alcohol:
(14 mg/mL) x (1.429 mL) = 20.006 mg
Converting mg to grams:
20.006 mg x (1 g / 1000 mg) = 0.020006 g
So, at 0.0 degrees Celsius, 0.020006 grams of sulfanilamide can be dissolved in the mother liquor.
To find out how many grams of sulfanilamide will remain dissolved after cooling to 0.0 degrees Celsius, subtract the mass of sulfanilamide dissolved at 0.0 degrees Celsius from the initial mass of sulfanilamide (0.300 g):
0.300 g - 0.020006 g = 0.279994 g
Therefore, approximately 0.279994 grams of sulfanilamide will remain dissolved in the mother liquor after the mixture has cooled to 0.0 degrees Celsius.
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.