Using the volume 1.42mL of 95% ethyl alcohol that's required to dissolve 0.3g at 78 C, how do i find how much sulfanilamide that will remain in the mother liquor after the mixture is cooled to 0 C? (at 0 degrees=14mg/mL is the solubility)
thanks
To find out how much sulfanilamide will remain in the mother liquor after the mixture is cooled to 0°C, we need to compare the solubility of sulfanilamide in ethyl alcohol at 78°C and 0°C.
1. First, calculate the amount of sulfanilamide that dissolved in the ethyl alcohol at 78°C:
Mass of sulfanilamide = 0.3g
2. Now, calculate the amount of sulfanilamide that dissolved in the 1.42mL of 95% ethyl alcohol at 78°C:
Volume of ethyl alcohol = 1.42mL
Concentration of ethyl alcohol = 95%
Mass of sulfanilamide dissolved = (0.3g / 1.42mL) * 95%
3. Next, compare the solubility of sulfanilamide at 78°C and 0°C:
Solubility at 78°C = unknown
Solubility at 0°C = 14mg/mL
Since we don't know the exact solubility at 78°C, we cannot directly calculate the amount of sulfanilamide that will remain. However, we can make an assumption based on the information given.
4. Assume that if the solubility of sulfanilamide at 78°C is greater than 14mg/mL, all the sulfanilamide will remain dissolved at 0°C. If the solubility at 78°C is less than 14mg/mL, some sulfanilamide will precipitate out as the mixture cools.
Therefore, if the solubility at 78°C is greater than 14mg/mL, the amount of sulfanilamide that will remain in the mother liquor at 0°C is 0g (since it all remains dissolved).
If the solubility at 78°C is less than 14mg/mL, the exact amount of sulfanilamide remaining cannot be determined without knowing the actual solubility at 78°C. However, you can assume that the remaining amount will be less than or equal to the amount of sulfanilamide initially dissolved at 78°C.
Please note that this is a theoretical explanation based on assumptions, and the actual result may vary depending on the specific solubility of sulfanilamide at 78°C.