According to the source below, the solubility of sulfanilamide in 95% ethyl alcohol at 78°C is 210 mg/mL. Since 0.1 g = 100 mg, we can set up a proportion:
(210 mg) / (1 mL) = (100 mg) / (x mL) Solving, x = 0.48 mL of 95% ethyl alcohol will be required.
I do not know previously the solubility of sulfanilamide in 95% ethyl alcohol. Let us accept the solubility you quoted here.
100/210 = 0.47619047619.. ≈ 0.48 (ml)
at 0C, the amount of sulfanilamide remains in the solution is: 14*(100/210) = 6.67 (mg), since you only have 0.48 ml solution.
The volume of the solution will change a little by cooling from 78C to 0C. You may also consider this volume change if you have data.
Answer:
The reason is that sodium attaches itself very strongly to other elements. Its compounds are very difficult to break apartand also because it is so reactive.
A line that stay s straight across the graph
To find the empirical formula you would first need to find the moles of each element:
58.8g/ 12.0g = 4.9 mol C
9.9g/ 1.0g = 9.9 mol H
31.4g/ 16.0g = 1.96 O
Then you divide by the smallest number of moles of each:
4.9/1.96 = 2.5
9.9/1.96 = 6
1.96/1.96 = 1
Since there is 2.5, you find the least number that makes each moles a whole number which is 2.
So the empirical formula is C5H12O2.
