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Alex
3 years ago
12

Calculate how much 95% ethyl alcohol will be required to dissolve 0.1 g of sulfanilamide at 78 c using volume calculate how much

sulfanilamide will remain dissolved after mixture is cooled to 0
Chemistry
1 answer:
Oduvanchick [21]3 years ago
6 0
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.
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2) S-H;

               E.N of Sulfur       = 2.58

               E.N of Hydrogen = 2.20

                                           ------------

               Difference               0.38


3) H-Br;

               E.N of Bromine   = 2.96

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               Difference              0.76


4) H-Cl;

               E.N of Chlorine   = 3.16

               E.N of Hydrogen = 2.20

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               Difference               0.96


Hence it is proved that the greatest electronegativity difference is found between H and Chlorine in H-Cl, therefore it is highly polar bond and vice versa.

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