Question

Calculate how much 95% ethyl alcohol will be required to dissolve 0.75 g of sulfanilamide at 78°c

Answer 1

Missing question: sulfanilamide has a solubility in 95% ethanol of 210 mg/ml at 78 degree Celsius.
m(sulfanilamide) = 0,75 g = 0,75 · 1000 = 750 mg.
210 mg : 1 ml = 750 mg : V(ethyl alcohol).
210 mg · V(ethyl alcohol) = 750 ml · mg.
V(ethyl alcohol) = 3,57 ml.
Answer is: 3,57 ml of 95% ethyl alcohol will be required.

Answer 2

The amount of 95% ethyl alcohol required to dissolve in 750 mg of sulfanilamide at 78°c is 3.57 mL

Sulfanilamide solubility in 95% ethyl alcohol (ethanol) is known to be 210 mg/ml.

Given that:

• Mass of sulfanilamide = 0.75 g  

= 0.75 × 1000 mg

= 750 mg

• Temperature = 78° C

Thus;

If 1 mL ethyl alcohol dissolves in 210 mg/ml

In 95% ethyl alcohol, the solubility of sulfanilamide concentration in mass 750 mg can be computed as follows:

1 mL = 210 mg/mL

x mL = 750 mg

∴

$\mathbf{x (mL) = \dfrac{750 \ mg \times 1 \ mL}{210 \ mg/mL}}$

x = 3.57 mL

Therefore, we can conclude that the amount of 95% ethyl alcohol required to dissolve in 750 mg of sulfanilamide at 78°c is 3.57 mL

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