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how much alum product would be lost to the crystallization solution if you had 42.5 ml of solution after filtration and the solubility of alum is approximately 2.63 g alum in 100 ml of 0 degrees celcius acidic water

Sagot :

Answer:

Total amount of alum lost = 0.5122 grams

Explanation:

Let the total volume of the solution be 100 mL

In 100 mL of solution, there is 2.63 gram of alum.

Out of this 100 mL solution, 42.5 mL is remaining.

Amount of alum in 42.5 mL solution is

[tex]\frac{42.5}{100} * 2.63 = 1.117\\[/tex] grams

Now the amount of alum lost is [tex]2.63 -1.117 = 0.5122[/tex] grams

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