Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.