Answer to Question #139368 in Organic Chemistry for Lyza

Question #139368

A certain compound has a solubility of 169 g/L in a solvent. If the mixture contains 50 g of

solute in a 300 cm3

of solution, is the solution unsaturated? If yes, how many grams of

solute should be added to make a saturated solution?

Expert's answer

If the solubility is 169 g/ L, then 300 cm³ are able to dissolve "\\frac{169g\\times300mL}{1000mL}=50.7g" of a solute.

So the solution is unsaturated. There should be 50.7 - 50 = 0.7 g of a solute added to make this solution saturated.

Answer: 0.7 g

Learn more about our help with Assignments: Organic Chemistry

Comments

Assignment Expert

21.10.20, 20:46

Dear Peter, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

Peter

20.10.20, 03:51

Unsaturated. Since we are given solubility of 169 g in 1000 ml, but we have 50 g in 300 ml, you need 295.9 ml to dissolve 50 g, but we have more. (50 x 1000 ml / 169 = 295.9 ml). You need an additional 0.7 g to make the solution in 300 ml with the same solubility as given ie. 169g/L. Hence, take 50.8 g in 300 ml. Hence the solution will be saturated since it will have the maximum number of grams that the solute can dissolve.

Answer to Question #139368 in Organic Chemistry for Lyza

Comments

Leave a comment

Related Questions