<u>Given:</u>
Initial concentration of potassium iodate (KIO3) M1 = 0.31 M
Initial volume of KIO3 (stock solution) V1 = 10 ml
Final volume of KIO3 V2 = 100 ml
<u>To determine:</u>
The final concentration of KIO3 i.e. M2
<u>Explanation:</u>
Use the relation-
M1V1 = M2V2
M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M
Ans: The concentration of KIO3 after dilution is 0.031 M
Answer:
Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:
[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.
So, when you compare with the solutiblity, 15 g of solute / 100 g of water, you realize that the solution has more solute dissolved with means that it is supersaturated.
To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.
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