Problem+Set+16

1. Solubility-Amount of a substance that will dissolve in a given amount of solvent Solution- a homogeneous mixture of two or more substances, may exist in any phase Unsaturated-for solutions: being able to dissolve more solute, for organic compounds: containing double or triple carbon-carbon bonds Saturated-the theoretical max. of stuff you can have in stuff Supersaturated- containing more solute than a saturated solution and therefore not in equilibrium (Matt C.)

2, Using a solubility graph answer the following: a. How many grams of lead(II) nitrate can dissolve in 100mL of water at 40 degrees Celsius? Using the graph, I easily found lead (II) nitrate which is Pb(NO3)2 I read the answer as 75 grams. (Sam Crowe)

b.How many grams of potassium chloride (KCl) can dissolve in 200 ml of water at 60 degrees Celsius? Using the graph to find how much grams that could be dissolved we found the answer was around 92 grams. Josef Niemann

c.Suppose 100mL of water at 20 degrees Celsius is saturated with calcium chloride. This solution is then allowed to cool to 0 degrees Celsius. How many grams of calcium chloride will precipate out of the solutiong? Using the graph again, I found the line for CaCl2 and found the amount in grams of calcium chloride will dissolve in 100mL of water at 0 degrees and 20 degrees Celsius. I read 76 grams at 20 degrees Celcius and 58 grams at 0. Then I subtracted 58 from 76 to get 18 grams of calcium chloride that would precipate from the solution. (Sam Crowe)

d. How many grams of sodium chloride can be dissolved in 200 mL of water at 60 degrees Celsius? Again, using the graph, we found that the amount of sodium chloride (NaCl) dissolved in 200 mL of water at 60 degrees Celsius was 18 g. (Jeremy Chin)

3. Molarity: The number of moles of a solute dissolved in one liter of a solution.

Percent Solution: The concentration of a solution in percent.

%(m/m): (mass of solute)/(mass of solution) X 100%

%(v/v): (volume of solute)/(volume of solution) X 100%

(Ethan Matlin)

5. Calculate Molarity a]2.00 moles of sodium chloride in 2.00 liters of solution. 2.00mol/2.00L = 1 M b] 35.0 grams of lead (II) nitrate in 1.25 liters of solution. 331.2 = molar mass of lead II nitrate 35.0g/331.2 = 0.106 mol 0.106mol/1.25L = 0.085 M c] 150.25 grams of potassium chloride in 750.0 milliliters of water. 74.551 = molar mass of KCl 150.25g/74.551 = 2.02 mol 2.02mol/0.75L = 2.69 M

(marielle billig)