63 g / 100 g H2O 50 g / 100 g H2O What is the solubility of potassium nitrate at 40ºC?  63 g / 100 g H2O How many grams of potassium chloride are needed to make a saturated solution at 80ºC? 50 g / 100 g H2O

What type of solution (saturated, unsaturated, or supersaturated) is 32 g of NH4Cl in 60 g of water at 50ºC? 32 = x 60 100 x = 53 g The solubility at 50ºC is 50g so the solution is supersaturated What is the mass percentage of NH4Cl at 10ºC? 33 g / 100 g H2O (33 g / 133 g )x 100 = 25%

47 = x 100 15 x = 7.1 g have 3 so need (7.1 – 3 ) = 4.1 g more. A solution has a 3 g KClO3 in 15 g H2O at 50ºC. How many more grams dissolve at 90ºC? 47 = x 100 15 x = 7.1 g have 3 so need (7.1 – 3 ) = 4.1 g more.

15.0 g x 1 mole = 0.0438 mole 342.3 g 0.0438 mole = 0.219 M Al2(SO4)3 What is the molarity of 15.0 g of Al2(SO4)3 in 200. ml solution? 15.0 g x 1 mole = 0.0438 mole 342.3 g   0.0438 mole = 0.219 M Al2(SO4)3 0.200 L

7. How many grams of Na2C2O4 are needed to make 80.0 ml of 0.90 M solution? 0.90 mole/L x 0.0800 L = 0.072 mole 0.072 mole x 134.0 g = 9.6 g Na2C2O4 1 mole

  How many grams of aluminum sulfate precipitate are made by the reaction of 60.0 ml of 0.54 M sodium sulfate solution with 100.0 ml of 0.080 M aluminum nitrate solution? Write the balanced equation and then show all work. 3 Na2 SO4 + 2 Al(NO3)3 → Al2(SO4)3 + 6 NaNO3

3 Na2 SO4 + 2 Al(NO3)3 → Al2(SO4)3 + 6 NaNO3 Find moles of each reactant using: molarity x volume (L) = moles 0.54 moles x 0.060 L = 0.032 moles Na2 SO4 L 0.080 moles x 0.100 L = 0.0080 moles Al(NO3)3 L

3 Na2 SO4 + 2 Al(NO3)3 → Al2(SO4)3 + 6 NaNO3 Next use mole ratios to find moles of product and grams of product. 3 Na2 SO4 = 0.032 mole 3x = 0.032 1 Al2(SO4)3 X x = 0.011 moles 2 Al(NO3)3 = 0.0080 x = 0.0040 moles 1 Al2(SO4)3 X Al2(SO4)3 0.0040 moles x (342.3g) = 1.4 g Al2(SO4)3 1 mole