Chemistry I-Honors
Solution Concentration Problem Set #3
Solution Set
1.When 25.0 grams of beryllium hydroxide is added to 5.00 liters of a 0.265 M solution of phosphoric acid, how many equivalents of the acid are used in the reaction?
Be(OH)2 + H3PO4
(25.0 g / 43.0 g/mol) (5.00 liters)(0.265 moles/liter)
= 0.581 moles cpd = 1.33 moles cpd
= 1.16 moles OH-1 = 3.98 moles H+ Base is the L.R, so only 1.16 moles of H+ react.
2.How many equivalents are there in 7.00 grams of chromium (III) oxide if the compound is completely reduced to oxygen gas and the elemental metal?
+6 0
(ie. Cr2O3 ------> 2 Cr(s) + 3O2 - no need to balance)
(7.00 g)( 1 mol cpd / 152.0 g cpd)(6 moles e-1 / 1 mole cpd) = 0.276 eq
3.If 3.35 grams of KOH are required to neutralize 55.0 ml of a solution of telluric acid, what was the molarity of the original acid solution? H2TeO4
3.35 g KOH / 56.1 g/mol = 0.0597 mol OH-1 = 0.0597 mol H+1
= 0.0299 mol H2TeO4 / 0.0550 liter = 0.543 M
4.When 400.0 ml of a 3.55 N solution phosphoric acid is diluted with 500.0 ml of water, what is the molarity of the final, diluted solution? N1V1 = N2V2
N2 = ( 3.54 N)(400.0 ml) / 900.0 ml = 1.57 N
1.57 eq / 1 liter x 1 mol / 3 eq = 0.524 M
5.What is the molecular weight of a compound if 6.00 grams of the compound, when dissolved in 45.0 grams of water, produces a 0.615-molal solution?
0.615 m = 6.00 g / M.W. / 0.0450 kg
M.W. = 217 g/mol
-2-
6.What is the mole fraction of the solute if 67.0 grams of lithium fluoride is dissolved in 100.0 grams of water?
(100.0 g / 18.0 g/mol)5.56 mol
Xwater = ------= ------= 0.682
(100.0 g / 18.0 g/mol) + (67.0 g / 25.9 g/mol) 8.15 mol
Then XLiF = 1- 0.682 = 0.318
7.What is the density of a sodium chloride solution that is 3.45 N and is 12.0% by weight of solute?
3.45 N = 3.45 M
100 g soln 58.5 g NaCl 3.45 mol NaCl
------x ------x ------= D = 1.68 g / ml
12.0 g NaCl 1 mol NaCl 1000 ml soln
8.What is the molarity of a solution if 333.5 grams of sodium phosphate is dissolved in enough water to produce 950.0 ml of solution?
(333.5 g / 164.0 g/mol)
------= 2.141 M
0.9500 liter
9.How many grams of calcium sulfate must be weighed out in order to end up with 67.0 grams of an aqueous solution that is 32.8% by weight?
(67.0 g soln)( 32.8 g CaSO4 / 100 g soln) = 21.976 g = 22.0 g CaSO4
10.How much water is needed if dissolving 50.0 grams of potassium chloride produces a 2.28-molal solution?
(50.0 g KCl )( 1 mol KCl / 74.6 g KCl )( 1 kg H2O / 2.28 mol KCl ) = 0.294 kg H2O
= 294 grams H2O