1. 18.9 grams of K2Cr2O7 are present in 50.0 mL of 0.360 M K2Cr2O7 solution. 2. The molarity of the (NH4)2SO4 solution formed from dissolving 4.28 grams of (NH4)2SO4 in 300 mL of water is 0.108 M. 3. 58.7 mL of 0.240M CuSO4 solution contains 2.25 grams of CuSO4 solute.

1. KEY
GENERAL CHEMISTRY-I (1411)
S.I. # 19
1. How many grams of solute are present in 50.0 mL of 0.360 M K2Cr2O7?
M=(mol/L)  mol = (MxL) (0.360 mol/1L)(0.05L)(294.2 g / mol) = 5.30 g K2Cr2O7
2. If 4.28 g of (NH4)2SO4 is dissolved in enough water to form 300 mL of solution,
what is the molarity of the solution?
(4.28 g)(1mol/132.2g) = 0.0324 mol / 0.3 L = 0.108 M (NH4)2SO4
3. How many milliliters of 0.240M CuSO4 contain 2.25 g of solute?
(2.25 g) (1 mol / 159.6 g) (1L / 0.240 mol)(1000 mL / 1L) = 58.7 mL solution
4. Which will have the highest concentration of potassium ion: 0.20 M KCl, 0.15 M
K2CrO4, or 0.080 M K3PO4?
KCl  K+ + Cl- ; 0.20 M KCl = 0.20 M K+
K2CrO4  2 K+ + CrO42- ; 0.15 M K2CrO4 = 0.30 M K+
K3PO4  3 K+ + PO43- ; 0.080 M K3PO4 = 0.24 M K+
0.15 M K2CrO4 has the highest K+ concentration.
5. Which will contain the greater number of moles of potassium ion: 30.0 mL of
0.15M K2CrO4 or 25 mL of 0.080M K3PO4?
K2CrO4: (0.30 M K+) (0.030 L) = 0.009 mol K+  greater
K3PO4: (0.24 M K+) (0.025 L) = 0.006 mol K+
6. Indicate the concentration of each ion or molecule present in the following
solutions:
a. 0.22 M NaOH
0.22 M Na+ and 0.22 M OH-
b. 0.16 M CaBr2
0.16 M Ca2+ and 0.32 M Br-
c. 0.15 M CH3OH
0.15 M CH3OH is a molecular solute
d. a mixture of 40.0 mL of 0.15M KClO3 and 35.0 mL of 0.22M Na2SO4.
M2 = M1V1/V2 where V2 is the total solution volume.
+
K : (0.15 M x 0.040 L) / (0.075 L) = 0.080 M
ClO3- : concentration ClO3- = K+ = 0.080 M
SO42-: (0.22 M x 0.0350L) / (0.075 L) = 0.1027 M SO42-
Na+ : concentration of Na+ = 2xSO42- = 0.21 M Na+

2. KEY
7. What mass of NaOH is needed to precipitate the Cd2+ ions from 25.0 mL of
0.500M Cd(NO3)2 solution?
Cd(NO3)2(aq) + 2NaOH(aq)  Cd(OH)2(s) + 2NaNO3(aq)
(0.025 L) (0.50mol Cd(NO3)2 / 1L) ( 2 mol NaOH / 1mol Cd(NO3)2)(40gNaOH/1mol) = 1g NaOH.
8. How many milliliters of 0.120M HCl are needed to completely neutralize 50.0 mL
of 0.101 M Ba(OH)2 solution?
2HCl(aq) + Ba(OH)2 (aq)  BaCl2(aq) + 2H2O (l)
[(0.101 mol Ba(OH)2 /1 L] (0.05 L Ba(OH)2 ) (2 mol HCl / 1 mol Ba(OH)2) x
(1L HCl / 0.120mol HCl) = 0.0842 L HCl soln.
9. How many milliliters of 0.125 M H2SO4 are needed to neutralize 0.200 g of
NaOH?
H2SO4 (aq) + 2NaOH (aq)  Na2SO4 (aq) + 2H2O (l)
(0.200 g NaOH) (1mol NaOH / 40.0 g ) (1mol H2SO4 / 2 mol NaOH) x
(1L H2SO4/0.125 mol H2SO4) = 0.0200 L or 20 mL H2SO4 soln.
10. If 55.8 mL of BaCl2 solution is needed to precipitate all of the sulfate ion in a
752-mg sample of Na2SO4, what is the molarity of the solution?
BaCl2 (aq) + Na2SO4 (aq)  BaSO4 (s) + 2NaCl (aq)
752 mg = 0.752 g Na2SO4
(0.752 g Na2SO4) (1 mol Na2SO4 / 142.1 g )(1mol BaCl2 / 1 mol Na2SO4) (1 / 0.0558L) =
0.0948 M BaCl2