Unit 3
Wkst 5 - Concentration Units
KEY
On a separate sheet of paper, answer the following:
1.
Write the equation for the dissociation in water of Al(NO3)3.
Answer:
Al(NO3)3(s) → Al3+(aq) + 3 NO3–(aq)
2.
How many moles of ions are in solution when one mole of cobalt(III) sulfate dissolves?
Answer:
Co2(SO4)3 = 5 ions
Therefore there are 5 moles of ions in one mole of Co2(SO4)3.
3.
Calculate the mass percent of water in a solution made up of 123 g of NaOH and 289 g of
water.
Answer:
289gH2O
100  70.8%
289gH2O 123gNaOH
4.
Methyl mercuric chloride, CH3HgCl, is a toxic pollutant usually found in waters around
industrial sides. If a sample of 250.0 mL of water (d = 1.00 g/mL) is found to contain 0.200
mg of CH3HgCl, what is the concentration of methyl mercuric chloride in parts per million?
Answer:
250.0mLH 2O  250.0gH2O

0.000200gCH3 HgCl
10 6  0.800 ppm
250.0gH2O  0.000200gCH3 HgCl
5.

Calculate the mole fractions of tertiary butyl alcohol, C4H9OH, and water, H2O, in a solution
made by adding 100. g of tertiary butyl alcohol to 50.0 g of water.
Answer:
100.gC4 H 9OH 1mol
 1.35molC4 H 9OH
74.1g
50.0gH 2O 1mol
 2.77molH2O
18.02g
 C4 H 9 OH 

1.35molC4 H 9OH
 0.327
1.35molC4 H 9OH  2.77molH2O
 H 2 O  1 0.327  0.673
6.
Calculate the molality of a solution made by adding 5.00 g of ethanol, C2H5OH, to 100.0 g of
water H2O.
Answer:
5.00gC2 H 5OH 1mol
 0.108molC2 H5OH
46.1g
100.gH 2O  0.100kgH 2O
0.108molC2 H5OH
m
 1.08m
0.100kgH 2O
7.


8.
How would you prepare 100.0 mL of a 0.200 M aqueous solution of NaOH from a bottle that
is labeled 0.4871 M?
Answer:
(0.1000L)(0.200M)  (0.4871M)V
V  0.0411L
Add 41.1 mL of the 0.4871 M NaOH to a 100.0 mL volumetric flask. Then, add enough
DI water to fill the flask to the 100.0 mL mark.
Calculate the molarity of a solution made by dissolving 22.0 g of hydrochloric acid HCl, in
88.0 g of water, H2O. The density of the resulting solution will be 0.990 g/mL.
Answer:
22.0 g HCl  1mol 

  0.603mol HCl
 36.46 g 
88.0 g H 2O  22.0 g HCl  110.0 g solution
110.0 g solution  1mL 

  111.1mL solution  0.1111L solution
 0.990 g 
0.603mol HCl
 5.43M
0.1111L solution
9.
A solution is prepared by adding 0.200 L of a 1.00 M aqueous solution BaCl2 to 0.300 L of a
1.00 M aqueous solution of NaCl. Calculate the final mol concentrations of Na+, Ba2+, and
Cl–, assuming that both initial solutes are strong electrolytes.
Answer:
If these salts are strong electrolytes, then they will dissociate 100%.
BaCl2(s) → Ba2+(aq) + 2 Cl–(aq)
NaCl(s) → Na+(aq) + Cl–(aq)
(0.200L)(1.00M)  0.200molBaCl2  0.200molBa 2  2(0.200molCl )
(0.300L)(1.00M)  0.300molNaCl  0.300molNa   0.300molCl
Vtotal  0.200L  0.300L  0.500L
0.200molBa 2
 0.400M
0.500L
0.300molNa 
Na  
 0.600M
0.500L
2(0.200molCl )  (0.300molCl ) 0.700molCl
Cl  

 1.40M
0.500L
0.500L
Ba 2 
10. An aqueous solution contains sucrose, C12H22O11, in water. The solution has a total mass of
738 g and the mass percent of sucrose in the solution is 38.4%. Calculate the molality of the
 solution.
Answer:
xgC12H22O11
38.4% 
100
738gC12H 22O11  H2O
x  283gC12H22O11
283gC12H22O11
1mol
 0.827molC12H22O11
342.34g
(738gC12H22O11  H2O)  283gC12H22O11  455gH2O  0.455kgH2O
0.827molC12H22O11
 1.82m
0.455kgH 2O
11. What is the mole fraction of bromine in a 0.40 molal solution of Br2 in carbon tetrachloride,
CCl4, solvent?
 Answer:
0.40molBr2
Assume0.40mBr2 
1kgCCl4
1kgCCl4  1000gCCl4
1000gCCl4
 Br2 
1mol
 6.50molCCl4
153.81g
0.40molBr2
 0.058
0.40molBr2  6.50molCCl4
12. A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final
volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molality of
 ethanol in this solution.
Answer:
1.00gC2 H5OH 1mol
 0.0217molC2 H5OH
46.08g




100.0gH2O 1mol
 5.549molH2O
18.02g
Molarity:
0.0217molC2 H 5OH
 0.215M
0.101Lsolution
MassPercent :
1.00gC2 H5OH
100  0.990%
1.00gC2 H5OH  100.0gH2O
MoleFraction:
0.0217molC2 H5OH
 C2 H 5 OH 
 0.00390
0.0217molC2 H5OH  5.549molH2O
Molality:
0.0217molC2 H5OH
 0.217m
0.100kgH2O
13. A 1.37 M solution of citric acid (H3C6H5O7) in water has a density of 1.10 g/mL. Calculate
the mass percent, molality, and mole fraction of the citric acid.

Answer:
1.37molH3C6 H 5O7 1.37molH3C6 H 5O7
Assume1.37MH 3C6 H 5O7 

1LH3C6 H 5O7  H 2O
1Lsolution
1Lsolution  1000mLsolution
1000mLsolution 1.10g
 1100gsolution  1100gH 3C6 H 5O7  H 2O
1mL
1.37molH3C6 H 5O7 192.14g
 263gH 3C6 H 5O7
1mol
(1100gH 3C6 H 5O7  H 2O)  263gH 3C6 H 5O7  837gH 2O




837gH 2O 1mol
 46.4molH2O
18.02g
MassPercent :
263gH3C6 H5O7
100  23.9%H3C6 H5O7
(1100gH3C6 H5O7  H2O)
Molality:
1.37molH3C6 H5O7
 1.64mH 3C6 H5O7
0.837kgH2O
MoleFraction:
1.37molH3C6 H5O7
 H 3 C6 H 5 O7 
 0.0286
1.37molH3C6 H5O7  46.4molH2O
14. Calculate the mole fraction of HCl in an aqueous solution of hydrochloric acid containing 36
percent HCl by weight.
Answer:
36gHCl
Assume 36% 
100
100gHCl  H 2O
36gHCl 1mol
 0.99molHCl
36.46g
(100gHCl  H 2O)  36gHCl  64gH 2O
64gH 2O 1mol
 3.6molH2O
18.02g
 HCl 
0.99molHCl
 0.22
0.99molHCl  3.6molH2O
15. Given that the density of a solution of 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) is
0.876 g/mL, calculate the concentration of the solution in molarity and mass percentage of
 solute.
Answer:
5.0gC7 H 8 1mol
 0.054molC7 H 8
92.15g
5.0gC7 H 8  225gC6 H 6  230.gsolution



230.gsolution 1mL
 263mLsolution
0.876g
Molarity:
0.054molC7 H 8
 0.206MC 7 H8
0.263Lsolution
MassPercentage:
5.0gC7 H8
100  2.2%C7 H8
230.gsolution