Molarity (M), molality (m) and the Colligative properties
The molarity of a solution is the number of moles of solute per liter of solution. The symbol for molarity is M. Thus a 3.0 molar solution of nitric acid, abbreviated 3.0 M HNO3, contains 3.0 moles of HNO3 per liter of solution.
Sample problem #1: What mass of sodium hydroxide, NaOH, is required to prepare 3.00 liters of 0.25 M solution ?
Solution: given volume of solution and molarity of solution find the mass of solute.
3.00 liters 0.25 moles NaOH 40.0 g NaOH = 30.0 g NaOH
1 liter 1 mole NaOH
Sample Problem #2: What volume of 1.25 M NaOH can be prepared using 60.0g of sodium hydroxide?
Solution: given mass of solute and molarity of solution find the volume of solution
60.0 g NaOH 1 mole NaOH 1 liter soln = 1.2 Liters of soln
40.0 g NaOH 1.25 moles NaOH
Molarity Problems (moles / liter of solution)
1) Calculate the molarity, M, of the following solutions:
a) 1.5 moles of NaC2H3O2 dissolved in 750 ml of solution.
b) 3.00 moles of H2SO4 dissolved in 1250 ml of solution.
c) 66 grams of NaC2H3O2 dissolved in 500 ml of solution.
d) 68.4 g of glucose, C6H12O6 in 500 ml of water solution.
2) What is the molarity of a solution that contains 125 g CH3OH in 0.25 L of solution?
3) What is the molarity of a solution that contains 85.0 grams of Na2SO4 in 325 ml of solution?
4) What is the molarity of a solution that contains 210 grams of Al2(SO4)3 in 2.75 liters of solution?
Solve for mass:
5) Determine the grams of solute required to prepare the following solutions:
a) 1500 ml of 3M KOH
b) 750 ml of a 0.50 M solution of HC2H3O2
c) 2 liters of a 2.5 M solution of C6H12O6
d) 250 ml of a 1.25 M solution of NaCl
6) What mass of K3PO4 is required to prepare 4.00 liters of 1.50M solution?
7) What mass of CH3OH is required to prepare 1.50 liters of 3.00M solution?
Solve for volume:
8) What volume of 0.750 M solution can be prepared using 90.0 grams of NH4Cl?
9) If a 0.75 M solution of NaOH is to be prepared using 18.5 g NaOH, how many ml of solution can be produced?
10) What volume of 1.40 M HC2H3O2 solution contains 0.400 mole of HC2H3O2?
The molality of a solution is the moles of solute dissolved in 1 kilogram (1000g) of solvent (often water). The symbol for molality is lower case “m”. Thus a 3.0 molal solution of nitric acid, abbreviated 3.0 m HNO3, contains 3.0 moles of HNO3 in 1 kg of water.
MOLAL PROBLEMS (moles / kg of solvent )
11) How many grams of AgNO3 are needed to prepare a 0.125 m solution in 250 grams of water?
12) What mass in grams of sucrose, C6H12O6, must be dissolved in 2000 grams of water to make a 0.1 molal solution?
13) Determine the molality (m) of a solution containing 42 grams of glycerin, C3H5(OH)3 in 750 grams of water.
14) A solution contains 85.0 grams of methanol, CH3OH, in 3000 grams of water. Calculate the molality of the solution.
FREEZING POINT DEPRESSION AND BOILING POINT ELEVATION
(for water B.P. elevation const. = 0.512 oC/m, F.P. depression const. = 1.86oC/m)
15) A solution contains 15 grams of sucrose, C12H22O11, in 250 grams of water. What is the freezing point of the solution? (determine the molality first)
16) What is the boiling point of the solution in problem 15 above?
17) Calculate the boiling point and freezing point of a solution that contains 30.0 grams of acetic acid, HC2H3O2, dissolved in 250 grams of water.
18) Calculate the freezing point of a solution containing 5.70 grams of sugar, C12 H22O11, in 50 grams of water.
19) Calculate the boiling point of the sugar solution in problem 18 above.
20) Calculate the freezing point of a solution containing 60. grams of NaOH in 500 grams of water.
21) Calculate the boiling point of the solution in the above problem.
22) How many grams of ethanol, CH3OH, must be dissolved in 500 grams of water to lower the freezing point to -6.51oC?
Extra credit:
How many grams of ethylene glycol, C2H4(OH)2, must be dissolved in 200 grams of water to lower the freezing point to -29.75oC?
If the density of ethylene glycol is 1.12 g/ml, how many milliliters of ethylene glycol are required?
Answers to:
Molarity (M), molality (m) and the Colligative properties problems
Molarity Problems (moles / liter of solution)
1)Calculate the molarity, M, of the following solutions:
a) 1.5 moles of NaC2H3O2 dissolved in 750 ml of solution.
1.5 moles 1000 m l = 2.0 moles = 2.0 M NaC2H3O2
750 ml 1 liter liter
b) 3.00 moles of H2SO4 dissolved in 1250 ml of solution.
3.0 moles 1000 m l = 2.4 moles = 2.4 M H2SO4
1250 ml 1 liter liter
c) 66 grams of NaC2H3O2 dissolved in 500 ml of solution.
= 1.6 M H2SO4
d) 68.4 g of glucose, C6H12O6 in 500 ml of water solution.
=.76 M C6H12O6
2) What is the molarity of a solution that contains 125 g CH3OH in 0.25 L of solution?
15.6 Molar
3) What is the molarity of a solution that contains 85.0 grams of Na2SO4 in 325 ml of solution?
85g(1 mole Na2sO4/142.06g Na2SO$)= .598 mole Na2 SO4
Molarity= (.598/.325)= 1.84 M
4) What is the molarity of a solution that contains 210 grams of Al2(SO4)3 in 2.75 liters of solution?
.22 M
Solve for mass:
5) Determine the grams of solute required to prepare the following solutions:
a. 1500 ml of 3M KOH 252g
b. 750 ml of a 0.50 M solution of HC2H3O2 23g (22.5g is okay)
c. 2 liters of a 2.5 M solution of C6H12O6 900g
d. 250 ml of a 1.25 M solution of NaCl 18.125g
6) What mass of K3PO4 is required to prepare 4.00 liters of 1.50M solution?
1.50 m= 39 (3) +31+16(4)=212(4)/ L= 1272 g
7) What mass of CH3OH is required to prepare 1.50 liters of 3.00M solution? 144g
Solve for volume:
8) What volume of 0.750 M solution can be prepared using 90.0 grams of NH4Cl?
90g NH4Cl * 1 mol / 14+4+35.5=1.68mol. 1.68/.750=2.24L
9) If a 0.75 M solution of NaOH is to be prepared using 18.5 g NaOH, how many ml of solution can be produced? 616 ml
10) What volume of 1.40 M HC2H3O2 solution contains 0.400 mole of HC2H3O2?
.400 mole HC2H3O2 *(1 Liter/ 1.40 Mole) HC2H3O2 = .286 l HC2H3O2
MOLAL PROBLEMS (moles / kg of solvent )
11) How many grams of AgNO3 are needed to prepare a 0.125 m solution in 250 grams of water?
5.31 g AgNO3
12) What mass in grams of sucrose, C6H12O6, must be dissolved in 2000 grams of water to make a 0.1 molal solution? 36g
13) Determine the molality (m) of a solution containing 42 grams of glycerin, C3H5(OH)3 in 750 grams of water.
.61 m
14) A solution contains 85.0 grams of methanol, CH3OH, in 3000 grams of water. Calculate the molality of the solution. .89 m
FREEZING POINT DEPRESSION AND BOILING POINT ELEVATION
(for water B.P. elevation const. = 0.512 oC/m, F.P. depression const. = 1.86oC/m)
15) A solution contains 15 grams of sucrose, C12H22O11, in 250 grams of water. What is the freezing point of the solution? (determine the molality first) -0.325oC
16) What is the boiling point of the solution in problem 15 above? 100.09o C
17) Calculate the boiling point and freezing point of a solution that contains 30.0 grams of acetic acid, HC2H3O2, dissolved in 250 grams of water. 2.0m ;
1.0oC elevation to 101oC
3.72oC depression to -3.72oC
18) Calculate the freezing point of a solution containing 5.70 grams of sugar, C12 H22O11, in 50 grams of water.
-.62 degrees Celsius
19) Calculate the boiling point of the sugar solution in problem 18 above.
100.171 degrees Celsius
20) Calculate the freezing point of a solution containing 60. grams of NaOH in 500 grams of water.
1.5 mol NaOH/.5mol NaOH = 3m NaOH. -1.86 * 3 = -5.58. Therefore, freezing pt = -5.58 degrees C (-11.16oC when you recognize that NaOH breaks into 2 ions and the solution is 6m in parts)
21) Calculate the boiling point of the solution in the above problem.
3m * .512 degrees C/m + 100 = 101.54 degrees C (103.08oC when you recognize the two ions present in NaOH)
22) How many grams of ethanol, CH3OH, must be dissolved in 500 grams of water to lower the freezing point to -6.51oC? 56g CH3OH
Extra credit:
How many grams of ethylene glycol, C2H4(OH)2, must be dissolved in 200 grams of water to lower the freezing point to -29.75oC? 204.73 g
If the density of ethylene glycol is 1.12 g/ml, how many milliliters of ethylene glycol are required? 182.79 mL, (198.4g 177mL)
ANSWERS TO THE SOLUBILITY CURVE WORKSHEET
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ANSWERS TO THE SOLUBILITY CURVE WORKSHEET
1) NH3, and Ce2(SO4)3
2) Temperature; more
3) NaCl
4) 20g more
5) About 95oC
6) 10g more
7) 35 g
8) 200g; (5 x 40g)
9) KClO3
10) 10g
11) KNO3
12) 160g NaNO3
13) NH4Cl (55g/100g water)
14) About 5 grams
15) 300g H2O (85/100 = 255/x; x=300)
16) 50 g water (23/x = 46/100; x=50)
17) A) SATURATED
B) UNSATURATED
C) SUPERSATURATED