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Chemistry 1°

• To become familiar with colligative properties
• To be able to use these properties to determine the molar mass of a substance
• To determine the freezing-point depression of a naphthalene solution with a known concentration

A solution consists primarily of solvent and therefore, most of the solution’s properties reflect the solute’s properties. The physical properties that the solution and solute do not share are known as colligative properties and they depend solely on the solute concentration. Some of these properties include vapor pressure lowering, boiling-point elevation, freezing point lowering, and osmotic pressure. The solvent boils when the vapor pressure, or tendency of solvent molecules to escape, is equivalent to the atmospheric pressure. At this moment, the gaseous and liquid states of the solvent are in dynamic equilibrium and the molecules change from the liquid to the gaseous states and from the gaseous to liquid states at equal rates. The dissolution of a solute with very low vapor pressure, or a nonvolatile solute, raises the boiling point and lowers the freezing point. Similarly, anti-freeze lowers the freezing point and lowers the boiling point. The colligative-property law describes these effects, stating that the "freezing point and boiling point of a solution differ from those of the pure solvent by amounts that are directly proportional to the molal concentration of the solute" (Brown, 203-204). The colligative-property law can be expressed using the equation: D T = Km, where D T is the change in freezing or boiling point, K is a solvent-specific constant, and m is the solution’s molality.

1. Distinguish between solute and solvent.

Solute is the lesser component and solvent the greater component in a solution.

2. List three colligative properties and suggest a rationale for the choice of the word colligative to describe these properties.

The three colligative properties are boiling point, freezing point, and vapor pressure. They are called colligative properties because they are related to the number and energy of collisions between particles and not to what the particles are.

3. Distinguish between volatile and nonvolatile substances.

A volatile substance ha a high vapor pressure and a nonvolatile substance has a low vapor pressure at room temperature. Obviously, volatility is a relative term and depends upon temperature and pressure. Volatility increases with increasing temperature and decreases with increasing pressure.

4. What affect does the presence of a nonvolatile solute have upon:

1. the vapor pressure of a solution

A nonvolatile solute lowers the vapor pressure.

2. the freezing point

A nonvolatile solute lowers the freezing point.

3. the boiling point

1. What is the molality of a solution that contains 1.5 g urea (molecular weight = 60 amu) in 200 g of benzene?

(1.5 g urea)/( 60 g urea/mol urea) = .025 mol urea

.025 mol urea / .200 kg benzene = .125 m

2. What is supercooling? How can it be minimized?

Supercooling involves lowering of the temperature of a substance below its normal freezing point without the solidification of the substance. Supercooling can be minimized by cooling slowly with rapid stirring.

3. Calculate the freezing point of a solution containing 6.50 g of benzene in 160 g of chloroform.

Mole of Benzene: (6.50 g)(78 g/mol) = .083 mol

(.083 mol)/(.160 kg) = .520 molal

D T = (4.68ºC/m)(.520 molal) = 2.43ºC

-63.5C - 2.43ºC = 61.07ºC = 61.1ºC

 

 

4. A solution containing 1.00 g of an unknown substance in 12.5 g of naphthalene was found to freeze at 75.4°C. What is the molar mass of the unknown substance?

D T = 80.6ºC - 75.4ºC = 5.2ºC

5.2ºC = (6.9ºC/m)(x molal)

x = .75 molal

.75 molal = x mol / .0125 kg naphthalene

x = .0094 mol

1.00g / 0094 mol = 106.38 g/mol = 110 g/mol

5. How many grams of NaNO3 would you add to 250 g of H₂O in order to prepare a solution that is .200 molal in NaNO₃?

.200 molal = x mol / .250 kg H20

x = .0500 mol

(.0050 mol)/(85.00 g/mol) = .425g

6. Define the terms molality and molarity

wire stirrer

weighing paper

ring stand and ring

two-hole rubber stopper with slit

naphthalene

balance

1. Place a wire mesh on top of the ring attached to a ring stand.
2. On top of the wire mesh, place a 600 mL beaker.
3. Attach a test tube clamp above the ring on the ring stand.
4. Place a test tube in the clamp.
5. Attach a thermometer to a size five two-hole rubber stopper with slit by placing it in the slit.
6. In the other hole of the stopper, put a wire stirrer with a round hole at the bottom.
7. Place the stopper in the test tube.
8. A Bunsen burner should be placed under the wire mesh.
9. The apparatus should look as pictured below:

1. Weigh a large test tube to the nearest 0.01 g and record the mass in Data Table A.
2. Add about 15 to 20 g of naphthalene to the test tube and weigh again. Record the mass in Data Table A.
3. Insert the test tube into the beaker.
4. Make sure you have split your 2 hole size 5 rubber stopper Insert a thermometer into the hole that has been slit. Bend the stirrer so the loop encircles the thermometer.
5. Fill the 600 mL nearly full of water and heat till the water temperature is about 85 C.
6. After most of the naphthalene is melted, insert the stopper containing the thermometer and stirrer. Make sure the thermometer is not resting or touching the sides of the test tube.
7. Stop heating as soon as all the naphthalene melts.
8. Remove the test tube from the beaker and dry the outside off with a cloth towel.
9. Place the test tube in a wide mouthed bottle containing a piece of crumpled paper so that neither the bottle nor test tube break.
10. Using a timepiece and the thermometer measure the room temperature every 30 seconds and record it in Data Table A under "Cooling-Curve data".
11. Record the temperature until you have reached the freezing point, or when crystals start to form.

1. Measure about 1.2 g to 1.5 g of sulfur to the nearest 0.01 g using weighing paper. Record the mass in Data Table A. If you spill any sulfur, make sure you clean it up.
2. Replace the test tube in the water bath and heat until all the naphthalene has melted.
3. Gently remove the stopper making sure no naphthalene is lost. Carefully add the sulfur to the test tube.
4. Replace the stopper and stir gently until all the sulfur has dissolved.
5. Remove the test tube from the beaker and dry the outside off with a cloth towel.
6. Place the test tube in a wide mouthed bottle containing a piece of crumpled paper so that neither the bottle nor test tube break.
7. Using a timepiece and the thermometer measure the room temperature every 30 seconds and record it in Data Table A under "Cooling-Curve data".
8. Record the temperature until you have reached the freezing point, or when crystals start to form.

Clean-up

1. Cautiously heat the test tube in a water bath until the naphthalene melts.
2. Do not heat the thermometer beyond its temperature range. Also, take care because naphthalene is flammable.
3. Remove the stopper and remove the molten naphthalene onto a crumpled wad of paper.
4. When the naphthalene has solidified, throw both the paper and solid into the designated waste receptacle.
5. Make sure you do not pour the naphthalene in the sink.

1. Mass of test tube + naphthalene 48.590 g
2. Mass of test tube 42.585 g
3. Mass of naphthalene 16.005 g
4. Mass of paper + sulfur 1.835 g
5. Mass of paper .433 g
6. Mass of sulfur 1.402 g

Cooling-Curve data

Pure naphthalene Naphthalene & sulfur

Time Temp. Time Temp.

0	100		0	98
30	97		30	96
60	94		60	91
90	92		90	88
120	90		120	86
150	88		150	85
180	86		180	82
210	84		210	80
240	82		240	78
270	81		270	78
300	81		300	77
330	80		330	77
360	80		360	77
390	80		390	77
420	80		420	77
450	79		450	77
480	79		480	77
510	79		510	77
540	79		540	77
570	79		570	77
600	79		600	77
630	79		630	77
660	79		660	77
690	79		690	77
720	79		720	77
750	79		750	77
780	79		780	76
810	79		810	76
840	79		840	76
870	79		870	76
900	79		900	76
930	79		930	76
960	78		960	76
990	78		990	76
1020	78		1020	76
1050	78		1050	76
1080	78		1080	76
1110	78		1110	75
1140	77		1140	75
1170	77		1170	75
1200	76		1200	75
1230	76		1230	75
1250	75		1250	74
1280	75		1280	74
1310	74		1310	74
1340	74		1340	74
1370	74		1370	74
1400	73		1400	73
1430	71		1430	73
1460	69		1460	73
1490	68		1490	72
1520	66		1520	71
1550	63		1550	71
1580	61		1580	70
1610	59		1610	70
			1640	69
			1670	68
			1700	67
			1730	66
			1760	65
			1790	65
			1820	63
			1850	63
			1880	62

7. Freezing point of pure naphthalene, from cooling curve ___79°C____
8. Freezing point of solution of sulfur in naphthalene ___76°C____
9. Molality of sulfur or unknown (show calculations) ___.69 molal_

D T = 80.6ºC - 76ºC = 4..6º C = (6.9º C/molal) ( x molal)

x = .67 molal

.67 molal = x / .016005kg

x = .011 mol

.011 mol/ .016005 kg = .687 molal = .69 molal

10. Molar mass of sulfur (show calculations) __127 g/mol__

D T = 80.6ºC - 76ºC =

4..6º C = (6.9º C/molal) ( x molal)

x = .67 molal

.67 molal = x / .016005kg

1.402g/.011 mol = 127 g/mol

1. What are the major sources of error in this experiment

There are multiple sources of error in this experiment. Human error is the largest error. By not measuring substances accurately, the calculations could be a bit off. As well, by not stirring enough to prevent supercooling, the freezing point could have been affected.

2. Suppose your thermometer consistently read a temperature 1.2° lower than the correct temperature throughout the experiment. How would this have affected the molar mass you found?

The molar mass would not be affected because the change in temperature would always be the same.

3. If the freezing point of the solution had been incorrectly read 0.3° lower than the true freezing point, would the calculated molar mass of the solute be too high or too low? Explain your answer.

The molar mass would be lower because if the freezing point was 0.3° lower, then there would be a greater change in temperature, which would result in a larger molality and more moles. There would also be a smaller molar mass.

4. Arrange the following aqueous solution in order of increasing freezing point (lowest to highest temperature): 0.10m BaCl₂, 0.20m NaCl, and 0.20m Na₂SO₄.

0.10m BaCl2 - lowest amount of molal, lowest change

0.20m Na2SO4 - second lowest amount of molal, medium change

0.20m NaCl molal doubled, largest change

The larger the change in temperature, the lower the freezing point will be.

5. What mass of NaCl is dissolved in 150 g of water in a 0.050 m solution?

.050 m = x mol / 150 g H20

x = 7.5 mol

(7.5 mol)/(58.44 g/mol) = 438.3 g = 440 g

6. Calculate the molalities of some commercial reagents from the following data:

HCl HC₂H₃O₂ NH₃ (aq)

Formula weight (amu) 36.465 60.05 17.03

Density of solution (g/mL) 1.19 1.05 0.90

Weight (%) 37.2 99.8 28.0

Molarity 12.1 17.4 14.8

 

a) HCL

1.19 g

mL

1190 g

L

= 36.46g * .372 = 442.68 g/L

 

1190 g/L – 442.68 g/L = 747.32 g H₂ O

442.68 g

747.32 g

x

1000 g

= 592.28 g solute

592. 28 g	1 mol HCL
	36.46 g HCL

= 12.14 mol

 

12.14 g solute

.74732 kg solvent

m = 16.2 m HCl

 

b) HC₂H₃O₂

1.05 g

mL

1050 g

L

= 60.06g * .998 = 1047.9 g/L

 

1050 g/L – 1047.9 g/L = 2.1 g H₂ O

1047.9 g

2.1 g

x

1000g

= 499,000 g solute

499000 g solute

60.05 g/mol solvent

= 8310 M

 

8310 moles

1 kg

= 8310 m

c) NH₃

0.90 g

mL

900 g

L

= 17.04 g * .280 = 252 g/L

 

900 g/L – 252 g/L =648 g H₂ O

252g

648 g

x

1000g

= = 388.88 g solute

388.88 g	1 mol NH3
	17.04 g NH3

= 22.82 mol NH₃

22.82 moles

1 kg

= 22.8 m NH₃

 

 

7. A solution of 2.00 g of para-dicholorobenzene (a clothes moth repellant) in 50.0 g of cyclohexane freezes at 1.05 °C. What is the molar mass of this substance?

x = .27 molal

.27 m = x mol / .0500 kg cyclohexane

x = .014 mol

2.00 g/ .014 mol = 140 g/mol