Milwaukie HS Chemistry Linman. Period Date / /

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1 Milwaukie HS Chemistry Linman A701 Name Solutions Classify the following mixtures as homogeneous or heterogeneous: A: Freshly squeezed orange juice B: Tap water C: Human Blood D: Sand 1. Does a solution have to be a liquid mixture? Explain why or why not. 2. Name the solute and the solvent in aqueous sodium chloride solution, NaCl(aq). 3. How is distillation used to prepare pure water from tap water 4. Why are gas bubbles not visible in an unopened bottle of soda? 5. Explain what happens to gases dissolved in the blood when a diver returns to the water s surface very quickly 6. How does rising temperature affect the solubility of gases in water? 7. Why do bubbles escape from soda that has just been opened? 8. Name two factors that affect solubility.

2 Milwaukie HS Chemistry Linman A702 Name Solutions: Molarity and Solubility 1. What is the molarity of solution made from the equivalent of 125 g of sodium nitrate in 1.0 L of solution? Answer: 2. What final volume in liters of a 2.5 M solution of sodium nitrate can be made using 25.0g of solute? Answer: 3. What mass in grams of sodium nitrate, is needed to make 85.0 ml of a M solution? Answer: 4. A saturated sodium nitrate solution at 10 o C will have how many grams of sodium nitrate dissolved in 100 grams of water? Use Table a. Answer: 5. If there are 90g of sodium nitrate are in the container at 10 o C, then how many grams of sodium nitrate will remain undissolved in 100 grams of water? Use Table a. Answer: 6. Raising the temperature of the mixture from 5 o C to 20 o C will change the amount of dissolved sodium nitrate by how much? Use Table a. Answer: 7. What must the temperature be in order to dissolve the entire 85g sodium nitrate can be dissolved? Use Table a. Answer: Molarity(M) = Moles of Soute/Liter of Solution / Molality (m) = Moles of Solute/kilogram solvent 1. Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water? 2. What is the molarity of g of H 2 SO 4 dissolved in 1.00 L of solution? 3. What is the molarity of 5.30 g of Na 2 CO 3 dissolved in ml solution? 4. What is the molality of 5.00 g of NaOH in g of water? 5. How many moles of Na 2 CO 3 are there in 10.0 L of 2.0 M soluton? 6. How many moles of Na 2 CO 3 are in 10.0 ml of a 2.0 M solution?

4 Milwaukie HS Chemistry Linman A703 Name Solutions: Colligative Properties (FPD/BPE) 1. A saturated KCl solution at 10 o C will have how many grams of KCl dissolved in 100 grams of water? Answer: 2. Raising the temperature of the mixture to 30 o C will increase the amount of dissolved KCl by how much? Answer: 3. What would the boiling point of a 0.75 m BeF 2 solution be? K b (H 2 O) = C/m. Answer: 4. If I add 45 grams of sodium chloride to 500 grams of water, what will the melting and boiling points be of the resulting solution? K b (H 2 O) = C/m and K f (H 2 O) = C/m. Answer: 5. What is the formula mass for a 1.33 g sample of solute in g of water if the solution freezes at a temperature of C. 6. Calculate the freezing and boiling points to a solution made using the following data: 24.92g AlCl 3 in 470.5g water 7. In general, will the addition of an electrolytee have a greater effect on the colligative properties of a solution than the addition on a nonelectrolyte will? Explain why or why not.

5 Milwaukie HS Chemistry Linman L701 Name Mass by Freezing Point INTRODUCTION: The vapor pressure of a pure liquid at a given temperature is a characteristic property of that liquid. However, when a nonvolatile solute is dissolved in the liquid, the vapor pressure of the liquid is reduced. This lowering of the vapor pressure causes a change in the melting point, boiling point, and osmotic pressure of the liquid. The magnitude of the change in these properties depends upon the number of solute particles dissolved in a given amount of the solvent, but not upon the nature of the particles (their identity). Such properties are called colligative properties. The addition of ethylene glycol to the water in a car radiator in order to raise its boiling point or the use of salt to lower the melting point of ice on a sidewalk are some everyday applications of colligative properties. In this experiment we will investigate the phenomenon of freezing-point depression and determine the molar mass of an unknown solute. The relationship between the lowering of the freezing point and the concentration of a solution is given by the following: ΔT f = T f (pure solvent) - T f (solution) = K f m (1) where T f is the freezing temperature, K f is the molal freezing-point depression constant (a property of a given solvent), and m is the molality of solute in the solution. Molality of a solute is defined as follows: m = moles of solute / kg of solvent (2) One application of freezing-point depression is in the determination of the molar mass of an unknown solute. A weighed amount of the solute is dissolved in a known mass of solvent. The freezing point of the solvent (the temperature at which solid and liquid phases are in equilibrium) is determined by cooling the solution and plotting a graph of temperature as a function of time. A horizontal portion of the graph indicates that a pure liquid will freeze at a constant temperature. However, a solution will freeze over a range of temperatures and will not exhibit the constant freezing point of the pure liquid; instead of exhibiting a horizontal portion, the plot will show a change of slope when solid solvent begins to form. The concentration of dissolved solute then steadily increases as the solvent freezes, causing the freezing point to continually decrease. (The solid forming is pure solvent.) The temperature at which freezing begins is determined by the intersection of the two extrapolated straight lines. In practice, it is necessary to first determine the freezing point of the solvent alone, since the solvent may not be pure. This step also makes prior calibration of the thermometer unnecessary. With the change in freezing point and the value of K f, it is then possible to calculate the approximate molar mass of the solute. The phenomenon of supercooling occurs when a liquid cools below its freezing point without crystallizing. A supercooled liquid is in an unstable condition, and any disturbance such as vibration will cause crystallization to begin, with a consequent rise in temperature to the actual freezing point. The dip in the graph due to supercooling should be ignored and the two straight portions extrapolated to their intersection to find the freezing point. EQUIPMENT:: Thermometer 600 ml plastic beaker 50 ml flask MATERIALS: Ice distilled water rock salt methanol(ch3oh) PROCEDURE: Part Place 50.0 ml of distilled water in a 50-ml flask. Place about 250 ml of crushed ice in a 600-ml plastic beaker. Set the distilled water flask on the ice. Add more ice around the flask. On top of the ice layer about 50 ml of rock salt. Avoid getting any ice or salt in the flask. 2. Rotate the flask of distilled water to speed up the freezing process. When some ice has formed on the flask, measure the temperature of the distilled water. Record the freezing point of the distilled water in the date table.

6 Part Dry the flask. Place the flask on the balance and weight the flask on the balance. Record the mass of the flask in the data table. Add 50.0mL of distilled water to the flask and weight the flask containing the water record the mass in the data table. Add between 1.00g and 1.10g of methanol to the flask. Weigh and record in the mass data table. Freeze the solution in the same manner as the distilled water. Record the freezing point of the solution in the data table. Part Mass Empty Flask Mass Flask+Water Mass Flask+Water+Methanol Freezing Temperature CALCULATIONS AND CONCLUSIONS: 1. Calculate the freezing point depression mass of your solute using the following equations. Answer: 2. Calculate the chart mass of your solute. Answer: 3. Calculate the experimental error. E is from problem #1A is from problem #2 Answer: 4. Why is it important to determine the freezing point of your solvent? 5. Why must salt be placed in a hand-crank ice cream freezer? 6. What are some sources of error in this lab? 7. What is the formula mass for a 1.33 g sample of solute in g of water if the solution freezes at a temperature of C? Answer: 8. Why will salt on an icy sidewalk make the sidewalk safer?

7 Milwaukie HS Chemistry Linman Name L702 Gravimetric Copper (II )Chloride OBJECTIVE:.Experimentally determine the molar concentration of copper(ii) chloride solution. MATERIALS:. three ungalvanized nails, 8d balance one 18mmx185mm test tube test tube clamp copper(ii)chloride solution HAZARDS:.Avoid skin contact when using Copper (II) Chloride solution. INTRODUCTION: Molar concentration is defined as the number of moles of solute dissolved in a liter of solution, M=n/V. You have calculated the molar concentration of solutions based on the number of grams of solute dissolved in a given quantity of solution. In this lab you will measure out a given volume of copper(ii) chloride solution. Therefore, you will have the volume (V) of the copper(ii)chloride that you will using. However, you will not have the number of moles(n) of copper(ii) chloride that is in the solution. You must experimentally determine the number of moles by reaction the copper(ii) chloride with a common iron nail. EXAMPLE CALCULATION: A 14.9ml sample of iron(iii) chloride in a large test-tube was reacted with three clean copper nails having a total mass of 25.91grams. The nails and iron(iii) chloride were agitated to speed up reaction. After the reaction, the nails were cleaned, dried and weighed. The mass after reaction was recorded as 24.02grams. a) Calculate the mass change of the copper nails g g 1.89g b) Calculate the moles of copper consumed. 1.89g 63.5 g/mol=0.0298mol c) Calculate the number of moles of iron(iii)chloride that reacted with the copper. 2FeCl 3 + Cu 2FeCl 2 + CuCl 2 2FeCl molCu x = FeCl 3 mol 1m Cu d) Calculate molar concentration. M=n/V M=( FeCl 3 mol)/(0.0149l) M=4.00M FeCl 3 PROCEDURE 1). Clean three new, ungalvanized, 8d, iron nails until they shine. 2). Mass the three nails to the nearest 100 th of a gram and record the mass in the table. 3). Obtain about 10 to 15 ml of copper(ii) chloride solution; record the volume to three significant digits:. 4). Place the nails in the test tube, pour the copper(ii) chloride solution in with the nails. 5). Continuously swirl the test tube: this will keep fresh solution near the surface of the nails. 6). Continue swirling until the solution is no longer blue-green. 7). Pour the solution into a small beaker. Rinse and clean off all of the plated copper on the nail, using a paper towel. 8). Dry and weigh the nails. Record this weight in the table. 9). Pour the solution into the labeled recycle beaker. DO NOT POUR DOWN DRAIN!! 10). Put used nails in beaker labeled "used nails". Before After DATA TABLE Mass of nails

8 CALCULATIONS & QUESTIONS: 1) Write the balanced chemical reaction of solid iron reacts with copper(ii) chloride solution to form iron(ii) chloride solution and copper solid. 2) Calculate the mass change of the iron nails. 3) Calculate the moles of iron consumed. 4) Calculate the moles of Copper (II) Chloride that reacted with the iron. 5) Calculate the molar concentration of the Copper (II) Chloride solution. 6) If you had added water to your Copper (II) Chloride solution prior to adding the nails, would this have altered your reaction? How? (Explain)

9 Milwaukie HS Chemistry Linman A704 Name Molarity, Molality and 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 HNO 3, contains 3.0 moles of HNO 3 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 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 NaC 2 H 3 O 2 dissolved in 750 ml of solution. b) 3.00 moles of H 2 SO 4 dissolved in 1250 ml of solution. c) 66 grams of NaC 2 H 3 O 2 dissolved in 500 ml of solution. d) 68.4 g of glucose, C 6 H 12 O 6 in 500 ml of water solution. 2) What is the molarity of a solution that contains 125 g CH 3 OH in 0.25 L of solution? 3) What is the molarity of a solution that contains 85.0 grams of Na 2 SO 4 in 325 ml of solution? 4) What is the molarity of a solution that contains 210 grams of Al 2 (SO 4 ) 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 HC 2 H 3 O 2 c) 2 liters of a 2.5 M solution of C 6 H 12 O 6 d) 250 ml of a 1.25 M solution of NaCl 6) What mass of K 3 PO 4 is required to prepare 4.00 liters of 1.50M solution? 7) What mass of CH 3 OH is required to prepare 1.50 liters of 3.00M solution? Solve for volume: 8) What volume of M solution can be prepared using 90.0 grams of NH 4 Cl? 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 HC 2 H 3 O 2 solution contains mole of HC 2 H 3 O 2?

10 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 HNO 3, contains 3.0 moles of HNO 3 in 1 kg of water. MOLAL PROBLEMS (moles / kg of solvent ) 11) How many grams of AgNO 3 are needed to prepare a m solution in 250 grams of water? 12) What mass in grams of sucrose, C 6 H 12 O 6, 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, C 3 H 5 (OH) 3 in 750 grams of water. 14) A solution contains 85.0 grams of methanol, CH 3 OH, in 3000 grams of water. Calculate the molality of the solution. FREEZING POINT DEPRESSION AND BOILING POINT ELEVATION (for water B.P. elevation const. = o C/m, F.P. depression const. = 1.86 o C/m) 15) A solution contains 15 grams of sucrose, C 12 H 22 O 11, 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, HC 2 H 3 O 2, dissolved in 250 grams of water. 18) Calculate the freezing point of a solution containing 5.70 grams of sugar, C 12 H 22 O 11, 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, CH 3 OH, must be dissolved in 500 grams of water to lower the freezing point to o C? Extra credit: How many grams of ethylene glycol, C 2 H 4 (OH) 2, must be dissolved in 200 grams of water to lower the freezing point to o C? If the density of ethylene glycol is 1.12 g/ml, how many milliliters of ethylene glycol are required?

Name Date Class PROPERTIES OF SOLUTIONS

Name Date Class PROPERTIES OF SOLUTIONS 16.1 PROPERTIES OF SOLUTIONS Section Review Objectives Identify the factors that determine the rate at which a solute dissolves Identify the units usually used to express the solubility of a solute Calculate