118 ChemQuest 39 Name: Date: Hour: Information: Molarity Concentration is a term that describes the amount of solute that is dissolved in a solution. Concentrated solutions contain a lot of dissolved solute, but dilute solutions contain only a little. 1. Consider the terms "concentrated" and "dilute". Are these qualitative or quantitative terms? These are qualitative terms: very general and do not include the magnitude or quantity.. One way of quantitatively measuring solution concentration is with units of molarity, symbolized by M. You see 1.7 liters (L) of a sodium chloride and water solution. The label on the bottle reads "1.5 M ". You don't know what molarity is, but you decide to find out. After evaporating the water out of the solution you discover that there are about 149 grams of salt. Using this information, which of the following formulas is/are correct for finding molarity? grams of solute moles of solute A) Molarity B) Molarity moles of solute liters of solution 149g 58.5g/mol.547 mol 1.5.547 mol 1.7 L 3. Using the equation you discovered in question two, calculate the molarity of each of the following solutions. A) A solution is prepared by dissolving 4.9 g of CaCl in 10 ml (which 0.10 L) of solution. 1.07 M Change g to mol: 4.9g 111.1g/mol 0.4 mol Molarity mol L 0.4 0.10 1.07 M B) A solution contains 1.9 g of Na SO 4 in 35 ml of solution. 0.79 M Change g to mol: 1.9 14.1g/mol 0.09078 mol Molarity mol L 0.09078mol 0.35L 0.79 M 4. Verify that I need.15 moles of Ca(NO 3 ) to make 358 ml of a 6.00 molar solution. mol (Molarity)(Liters) (6.00M)(0.358L).15 mol 5. Verify that it takes 80.8 g of sodium chloride to make 45 ml of a 3.5 M solution. mol (Molarity)(Liters) (3.5M)(0.45L) 1.38 mol (1.38 mol)(58.5g/mol) 80.7 g
119 6. Consider 670 ml of a 4.10 M solution of Mg(NO 3 ) setting in a beaker. If you evaporate all 670 ml of the solution, how many grams of solute would be left in the beaker? 407.8g mol (Molarity)(Liters) (4.10M)(0.670L).747 mol convert mol to g: (.747mol)(148.3g/mol) 407.8g Information: Molality Molality is another way of expressing solution concentration. The symbol for molality is m. Whereas molarity (M) represents the ratio of moles solute to liters of solution, the molality (m) is the ratio of moles solute to kilograms of solvent. It can be expressed using the following formula: moles of solute molality kg solvent 7. Consider a solution that is prepared by adding 1.34 moles of sodium nitrate to.5 kg of water. What is the molality of the solution? 0.536 m m 1.34mol.5kg 0.536 m 8. Considering the data given in question 7, is this enough data to find the molarity? If so, calculate the molarity. If not, explain why not. No, because we don t know the total volume of the solution. 9. What is the molality of a solution that is made by dissolving 3.6 g of Na SO 4 in 475 g of water? 0.483 m Convert g to mol: 3.6g 14.1g/mol 0.9 mol molality mol/kg 0.9mol 0.475kg 0.483 m 10. Consider.35 moles of sodium chloride are dissolved in 1.1 kg of solution to make 1.9 liters. Calculate and compare the molarity and molality. Molarity: 1.8 M molality: 1.94 m M mol L.35mol 1.9 L 1.8 M m mol kg.35mol 1.1 kg 1.94 m 11. If 6.45g of Na SO 4 are dissolved in 1.10 kg of solution to make 1.4 L, calculate both the molarity and the molality of the resulting solution. Convert g to mol: 6.45g 14.1g/mol 0.186mol Molarity: 0.150 M M mol L 0.186mol 1.4 L 0.150 M molality: 0.169 m m mol kg 0.186mol 1.10kg 0.169 m Copyright 00-004 by Jason Neil. All rights reserved. To make copies permission must be obtained from www.chemistryinquiry.com.
10 Information: Mole Fraction Another way of expressing solution concentration is called mole fraction. The mole fraction (symbolized by X) of the solute or of the solvent can be calculated using the following equations: molsolute molsolvent X solute X solvent (molsolute + molsolvent ) (molsolute + molsolvent ) Note: both the solute and the solvent must be converted to moles when finding the mole fraction! 1. Prove that the mole fraction of salt (X ) equals 0.049 when 14.5 g of is dissolved in 85.0 g of H O. 14.5g 58.5g/mol 0.436 mol 85.0g 18.0g/mol 4.7 mol H O mol 0.436 X 0.049 (mol + mol ) (0.436 + 4.7) 13. Find the mole fraction of water (X water ) for the solution described in question 1. mol H 4.7 O X 0.951 (mol + mol ) (0.436 + 4.7) 14. Prove that X solute + X solvent 1. The answer from question 1 + answer to question 13 1 15. In a certain salt water solution, the mole fraction of salt is 0.18. Find the mole fraction of water. 1.00 0.18 0.8 Information: Mass Percent Composition Mass percent composition is similar to the mole fraction except the amounts of solute and solvent are in grams instead of moles. Here is the formula for finding the mass percent of a solute: mass solute mass% solute 100 (mass + mass ) 16. Prove that the mass percent of salt is 14.36% in the solution described in question 1. mass 14.5 mass% 100 100 14.36% (mass + mass ) 14.5 + 85.0 solute 17. Calculate the mass percent of sodium phosphate if 1.5g of it are dissolved in 50 ml of water. (Note: 1 ml of water has a mass of 1 g.) Because 1 ml 1g, the mass of H O 50 g mass Na 3PO4 1.5 mass% Na PO 100 100 4.76% 3 4 (mass + mass ) 1.55 + 50 Na PO 3 4 solvent
11 ChemQuest 40 Name: Date: Hour: Introduction Question: Melting Ice 1. In colder climates during the winter, people put salt on the roads and walkways to melt ice. Why do people do this? Why does salt melt the ice? Salt actually lowers the freezing point of water so that water will not freeze until much lower temperatures. Salt water remains a liquid at 0 o C Information: Dissociation and Total Molality of Particles When you dissolve a solute in a solvent, the resulting solution has slightly different properties than the original solvent. For example, salt water has a different freezing point and boiling point than pure water. The salt interferes with water s ability to freeze and boil. When ionic compounds dissolve, they dissociate. When an ionic compound dissociates that means that it breaks up into ions. For example, salt (sodium chloride) breaks up into sodium ions and chloride ions. This process is represented in the following balanced equation: Na + + Cl - Note for the above equation that Cl - does not need to be written as Cl because Cl - is a chloride ion and not a lone chlorine atom. Since calcium nitrate is an ionic compound it also dissociates as shown below: Ca(NO 3 ) Ca + + NO 3 - Covalent molecules do not dissociate. Although they may dissolve, they do not break up into ions.. Write the balanced equation for the dissociation of magnesium chloride. MgCl Mg + + Cl - 3. Write the balanced equation for the dissociation of ammonium sulfate. (NH 4 ) SO 4 NH + - 4 + SO 4 Copyright 00-004 by Jason Neil. All rights reserved. To make copies permission must be obtained from www.chemistryinquiry.com.
1 4. Consider calcium nitrate. Each calcium nitrate breaks up into one calcium ion and two nitrate ions according to the balanced equation given in the information section. If you take one mole of calcium nitrate and put it in water, it will dissociate. a) How many moles of calcium ions and how many moles of nitrate ions will there be in the solution? One mole of calcium ions and two moles of nitrate ions. b) What is the total number of moles of all ions in the solution? 3; one mole of calcium ions plus two moles of nitrate ions equal a total of three moles. 5. A solution is made so that it is.5 M Ca(NO 3 ). Therefore the concentration of Ca + is.5 M and the concentration of NO 3 - is 5.0 M. The total concentration of all particles is 7.5 M. Explain. Since each mole of Ca(NO 3 ) breaks up into 1 mole of Ca +, the concentration of Ca + would be the same as the concentration of Ca(NO 3 ). Since each mole of Ca(NO 3 ) breaks up into moles of NO 3 -, the concentration of NO 3 - is twice the concentration of Ca(NO 3 ). 6. A solution is made so that it the concentration is 3.0 m MgCl. What is the molality of Mg + and Cl - ions? What is the total molality of all particles in the solution? Mg + 3.0m Cl - 6.0m Total molality of all particles: 9.0m 7. A solution is prepared by dissolving 45.7 g of sodium carbonate in 00 g of water. a) What is the molality of the sodium carbonate? 45.7g 0.431mol 0.431mol.16m 106g / mol 0.00kg b) What is the total molality of all particles in the solution? 6.48m; Na CO 3 Na + + CO - 3 ; Since Na CO 3 breaks into 3 total ions, the total molality is 3(.16) 6.48m 8. Consider sugar (C 6 H 1 O 6 ), a covalent molecule. If a solution is made so that the concentration is 3.5 m in sugar, then what is the total molality of particles? 3.5m; Covalent compounds don t break up into ions so the total molality is still 3.5m. Information: Total Molality of Particles and Changes in Boiling/Freezing Points You may be wondering how all of this ties together. We have seen that adding a solute changes the boiling and freezing points of solvents. The amount of the change depends on how much solute is added. Equations relating the change in boiling or freezing point and the molality is shown below: T bp (m T )(K bp ) for boiling point T fp (m T )(K fp ) for freezing point Note: m T is the total molality of particles. K bp and K fp are constants called the molal boiling point elevation constant and the molal freezing point depression constant respectively. K bp for water is 0.515 o C/m and K fp for water is 1.853 o C/m.
13 9. What is the freezing point of a.5 m solution of salt water. Hints: first find T fp and then subtract the change from the original freezing point (0 o C for water). Also, remember m T is not.5 m in this problem. m T.5() 5.0 (recall that breaks into Na + and Cl - ) T fp (m T )(K fp ) (5.0)(1.853) 9.3 o C T fp 0 o C 9.3 o C 9.3 o C 10. Find the boiling point of a 3.7 m solution of calcium chloride. m T (3.7)(3) 11.1 (recall that CaCl breaks into Ca + and two Cl - ) T bp (m T )(K bp ) (11.1)(0.515) 5.7 o C T bp 100 o C + 5.7 o C 105.7 o C 11. What is the freezing point of a sugar solution in which the concentration of sugar is.5m? Note: sugar is covalent so it dissolves but it does not dissociate. T fp (m T )(K fp ) (.5)(1.853) 4.17 o C T fp 0 o C 4.17 o C 4.17 o C Information: Raoult s Law A solution will almost always have a lower vapor pressure than the pure solvent. For example, salt water will have a lower vapor pressure than pure water. The vapor pressure of a solution (P solution ) is related to the vapor pressure of the pure solvent (P solvent ) by the mole fraction of the solvent (X solvent ) in an equation known as Raoult s Law: P solution (X solvent )(P solvent ) 1. What is the vapor pressure of water at 0 o C is.3 kpa. What is the vapor pressure of a solution formed by dissolving 1.5g of LiCl in 84.3g of H O? 1.5g 84.3g 0.507mol 4.68mol 4.44g / mol 18.0g / mol x solvent 4.68 0.90 (4.68+ 0.507) P solution (X solvent )(P solvent ) (0.90)(.3).08 kpa Copyright 00-004 by Jason Neil. All rights reserved. To make copies permission must be obtained from www.chemistryinquiry.com.
14 ChemQuest 41 Name: Date: Hour: Information: Molar Mass As you know, the molar mass of any substance is how much mass (measured in grams) one mole of a substance has. The units for molar mass are g/mol. If you know that.0 moles of water have a mass of 36.0 g, you can find the molar mass of water by dividing 36.0 g by.0 moles to get 18.0 g/mol. 1. If.75 moles of a certain compound has a mass of 15.9 g, what is the molar mass of the compound? 15.9g.75mol 45.8 g/mol. If you put.53x10 4 molecules of an unknown compound on a balance you will discover that the mass is 757.8 g. What is the identity of the unknown compound? (Hint: find the molar mass and then calculate the molar masses of the following compounds.) A) C 10 H 1 O B) C 6 H 1 O 6 C) C 8 H 18 D) C 10 H 8 NO.53x10 6.0x10 4 3 757.8g 4.0 mol 4.0mol 180.3 g/mol B has a molar mass of 180 g/mol also Information: Relating Molar Mass and Colligative Properties One of the ways to determine the molar mass of a compound is by experiments involving colligative properties. By measuring the temperature changes of solutions, it is possible to calculate the molality of the solution and from the molality you can determine the molar mass. The following calculations walk you through the process. Please note that the process used here is valid only for covalent compounds.
15 3. When 135 g of an unknown covalent compound is dissolved in 450 g of water, the freezing point of the solution is 3.1 o C. Find the molar mass of the covalent compound. Follow these steps a) What is the molality of the solution? (Use T fp m T K fp and solve for m T. Note that since this is a covalent compound m T equals the molality of the solute because covalent compounds don t dissociate.) T fp m T K fp m T T fp K fp (3.1) 1.853 1.73 m b) How many moles of solute were dissolved? (Multiply your answer to part a by the kilograms of solvent.) (1.73m)(0.450kg) 0.779 mol c) Calculate the molar mass of the compound. (Take the mass of the solute given in the problem and divide by your answer to part b.) 135g 0.779mol 173 g/mol 4. Find the molar mass of a covalent compound if 10 g of the substance is dissolved in 810 g of water changes the boiling point of the solution to 101.3 o C. T bp m T K bp m T T bp K bp (1.3) 0.515.5 m (.5m)(0.810kg).04 mol 10 g.04 mol 103 g/mol 5. Find the molar mass of a covalent compound if 38.5 g dissolves in 50 g of water to give a freezing point of.15 o C. T fp m T K fp m T T fp K fp.15 1.853 1.16 m (1.16m)(0.50 kg) 0.603 mol 38.5g 0.603 mol 63.8 g/mol Copyright 00-004 by Jason Neil. All rights reserved. To make copies permission must be obtained from www.chemistryinquiry.com.