# CHEM125 – CHAPTER 13 – BASIC HOMEWORK

## (NEW) CHAPTER 13 BASIC HOMEWORK

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### Question 1

What is the molarity of a sodium chloride solution, if 8.15 g of NaCl are dissolved in enough water to make 0.600 L of solution? ### Explanation

8.15g / 58.44(g/mol) = .13946 mol

.13946 mol/ 0.6L = .232 mol/L

### Question 2

What is the molality of a sodium chloride solution, if 8.49 g of NaCl are dissolved in 0.400 kg of water? Hint given in general feedback ### Explanation

Molality = mol/kg 8.49g / 58.44(g/mol) = .145277 mol .145277 mol /.400kg = .363 mol/kg

### Question 3

What is the mole fraction of a lithium fluoride solution prepared by dissolving 65.7 g of LiF in 0.200 kg of water. ### Explanation

Mole Fraction = n(solutes)/ [ n(solutes) + n(solvents)] n= moles 65.7 g / 25.94 (g/mol) = 2.53 mols LiCl 200g / 18.016 (g/mol) = 11.1 H2O = 2.53 mol / (11.1 + 2.53) = 0.186

### Question 4

What is the mass percent of a lithium fluoride in a solution prepared by dissolving 24.7 g of LiF in 0.100 kg of water. ### Explanation

24.7g / ( 24.7g + 100g ) = .198 .198 x 100= 19.8 %

### Question 5

Calculate the freezing point for a 1.42 m solution of CCl4 in benzene. ### Explanation

ΔT = m Kf ΔT = (1.42 m) (4.90 (C/m) ΔT = 6.96 ΔT = Tf – Tunkown 6.96 = 5.51 – Tunkown -1.45 = Tunkown

### Question 6

Calculate the increase in the boiling point,ΔT, for a 0.86 m solution of sucrose (dissolves as a molecule) in water. ### Explanation

Tb = .512 (water) ΔT = m Kf (solvent) ΔT = (.86 m) ( .512 C/m)

### Question 7

Sugar is dissolved in hot water at 45oC until the mole fraction of water decreases to 0.870. The vapor pressure of pure water at this temperature is 72 torr. What is the vapor pressure (in torr) of water for this solution? ### Explanation

Psolution = xsolvent Psolvent (pure) Psolvent (pure) = 72 torr  xsolvent = .870 Psolution = (.870)(72 torr) = 62.6

### Question 8

44.1 g of the nonvolatile solute urea (M.W. = 60.06 g/mol) is dissolved in 261 g of water at 60oC. The vapor pressure of pure water at this temperature is 149 torr. What is the equilibrium vapor pressure (in torr) of water for this solution? ### Explanation

Nonvolatie solute urea = 44.1 g  Nonvolatie solute urea (molar mass) = 60.6 (g/mol) Water = 261g 44.1 g / 60.6 (g/mol) = .72607 mol Nonvolatie 261 g/ 18.016 (g/mol) = 14.487 mol Water  x = 14.487 /(.72607 + 14.487) = 0.9523 Psolution = xsolvent Psolvent Psolution = 0.9523 x 149 torr = 142 torr

### Question 9

Which 0.06 m aqueous solution (in water) should have the greatest increase in boiling point?
Select one: ### Explanation

Large amount of heat which is needed to break the strong ionic bond.

### Question 10

Calculate the freezing point for a 3.93 m aqueous solution of NaCl.  ### Explanation

ΔT = m Kf (solvent) (water) ΔT = (3.93 m) (1.86 C/m) ΔT = 7.3098 x 2 mol = 14.6 Tf = 0 – 14.6 = -14.6

### Question 11

A 0.200 g sample of a nonvolatile yellow crystalline solid is dissolved in 18.2 g of benzene, producing a solution that freezes at 5.00oC. Find the molar mass of the yellow solid in g/mol.  ### Explanation

Freezing Point of Benzene = 5.51 C Kf Benzene = 4.90 C/m ΔT = 5.51 – 5 = .51 C nonvolatile = 0.200 g  benzene = 18.2 g ΔT = m Kf(solvent) (benzene .51 C = m (4.90 C/m) .104 mol /kg = m .200g (yellow solid) / 18.2g (Beneze) = .010989 (.010989) x (1000/.104) = 106

### Question 12

A 3.90 g sample of a nonvolatile nonelectrolyte blue crystalline solid is dissolved in 35.3 g of acetic acid, producing a solution that bolis at 119.9oC. Find the molar mass of the blue solid in g/mol.  ### Explanation

Boiling Point of Acetic Acid = 117.90 C Kb Acetic Acid = 3.07 C/m ΔT = 119.90 – 117.90= 2 C ΔT = m Kf(solvent) (benzene  2 C = m (3.07 C/m) .651 mol /kg = m 3.90g (blue crystal) / 35.3g (Acetic Acid)= .11048 (.11048) x (1000/ 0.651 (mol /kg) ) = 169.7 = 170

### Question 13

At a certain temperature the vapor pressure of pure benzene and pure toluene are 380 torr and 130 torr, respectively. If the mole fraction of benzene is 0.26, what is the total vapor pressure. ### Explanation

Psolution = xsolvent Psolvent   PBenzene = (0.26) (380 torr) = 98.8 torr  Ptoluene = (1 – 0.26) (130 torr) = 96.2 torr P total = 98.8 + 96.2 = 195 torr

### Question 14

At a certain temperature the vapor pressure of pure benzene and pure toluene are 380 torr and 130 torr, respectively. If 7.58 mol benzene is mixed with 4.38 mol toluene, what is the total vapor pressure. ### Explanation

Psolution = xsolvent Psolvent  x benzene  =  7.58 / (7.58 + 4.38) = .6338 x toulene =  4.38 / (7.58 + 4.38 = .3662  PBenzene = (.6338) (380 torr) = 240.8 torr  Ptoluene = (.3662) (130 torr) = 47.6 torr P total = 240.8 + 47.6 = 288 torr

### Question 15

Which of the following pairs of molecules are expected to obey Raoult’s law?

Select one or more:   ### Explanation

Raoult’s solution is based on ideal solution B does not work because it is an endothermic reaction

### Question 16

What is the osmotic pressure in atm of a 0.378 M solution of sugar at 25oC? ### Explanation

π = m R T π = (0.378 M) (.08206 ) (298 K)

### Question 17

What is the osmotic pressure in atm of a 0.119 M solution of NaCl at 0oC? Hint given in feedback ### Explanation

π = i m R T π = (2) (.119 M) (.08206) (273 K) π = 5.33