Answer Key – Chapter 7

7.1 How many grams of solute are present in the following solutions? a) 25.2 g of a 4.25 % w/w solution of glucose

25.2 g solution x 4.25 g solute = 1.07 g solute 100 g solution b) 125 g of a 6.55% w/w solution of sodium sulfate

125 g solution x 6.55 g solute = 8.19 g solute 100 g solution

7.3 How many grams of KHCO3 must you add to 250 mL of water to prepare a 0.600% w/v solution?

0.600% w/v solution means grams of solute = 0.600 g = 0.006 mL of solution 100 mL

x g KHCO3 = 0.006 so x = 1.50 g KHCO3 250 mL solution

7.6 What mass of glucose is required to prepare 210 g of a 8.50% w/w solution of glucose?

8.50 % w/w solution means grams of solute = 8.5 g = 0.085 grams of solvent 100 g

x g glucose = 0.085 so x = 17.9 g glucose 210 g solution

7.11 What is the concentration in mg% of a uric acid solution that contains 0.230 mg uric acid in 19.0 mL of solution?

mg % solution means mg of solute x 100 mL of solution

0.230 mg x 100 = 1.21 mg % 19.0 mL

7.14 What is the volume of a 2.8 M solution that contains 0.70 mol of solute?

Molarity - M = moles of solute L of solution

2.8 M = 0.70 moles of solute so x = 0.25 L x L of solution

7.16 What are the molarities of solutions containing
a) 0.60 mol in 0.80 L

M = 0.60 mol = 0.75 M 0.80 L

b) 0.75 mol in 1.5 L M = 0.75 mol = 0.5 M 1.5 L

7.18 Calculate the number of moles contained in
a) 0.200 L of a 0.700 M solution

0.200 L x 0.700 mol = 0.14 mole L

b) 2.60 L of a 0.750 M solution

2.60 L x 0.750 mol = 1.95 mole L

7.22 What is the molarity of a solution containing 21.55 g of KCL in 643 mL?

first find the number of moles KCl 21.55 g x 1 mole = 0.286 mole KCl 74.6 g

next calculate M in mol/L = 0.286 mol = 0.45 M 0.643 L

7.28 Given 20.0 M of a solution of 2.50 M HCl, to what volume must we dilute it to obtain a concentration of 0.200 M?

M1V1 = M2V2

so V2 = M1V1 = 0.020 L x 2.5 M = 0.250 L M2 0.2 M

7.30 a) use a 12 M H2SO4 solution to prepare 2.0 L of 1.5 M H2SO4

M1 x V1 = M2 x V2 V1= ? , M1 = 12 M, M2= 1.5 M, V2 = 2.0 L

12 M x V1 = 1.5 M x 2.0 L so V1 = 0.25 L b) V1= ? , M1 = 2.0 M, M2= 0.5 M, V2 = 0.200 L

2.0 M x V1 = 0.5 M x 0.200 L so V1 = 0.05 L

7.33 From a 0.80L glucose stock solution (C6H12O6 MW = 180 g/mol) prepare a diluted solution:

M1 x V1 = M2 x V2 V1= 0.80 L , M1 = ?, M2= 1.8 % w/v, V2 = 4.0 L

First find Molarity of M2 in moles/L

1.8 % w/v = 1.8 g x 1000 ml x 1 mol = 0.1 mol 100 mL 1 L 180 g L

Then use M1 x V1 = M2 x V2 M1 x 0.80 L = 0.1 M x 4.0 L , so M1 = 0.5 M

7.38 The solubility of aluminum sulfate in water at 20oC is 26.7 g/100 mL. Calculate it’s molarity.

M = mol/L MW of Al2(SO4)3 = 342 g/mol see pages 70-75 for a review of ions on an exam I would give you the formula for something complicated , but you should feel comfortable figuring it out for the straight-forward cases.

26.7 g x 1000 ml x 1 mol = 0.78 mol 100 mL 1 L 342 g L

alternatively you could do it stepwise:

1. figure out the number of moles : 26.7 g x 1 mol = 0.078 mol 342 g

2. divide by the volume in L, so 100 mL = 0.1 L

0.078 mol = 0.78 M 0.1 L

7.40 Calculate the osmolarity of the following aqueous solutions:

a) 0.40 M glucose, Osmolarity = i(M) for glucose (covalent compount) i = 1

So Osmolarity = 1 (0.40 M) = 0.40 osmol/L

b) CaCl2 ionizes to Ca2+ and 2 Cl- so i = 3

Osmolarity = i(M) = 3 (0.1 M) = 0.3 osmol/L

7.48 Calculate % v/v for the solutions: a) 15 mL x 100 = 7.5 % v/v 200 mL

b) 8.40 mL x 100 = 24.0 % v/v 35.0 mL

c) 10.6 mL x 100 = 21.2 % v/v 50.0 mL

7.67 75 mL propylene glycol is dissolved in 55 mL of water

Solution (present in greater amount) = propylene glycol Solute (present in lesser amount) = water

7.69 hydrostatic pressure at the artery (forcing out) = 100 torr osmotic pressure of blood is 15 torr

Does water move in or out of the arterial blood?

Which pressure is higher? The liquid will flow from higher to lower pressure. 100 torr > 15 torr ; therefore water moves out of the arterial blood.

7.70 How would you prepare an isotonic saline solution described by a medial dictionary as a 0.90% NaCl w/v solution?

0.90 % w/v NaCl means grams of NaCl = 0.90 g 100 mL of solution 100 mL

Therefore place 0.90 g of NaCl in a volumetric flask and add sufficient water to bring the total volume of the solution to 100 mL.