Solutions

Entry Task: Nov 21st Block 2

Question:

What is the molality of a 200 g solution with 32 grams of NaCl?

TURN IN ENTRY TASK SHEETS!!

You have 5 minutes!

Solutions

Agenda:

• Discuss Ch. 13 sec. 4• pHet Concentration and Molarity

Lab• HW: Solutions ws

Solutions

I can …

• Express a concentration of a solution in different ways- %, ppm, mole fraction, Molarity, & Molality

Solutions

Chapter 13Properties of Solutions

Solutions

Ways of Expressing

Concentrations of Solutions

Solutions

Mass Percentage

Mass % of A =mass of A in solutiontotal mass of solution

100

Solutions

Parts per Million andParts per Billion

ppm =mass of A in solutiontotal mass of solution

106

Parts per Million (ppm)

Parts per Billion (ppb)

ppb =mass of A in solutiontotal mass of solution

109

Solutions

13.3 problem(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water.

1.50 51.5g solution

X 100 = 2.91%

Solutions

13.3 problem(b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution?

X 2500g

36.2100

100 X = 90500

X = 90.5g

Solutions

moles of Atotal moles in solution

XA =

Mole Fraction (X)

• In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!

Solutions

mol of soluteL of solution

M =

Molarity (M)

• You will recall this concentration measure from Chapter 4.

• Because volume is temperature dependent, molarity can change with temperature.

Solutions

mol of solutekg of solvent

m =

Molality (m)

Because both moles and mass do not change with temperature, molality (unlike molarity) is not temperature dependent.

Solutions

Changing Molarity to Molality

If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

Solutions

13.4 problem• What is the molality of a solution made by

dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)?

35g

128.1 g = 0.28 mole

0.425 kg

1 mole= 0.670 kg

Solutions

13.5 problem• A commercial bleach solution contains 3.62 mass %

NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution.

3.62g

74 g = 0.0489 mole1 mole

3.62 % in water means 3.62 g in water3.62 g – 100 = 96.38 g of water

96.38g

18 g = 5.35 mole1 mole

0.0489 mole + 5.35 mole = 5.398 mole

0.0489 mole 5.398 mole

9.06 x 10-3

Solutions

13.5 problem• A commercial bleach solution contains 3.62 mass %

NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution.

3.62g

74 g = 0.0489 mole1 mole

3.62 % in water means 3.62 g in water3.62 g – 100 = 96.38 g of water

0.09638 kg= 0.507 m

0.0489 mole

Solutions

13.21 13.21 a) Calculate the mass % of Na2SO4 in solution containing 14.7 g Na2SO4 in 345g H2O. b) Ore containing 7.35 g of silver per ton of ore. What is the concentration of silver in ppm?

X 100 = 4.09%14.7 g14.7 + 345 g

7.35 g

1 g = 8.09 ppm

1.10 x 10-6 ton

Solutions

13.23 13.23 Calculate the mole fraction of methanol, CH3OH, in the following solutions: a) 7.5 g CH3OH in 245 gH2O; b)55.7 g CH3OH in 164g CCl4.7.5 g

32.01 g = 0.23 mole of methanol

1 mole

245 g

18.0 g = 13.6 mole of water

1 mole

0.23 mole of methanol13.6 + 0.23 mole solution

= 0.017 mole fraction

Solutions

13.23 13.23 Calculate the mole fraction of methanol, CH3OH, in the following solutions: a) 7.5 g CH3OH in 245 gH2O; b)55.7 g CH3OH in 164g CCl4.55.7 g

32.01 g = 1.74 mole of methanol

1 mole

164 g

153.81 g = 1.07 mole of CCl4

1 mole

1.74 mole of methanol1.74 + 1.07 mole solution

= 0.620 mole fraction

Solutions

13.25 13.25 Calculate the molarity of the following aqueous solutions: a) 10.5g KCl in 250.0 ml of solution; b) 30.7g LiClO4•3H2O in 125ml o solution; c) 25.0 ml 1.50M HNO3 diluted to 0.500L10.5 g

74 g = 0.14 mole

1 mole

0.250 L = 0.560 M

30.7 g

106.397 g + 3(18g) = 0.191 mole

1 mole

0.125 L = 1.53 M

(0.025L)(1.50) = (x)(0.500L) = 0.075 M

Solutions

13.27 13.27 Calculate the molality of each the following solutions; a) 13.0 g benzene, C6H6, dissolved in 17.0 g of CCl4; b) 4.75 g NaCl dissolved in 0.250 L of water whose density is 1.00g/ml.

13.5 g

78 g = 0.167 mole

1 mole

0.017 kg = 9.79 m

4.75 g

58.45 = 0.081 mole

1 mole

0.250 kg = 0.325 m

Solutions

13.29 13.29 A sulfuric acid solution containing 571.6g of H2SO4 per liter of solution has a density of 1.329 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H2SO4

571.6 g

1329 gX 100 = 43.0%

571.6 g

98 g = 5.83 moles of H2SO4

1 mole

757.4 g

18 g = 42.1 moles of H2O

1 mole

5.83 moles of H2SO4

5.83 + 42.1 = 47.93 moles = 0.122

Solutions

13.29 13.29 A sulfuric acid solution containing 571.6g of H2SO4 per liter of solution has a density of 1.329 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H2SO4

5.83 mol

0.7574 kg= 7.70 m

5.83 moles of H2SO4

1 L = 5.83M

Solutions

13.30 13.30 A solution containing 80.5 g of ascorbic acid, C9H8O6, dissolved in 210 g of water has a density of 1.22 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of C9H8O6

80.5g

210 g + 80.5gX 100 = 27.7%

80.5 g

176.1 g = 0.457 moles of C9H8O6

1 mole

210 g

18 g = 11.7 moles of H2O

1 mole

0.457 moles of C9H8O6

0.457+ 11.7 = 12.16 moles = 0.0376

Solutions

13.30 13.30 A solution containing 80.5 g of ascorbic acid, C9H8O6, dissolved in 210 g of water has a density of 1.22 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of C9H8O6

0.457 mol

0.210 kg= 2.18 m

0.457 moles of C9H8O6

0.238 L = 1.92 M

290.5 g

1.22 g = 0.238L

1 mL

1000 mL

1 L

Solutions

In-Class- pHet Molarity and

concentration

Solutions

HW: Ch. 13 sec. 5 reading notes