1. You need to dissolve CaCl2 in water to make a mixture that is 41.5% calcium c

Question

1. You need to dissolve CaCl2 in water to make a mixture that is 41.5% calcium chloride by mass. If the total mass of the mixture is 883.2 g, what masses of CaCl2 and water should be used?

Answer: _____ g CaCl2, and ____ g H2O.

2. What is the mole fraction of solute and the molal concentration for an aqueous solution that is 18.0% NaOH by mass? Answer: x= _____ (mole fraction) m=_____ (molality)

3. Express the concentration of a 0.0550 M aqueous solution of fluoride, Fâ, in mass percentage and in parts per million. Assume the density of the solution is 1.00 g/mL. Answer: ___% and ____ppm.

4. At 25 Ý°C and 765 Torr, carbon dioxide has a solubility of 0.0342 M in water. What is its solubility at 25 Ý°C and 1190 Torr? Answer: ____M

5. Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 Ý°C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 11.6 g of biphenyl in 27.0 g of benzene? Answer: ______ torr

Explanation / Answer

1. Mass of CaCl2 = 883.2 x 41.5/100 =366.528g

Mass water = 883.2 - 366.528 = 516.672g

2. 18% NaOH by mass means 18g NaOH in 100g solution:

solution consists of 18g NaOH and 82g H2O
Molar mass NaOH = 40g/mol
18g = 18/40 = 0.45 mol
Molar mass H2O = 18g/mol
82.0g = 82.0/18 = 4.555 mol

Total mol = 0.45+4.55 = 5.
Mol fraction NaOH = 0.45/5 = 0.09 ( note that mol fraction does not have units)
As a check, the total mol fractions must add up to 1.00
Mol fraction of water = 4.55/5. = 0.91

Total mol fractions: 0.09 + 0.91 = 1.00 so you know that you are correct.

Molality of the solution:
definition: molality = mol solute dissolved in 1kg of solvent.

You have 0.45mol NaOH dissolved in 87g solvent
mol dissolved in 1000g solvent = 0.45*1000/82 =5.487 m solution.

3. Assume that you have 1 L of solution.

moles F- = M F- x L F- = (0.0550)(1) = 0.0550 moles F-
0.0440 moles F- x (19.0 g F- / 1 mole F-) = 1.045 g F- in 1 L of solution

1 L solution x (1000 mL / 1 L) x (1.00 g / mL) = 1000 g solution

mass % F- = (g F- / g solution) x 100 = (1.045 / 1000) x 100 = 0.1045%

parts per million F - = mg F - / L = 1045 / 1 = 1045 ppm F-

5. We need to calculate the mole fraction of benzene after adding the biphenyl (molar mass = 154.2 g/mole).

11.6 g C12H10 x (1 mole C12H10 / 154.2 g C12H10) = 0.075 moles C12H10

27 g C6H6 x (1 mole C6H6 / 78.1 g C6H6) = 0.345 moles C6H6

Total moles = moles C12H10 + moles C6H6 = 0.075 + 0.345 = 0.420

mole fraction C6H6 = (moles C6H6 / total moles) = 0.345 / 0.420 =
0.821

What that means is that the vapor pressure will be 82.1% of the vapor pressure of pure benzene since the solution contains only 82.1 mole % benzene. The biphenyl contributes no vapor pressure (nonvolatile).

P = (x C6H6)(Po C6H6) where x C6H6 is the mole fraction of C6H6 and Po C6H6 is the vapor pressure of pure C6H6.

P = (0.821)(100.84 torr) = 82.789 torr.

4. 1. Mass of CaCl2 = 883.2 x 41.5/100 =366.528g

Mass water = 883.2 - 366.528 = 516.672g

2. 18% NaOH by mass means 18g NaOH in 100g solution:

solution consists of 18g NaOH and 82g H2O
Molar mass NaOH = 40g/mol
18g = 18/40 = 0.45 mol
Molar mass H2O = 18g/mol
82.0g = 82.0/18 = 4.555 mol

Total mol = 0.45+4.55 = 5.
Mol fraction NaOH = 0.45/5 = 0.09 ( note that mol fraction does not have units)
As a check, the total mol fractions must add up to 1.00
Mol fraction of water = 4.55/5. = 0.91

Total mol fractions: 0.09 + 0.91 = 1.00 so you know that you are correct.

Molality of the solution:
definition: molality = mol solute dissolved in 1kg of solvent.

You have 0.45mol NaOH dissolved in 87g solvent
mol dissolved in 1000g solvent = 0.45*1000/82 =5.487 m solution.

3. Assume that you have 1 L of solution.

moles F- = M F- x L F- = (0.0550)(1) = 0.0550 moles F-
0.0440 moles F- x (19.0 g F- / 1 mole F-) = 1.045 g F- in 1 L of solution

1 L solution x (1000 mL / 1 L) x (1.00 g / mL) = 1000 g solution

mass % F- = (g F- / g solution) x 100 = (1.045 / 1000) x 100 = 0.1045%

parts per million F - = mg F - / L = 1045 / 1 = 1045 ppm F-

5. We need to calculate the mole fraction of benzene after adding the biphenyl (molar mass = 154.2 g/mole).

11.6 g C12H10 x (1 mole C12H10 / 154.2 g C12H10) = 0.075 moles C12H10

27 g C6H6 x (1 mole C6H6 / 78.1 g C6H6) = 0.345 moles C6H6

Total moles = moles C12H10 + moles C6H6 = 0.075 + 0.345 = 0.420

mole fraction C6H6 = (moles C6H6 / total moles) = 0.345 / 0.420 =
0.821

What that means is that the vapor pressure will be 82.1% of the vapor pressure of pure benzene since the solution contains only 82.1 mole % benzene. The biphenyl contributes no vapor pressure (nonvolatile).

P = (x C6H6)(Po C6H6) where x C6H6 is the mole fraction of C6H6 and Po C6H6 is the vapor pressure of pure C6H6.

P = (0.821)(100.84 torr) = 82.789 torr.

4.Pressure and solubility are directly proportional

Given by P/C = Kh

Henry's Law is P = Kh*C

where Kh is the constant

Increase in pressure = increase in solubility

We're just finding end result using variable C2 = solubility CO2

P1/C1 = P2/C2

Plug in values:

765/ 0.0342 = 1190/ C2

Solve for C2:

765C2 = 40.698

C2 = 0.0532 M CO2

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.