Concentration Calculations
A solution of 20 grams of substance X (molar mass =212.7 g/mol) is dissolved in 250.0 mL of benzene (molar mass =80.2 g/mol; density: 0.876 g/mL). What is the molality of this solution? Assume the density of solution is the same as that of the pure solvent.
moles solute/ kg solvent
20g /212.7 g/mol = 0.094 moles solute
250 mL x0.876g/mL = 219 g
0.094 moles solute / 0.219 kg = 0.429 m
Write the units for each of these concentration units WITHOUT looking at notes:
(a) molality
(b) molarity
(c) percent by mass
molality: moles solute/kg solvent
molarity: moles solute/ liters solution
percent by mass: mass A/ total mass x 100
Which of the following are incorrect about solubility?
I. Solubility of gases decreases with decreasing temperature.
II. Solvents dissolve solutes that have similar IMFs
III. Pressure doesn’t majorly impact solubility of solids
IV. Heating an endothermic reaction causes more solute to dissolve.
V. Heating an exothermic reaction causes more solute to dissolve.
I, V
Which of the following is/are NOT colligative properties?
Vapor pressure lowering
Osmotic pressure
Henry’s Law
Boiling point elevation
3 ONLY
Determine the theoretical van't hoff factor for the following:
(a) sucrose
(b) MgBr
(c) Ammonium sulfate
(a) 1
(b) 3
(c) (NH4)2(SO4) - 3
Calculate the molality of a 8% aqueous NaBr solution.
0.845m NaBr
Assume 100g of NaBr solution.
8 g NaBr / 100g NaBr solution
We need moles NaBr / Kg NaBr solution.
8g x 1mole/(102.89g/mol) = 0.0778 moles NaBr
100g - 8g = 92g water = .092 kg
.0778 moles NaBr / .092 kg = 0.845 m NaBr
Heat + solute (undissolved) -> solute (dissolved)
In the above equation, how would increasing temperature change the amount of solute that is dissolved?
Reactants increase, so equilibrium shifts toward the products, increasing the amount of solute that is dissolved (=increasing solubility).
`Select all of the following solutes that are likely to dissolve in water:
NH3
MgO
SF6
CH3OH
NF3
1,4
Define a colligative property.
Properties that depend on the number of solute particles in solution, not on the nature of the solute particles.
Classify the following as weak electrolytes, non-electrolytes, or strong electrolytes:
1. Cs(OH)2
2. C2H5
1. Strong Base = Strong Electrolyte
2. Non-Electrolyte
Determine the molarity of a 8% HCl solution (MW = 36.46), Density = 2.56 g/mL.)
5.627 M
moles solute / 1L
8g HCl/ 100g solution = 8% HCl
8g x 1mol/36.46g = 0.22 moles HCl
100g solution/2.56g/mL = 39.06 mL /1000 = .0391 L
0.22 mols HCl / .0391 L = 5.627 M
FREE SPACE
What part of Genchem 2 is worrying you the most right now? Is there any resources you want me to make in order to help you succeed in this class?
How would you determine if a compound is water soluble or fat soluble?
Water soluble compounds would have hydrogen bonds in them whereas fat soluble compounds would likely be hydrocarbons and have only nonpolar bonds within them.
Which of the following are TRUE about colligative properties?
Increasing solute in a solution will lower its vapor pressure.
VP of pure solvent will be higher than VP of pure solute.
BP of solution is greater than BP of pure solvent.
FP of solution is greater than FP of pure solvent.
Osmotic pressure cannot be solved for.
1,2,3
Name the 3 classes of chemical compounds that are strong electrolytes.
Strong Bases
Strong Acids
Ionic Compounds (that are soluble)
You have a solution of 8.9 molal sodium dissolved in water. What is this concentration in molarity (M)? Density of solution = 2.9 g/mL.
8.9 moles/ 1kg solvent
1000g SOLVENT + (8.9 moles x 22.99 g Na = 203.81g SOLUTE) = 1203.81 g total /2.9 g/mL = 415.1 mL = 0.4151 L
8.9 moles solute/ 0.4151 L SOLUTION = 21.44 M
First team to go to the whiteboard and define the following wins the points:
(a) Henry's Law
(b) Raoult's Law
Henry's Law: solubility of gas is directly proportional to partial pressure of gas over the solution
Raoult's Law: Solves for Vapor pressure of solution, a colligative property
P solution = (Mole Fraction Solvent) (VP solvent)
Select all of the following that are TRUE of solubility:
I. Solubility of gases increases with increasing temperature
II. Solubility of gases increases as pressure increases
III. The way pressure impacts the solubility of gases is described by Henry's Law
IV. For solids, temperature's effect on solubility depends on if the solution is endothermic or exothermic
V. Pressure impacts the solubility of solids
II, III, IV
XeF4 has VP = 90.0 torr (18 ℃) (MW = 207.3 g/mol). It can dissolve wax (C22H46; MW = 311 g mol–1). What is the VP at 18 ℃ of a solution of 8.0 g wax dissolved in 40.0 g XeF4?
HINT: Use Raoult's Law
P solution = (X Solvent) (P solvent)
X solvent = mole fraction = (moles solvent) / (total moles)
(40/207.3) / [(40/207.3) + (8/311)] = 0.885
P solution = (0.885)(90 torr) = 79.7 torr
Determine the theoretical van't hoff factor (i) for the following:
(a) MgSO4
(b) MgCl2
(c) FeCl3
(d) HCl
(a) 2
(b) 3
(c) 4
(d) 2
Calculate the molality of a 8.23M aqueous NaCl solution. Density = 2.6 g/mL.
8.23 moles NaCl/ 1 L solution
1000 mL x 2.6g/mL = 2600g solution
8.23 moles NaCl x (22.99 + 35.45 g/mol) = 480.96 g
2600g solution - 480.96g SOLUTE = 2119.04 g solvent = 2.119 kg solvent
8.23 moles NaCl/2.119 kg solvent = 3.884 m
Select all of the following that are TRUE of Raoult's Law:
I. An increase in solute will bring the vapor pressure of solution closer to the vapor pressure of the pure solvent.
II. Vapor Pressure is proportional to the mole fraction of the solvent in a solution.
III. An ideal solution is one where all components obey Raoult's Law.
IV. Vapor pressure is independent of the amount of solvent in a solution.
V. Vapor pressure is not a colligative property.
II, III
Explain why heating endothermic solutions leads to a decrease in solubility, whereas heating exothermic solutions leads to an increase in solubility.
Endothermic
Heat + solute(undissolved) -> solute (dissolved)
Exothermic
solute (undissolved) -> solute (dissolved) + Heat
We know that equilibrium moves to the OTHER side when we add either a product or reactant.
Select all of the following that are TRUE of colligative properties:
I. A greater amount of solute decreases the VP of a solution.
II. The boiling point is the temperature where vapor pressure is greater than atmospheric pressure
III. Freezing point and boiling point must take into consideration the electrolyte qualities of the solute.
IV. The van't hoff factor accounts for effects of dissociation of electrolytes.
V. A greater amount of solute increases the freezing point.
I, III, IV
Which of the following solutions will have the lowest boiling point when dissolved in water?
0.1m MgCl2
0.1m NH4Cl
0.15m NaCl
0.15m MgCl2
0.25m MgF2
2
calculate (im) for each. Boiling point elevation, so lowest will be the one with the smallest change in temperature, so the smallest (im) value. Remember, NH4 = 1 for the electrolyte and SO4 = 1, all the polyatomic ions are only 1 (don’t dissociate)
1. 0.3
2. 0.2 = !!
3. 0.3
4. 0.45
5. 0.75