Chemistry 114
First Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. Write the following equations. For each equation also define all terms used in the
equation.
A. Equation used to determine the colligative molality of a solution.
Where i is the van’t Hoff factor and m is the molarity
B. Equation used to determine the volume of a unit cell from number of atoms in
unit cell.
n=n umber
of
atoms
d=density,N0=Avogadro’s number
C. Equation used to determine the vapor pressure of a solution when it has two
components, and both components have a vapor pressure (Raoult’s Law)
VP=vapor pressure, ÷=mole fraction
D. Equation used to determine the freezing point depression of a solution.
ÄT=KfAim Where ÄT= Normal freezing point - new freezing point, Kf = Freezing point
depression constant, I = van’t Hoff factor, and m=molality
E. The integrated rate equation for a second order reaction.
1/[A]t = 1/[A]0 +kt Where [A]t is the concentration of A at time t, [A]0 = initial
concentration of A, t=time, k=rate constant for this reaction.
F. Equation used to determine the half-life of a first order reaction.
t1/2=.693/k
Where t1/2 is the half-life, and k is the rate constant
2. As I wrote this test on Saturday, the temperature was -7oF or - 22oC. How much heat
energy is required to change a snowflake at -22oC to a liquid at room temperature (25oC)?
Some of the following data may be useful in your calculation:
Mass of a snowflake : 0.003 g
ÄH fus H2O = 6.01 kJ/mol
ÄH vap H2O = 40.66 kJ/mol
Specific Heat Capacity of ice : 2.09 J/g@oC
Specific Heat Capacity of water: 4.19 J/g@oC
Three steps
Bring the temp of the solid water up to 0
q=S.H.C. x mass x ÄT
=2.09 J/gAoC x .003 g x (0-(-22))0C
=.138 J
Melt the solid water at 0oC
q=ÄH fus x moles
moles = .003/18
q=6.01 x .003/18 = .001002 kJ = 1.002J
Bring the liquid water at 0 up to 25
q=S.H.C. x mass x ÄT
=4.19 J/gAoC x .003 g x (25-0)0C
=.314 J
Sum it all together = .138 + 1.002 + .314 = 1.454J
1
in
3. Zinc has a density of 7.14 g/cm3. Assuming it crystalizes in a body centered cubic
lattice, determine the volume of its unit cell.
Body centered cubic has 2 atoms per unit cell so n=2
=2(65.38)/7.14(6.022x1023)
=3.04x10-23 cm3
4. Vinegar is 5% acetic acid (CH3COOH) by mass.
A. What is the mole fraction of acetic acid in vinegar?
5% by mass says that for every 100 g of solution 5 g is acetic acid and 95 g is water
molar mass of acetic acid = 2(12) + 2(16) + 4(1) = 60
molar mass water = 16+2(1) =18
moles acetic acid = 5/60 = .0833 moles
moles water = 95/18 = 5.28
mole fraction acetic acid = .0822/(.0833+5.28) = .0155
B. What is the molality of acetic acid in vinegar?
Moles acetic acid/kg of water
=.0833/.095 kg = 0.877m
2
5. 1-Propanol has a molar mass of 60.1 g/mol and a vapor pressure of 1.99 kPa @20oC
Acetone has a molar mass of 58.08 g/mol amd a vapor pressure of 24.5 kPa @ 20oC.
A. If I mix 120.2 grams of 1-propanol and 58.08 g of acetone, what is the vapor
pressure of the solution?
120.2 g propanol/60.1 g/mol = 2 mol propanol
58.08 g acetone/58.08 g/mol = 1 mol acetone
VP = 1.99(2/3) + 24.5x(1/3)
=9.49 kPa
B. If the actual vapor pressure is higher than this value, what does it tell me about
the interactions between 1-Propanol and Acetone in the solution?
Higher vapor pressure is a positive deviation from Raoult’s law and this indicates
poor interaction between solvent and solute molecules.
6. Define the following terms
Enthalpy of fusion
Heat required to change 1 mole of solid when the temperature is at the melting point
of the material.
Normal boiling point
The temperature at which the liquid and the gas have the same vapor pressure, and
the vapor pressure is 1 atm.
Viscosity
Resistance to flow.
Triple point
The point on a phase diagram where three different phases of a material all coexist
at one temperature and pressure.
Osmotic pressure
The pressure required to prevent the flow of a solvent into a solution that is inside
a semi-permeable membrane.
Distillation
Heating a solution until it turns into a vapor, and then condensing the vapor back
into a liquid. This process serves to concentrate the component of the solution which has
the higher vapor pressure.
1st order reaction
A reaction in which the order parameter is 1
3
7. I have a first order reaction with a rate constant of 10 min-1. If I start with a .1M solution
of the reactant, how long will it take before 75% of the reactant has disappeared?
Ln[A]t = ln[A]0 -kt
[A]0 = .1M
75% has disappeared means 25% remains so
[A]t = .25(.1) = .025
ln(.025) = ln(.1) -(10)X
-3.689 = -2.303 -10X
-3.689 + 2.303 =-10X
-1.386 =-10X
-1.386/-10 =X
X = .139 minutes
8. A second order reaction has a half-life of 10 minutes when the reactant concentration
is .1M. What will be the half-life of the reaction when the reactant concentration is .03 M?
t1/2 = 1/(k[A]
10 = 1/k(.1)
k=1/[10x.1]
k=1
t1/2 = 1/1(.03)
=33.3 minutes
4