The Islamic University
of Gaza
Chemistry Department
General Chemistry B
1st Hour Exam
March 20th, 2011
KEY
Answer the Following questions:
1. Circle the compounds that are capable of hydrogen bonding with themselves:
C2H6, CH3OH, CF4, HF, CH3OCH3
2. Arrange the following in order of increasing boiling point (from left (lower boiling)
to right (higher boiling): CH3CH2OH, CH3CH2CH2CH3, CH3Cl, NaCl
CH3CH2CH2CH3, CH3Cl, CH3CH2OH, NaCl
3. Circle the species in each pair that you think it should have a higher surface
tension:
1) CH3CH2CH2CH3, CH3CH2CH2CH2CH2CH3
2) CH3OH and HOCH2 CH2OH
3) ICl and Br2 (both have same molecular mass)
4. Name the type(s) of intermolecular forces that exist between molecules in each of
these species:
1) CI4 (dispersion)
2) KBr (ionic + dispersion)
3) CH3F (dipole-dipole + dispersion)
5. In which of the following solvent pairs is each of the following solutes more
soluble (circle the better solvent):
1) I2 in water or CCl4
2) NaCl in water or CH3OH
3) C10H8 in HOCH2 CH2OH or Benzene (C6H6)
6. Arrange the following compounds in order of increased water solubility(from left
(lower solubility) to right (higher solubility):
CH3CH2OH, CH3CH2CH2CH3, CH3Cl, NaCl
CH3CH2CH2CH3, CH3Cl, NaCl, CH3CH2OH
7. Arrange the following 1.0 m solutions in terms of increasing freezing point
depression (from left (lower boiling) to right (higher boiling):
NaCl, glucose (non dissociable), acetic acid (a weak acid)
Glucose, acetic acid, NaCl
8. For the reaction:
N2 + 3H2  2NH3
The reaction rate can be written as::
rate = - [N2]/t = - [H2]/3t = [NH3]/2t
1
9. The vapor pressure of ethanol is 100 mmHg at 34.9 oC. What is its vapor pressure
at 63.5 oC? (Hvap for ethanol is 39.3 kJ/mol, R = 8.314 J K-1mol-1 )
ln (P1/P2) = (Hvap/R){(T1 –T2)/T1T2}
T1 = 34.9 + 273 = 307.9K, T2 = 63.5 + 273 = 336.5K
ln (100/P2) = (39.3*103/8.314){(307.9 –336.5)/ 307.9*336.5 }
ln (100/P2) = -1.3
P2 = 369 mmHg
10. Calculate the molarity of a 1.88 m glucose solution (FW=180g/mol) whose
density is 1.12 g/mL.
g glucose = 1.88 * 180 = 338.4 g
g solution = 1000 + 338.4 = 1338.4 g solution
mL solution = 1338.4 g soln * (mL soln/1.12 g soln) = 1195 mL
M = mol/L soln = 1.88/1.195 = 1.57 M
Answer either question 11 or 12 only
11. The vapor pressure of glucose (FW=180g/mol) in solution is 17.1 mmHg at 20
o
C, whilw that of pure water is 17.25 mmHg at the same temperature. Find the
molality of the solution. (glucose is a nonvolatile solute and FW of water = 18 g/mol)
12. A 100.00-mL sample of an aqueous solution contains 2.2 g of human serum
albumin, a blood plasma protein. The solution has an osmotic pressure of 6.04 torr at
298 K. What is the molar mass of albumin? R = 0.0821 L atm mol-1 K-1
Question No. 3
Question No. 4
Psoln = (1-Xsolute) Posolvent
17.1 = (1-Xsolute) * 17.25
Xsolute = 8.7*10-3
 = MRT
(6.04/760) = M * 0.0821 * 298
M = 3.2*10-4
No. mol = M * VL
No. mol = 3.2*10-4 * 0.100 L = 3.2*10-5
Xsolute = {nsolute/(nsolute + nsolvent)
Nsolvent = 1000/18 = 55.6
FW = {g/No. mol}
FW = {2.2/3.2*10-5}
FW = 6.9*104 g/mol
8.7*10-3= {nsolute/(nsolute +55.6)
Nsolute = 0.48
Since we considered 1000g solvent, the
molality = {mol solute/kg solvent)
m = 0.48 molal
2
13. Calculate the boiling point elevation of water when 478 g of ethylene glycol (a
nonvolatile solute, FW = 62 g/mol) is dissolved in 2500 mL of water. (Kb = 0.52
o
C/m)
Tb = Kb m
No. mol ethylene glycol = 478/62 = 7.71 mol
m = 3.08/2.500 = 3.08
Tb = 0.52 * 3.08 = 1.6 oC
14. In the reaction:
S2O82- + 3I-  2 SO42- + I3The following date was collected:
[S2O82-], M
Rate, M s-1
Experiment
[I-], M
1.
0.080
0.034
2.2x10-8
2.
0.080
0.017
1.1x10-8
3.
0.160
0.017
2.2x10-8
a. Find the rate law
From experiments 1 and 2, the rate with respect to I- is first order
From experiments 2 and3, the rate with respect to S2O82- is first order
Rate = k [S2O82-][I-]
b. Calculate the rate constant
From experiment 1, 2.2*10-8 = k * 0.080 * 0.034
K = 8.1*10-6 M-1S-1
15. The reaction:
2A  B
Is first order with a rate constant of 2.8*10-2 S-1 at 80 oC. How long would it take for
A to decrease from 0.44 M to 0.08M. Calculate the half life of the reaction.
ln[A]t - ln[A]0 = - kt
ln 0.08 – ln 0.44 = - 2.8*10-2 t
t = 61 s
t1/2 = (ln 2/k) = (0.693/2.8*10-2) = 24.7 s
16. NO2 reacts with CO to form products. The reaction is believed to occur in two
steps as below:
1) Write the overall balanced equation:
2) Identify the intermediate, if any. NO3
3) Write the rate law:
From the slow step: Rate = k[NO2]2
3
NO2(g) + CO(g)  NO(g) + CO2(g)