Questions...

A couple questions came in from Exam 1, Summer 2007:

12. You have prepared a solution by diluting 18.64mL of 1.16M potassium iodate solution to50.00mL with water. What is the concentration of iodate ions in the resulting solution?
a. 3.11 M
b. 0.216 M
c. 0.432 M
d. 0.242 M
e. 2.31 M
For dilutions, you can use the formula C1V1 = C2V2. C1 = 1.16M (the stock concentration); V1 = 18.64mL (the stock volume); C2 = the "new" concentration; V2 = 50.00mL (the diluted volume). Every potassium iodate unit contains 1 iodate ion, so the concentration of potassium iodate is the same as the concentration of iodate.
(1.16M)(18.64mL) = C2(50.00mL)
C2 = 0.432M

13. You have prepared a solution by dissolving 1.38 mols of sugar (C6H12O6) in 500.0g of water. What is the boiling point of the resulting solution?
a. 1.4ºC
b. 105.13ºC
c. 101.4ºC
d. 98.6ºC
e. 5.13ºC
Boiling point elevation uses molality, so:
1.38mols sugar / 0.5000kg solvent = 2.76m
{delta}T = (0.52)(2.76)(1)
NOTE: I haven't included units because they would be hard to type. 0.52 deg.C/m is the boiling point elevation constant for water, it will be given on the front of the exam. Sugar is a molecular solute, so the van't Hoff factor is 1. Solving, the boiling point will CHANGE by 1.4 deg.C, so the boiling point of the solution is 100.0+1.4 = 1.104 deg.C

14. A reaction is found to be second order with respect to reactant A and zero order with respect to reactant B. If [A]o = 0.942M, [B]o = 0.613M and k = 2.49x10-2 M-1min-1, what is the initial rate of the reaction?
a. 1.44x10-2 M/min
b. 4.31x10-2 M/min
c. 2.21x10-2 M/min
d. 9.36x10-3 M/min
e. 2.81x10-2 M/min
You don't need a time for this one, the problem is basically giving you all the parts of the rate law equation except the rate itself. If the reaction is second order w.r.t. A and zero order w.r.t. B, the rate law expression is:Rate = k [A]^2Plugging in the given values for k and [A], should give you the correct answer.

15. A reaction is found to be second order with respect to ammonium ion, a reactant. If [NH4+]o = 3.34M and k = 1.28 M-1sec-1, what will the concentration of ammonium ion be after 2.16minutes have passed?
a. 0.326 M
b. 3.02x10-72 M
c. 0.575 M
d. 6.02x10-3 M
e. 1.18 M

Use the second order integrated rate law. If you got the wrong answer on this one, check your time units, k is given in seconds and the time is given in minutes. Convert t to seconds before you start.
1/[NH4+]t = (1.28 M-1sec-1)(129.6sec) + 1/3.34 = 166.2 M-1
[NH4+]t = 6.02x10-3 M

Other questions, let me know...