Wednesday, January 5, 2011

Chemistry 11 Assignment

Today & Yesterday's assignment:
- Complete Review & Practice p.450 #1-4
- Complete Molarity Practice Problems
- Complete Dilution Worksheet



p.450
1.
a) 0.04 M
b) 0.011 M
c) 0.0080 M
d) 0.0020 M

2.
a) 0.083 L
b) 0.004 L
c) 0.0025 L
d) 0.005 L

3.
a) [Fe] = 0.12 M, [SO4] = 0.18 M
b) [Li] = 7.2 M, [PO4] = 2.4 M
c) [Ca] = 0.023 M, [OH] = 0.046 M
d) [NH4] = 5.1 M, [PO4] = 1.7 M
e) [Mg] = 0.42 M, [ClO3] = 0.84 M

4.
a) [Mg] = 0.20 M, [SO4] = 0.20 M, [K] = [I] = 0.04 M
b) [Na] = 0.36 M, [K] = 0.08 M, [OH] = 0.44 M
c) [Al] = 0.2 M, [Ba] = 0.2 M, [Br] = 1.2 M
d) [Fe] = 0.10 M, [NH4] = 0.12 M, [Cl] = 0.42 M
e) [Na] = 0.44 M, [K] = 0.28 M, [Cl] = 0.72 M

Molarity Practice
1. 69.1 g
2. 0.29 L
3. 3.51 M
4. 198 g
5. 0.85 M
6. 816 g
7. 1.11 L
8. 0.24 M
9. 0.45 L
10. 1021.9 g
11. 2.08 M
12. 0.38 mol
13. 31.8 M

Dilution Worksheet
1. 0.125 M
2. 0.100 M
3. 50, 000 mL (50 L)
4. 2.07 M
5. 1.2 L

2 comments:

  1. Hey
    On the Molarity Worksheet question 4:

    For some reason we keep getting 198.15 instead of 171.2

    first we figured out the formula (NH4)2SO4
    which has a molar mass of 132.1

    then we wrote the equation as 6M = ? over 0.25
    so we did 6 times 0.25 and got 1.5
    so the equation was 6M = 1.5 over 0.25
    to get the mass we then went 1.5 mol divided by 132.1 (molar mass of ammonium sulfate)and thats how we ended up with 198.15

    Anyways we were just wondering where we went wrong. sorry if it seems kinda confusing how we typed it.

    thanks

    ReplyDelete
  2. Doh, question 4 is worded wrong. It should have said

    "How many grams of Ammonium Sulphate are needed to make a 0.25L solution at 6M?"

    Your solution is correct by the way and I will correct the key.

    ReplyDelete