lab 7 - 4.3 Lab 7 Freezing Point Depression QuickTime and a...

4.3
Lab # 7
Freezing Point Depression
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Jennie Chang (odd )
Allison Yee (partner)
4/5/11

CALCULATIONS:
1. Molality
Equation: M = moles / kg solvent
Sample from data:
0.840g NaCl x 1 mol NaCl = 0.0144 mol NaCl
58.5g NaCl
m = 0.0144 mol NaCl = 2.47 m
0.00584 kg H
2
O
2. Calculate the freezing point of your solution
Equation: ∆T
FP
= m x K
FP
x ί
Sample from data:
∆T
FP
= 2.47 m x 1.86 °C/ m x 2 = -9.19 °C
3. Molality
Equation: m = ∆T
FP
/ (K
FP
x ί)
Sample from data:
m = -10°C / (1.86 °C/ m x 2) = 2.69 m
3. Percent error
Equation: E – T x 100%
T
2.69 – 2.47 x 100% = 8.91%
2.47
TABLE OF RESULTS:
Molality (Using data) 2.47 m

Freezing point -9.19°C
Molality (Using lowest
temperature)
2.69 m
Percent error 8.91%
DISCUSSION:
This lab was extremely fun and delicious! However, there were a few
sources of error. 1) At first, we couldn’t find the table spoon so we
decided to just go ahead and eyeball the vanilla extract. However, we
didn’t realize how big the hole was on the vanilla extract, so we
ended up pouring in way too much. A suggested improvement