4.3Lab # 7Freezing Point DepressionQuickTimeᆰ and adecompressorare needed to see this picture.Jennie Chang (odd) Allison Yee (partner) 4/5/11
CALCULATIONS:1. Molality Equation: M = moles / kg solventSample from data:0.840g NaCl x 1 mol NaCl= 0.0144 mol NaCl58.5g NaClm= 0.0144 mol NaCl= 2.47 m0.00584 kg H2O2. Calculate the freezing point of your solutionEquation: ∆TFP= mx KFPx ίSample from data:∆TFP= 2.47mx 1.86 °C/mx 2 =-9.19°C3. MolalityEquation: m= ∆TFP/ (KFPx ί)Sample from data: m= -10°C / (1.86 °C/mx 2) = 2.69 m3. Percent errorEquation: E – Tx 100%T2.69 – 2.47x 100% = 8.91%2.47TABLE OF RESULTS:Molality (Using data)2.47 m
Freezing point-9.19°CMolality (Using lowest temperature)2.69 mPercent error8.91%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