Question

Assume that you were assigned BaCl2 in lab. The water in your test tube weighed 23.651 g. Following the procedure in the lab manual, you determined that freezing point of water is 0.2oC. You weighed out 1.014 g of salt and added it to the original test tube, then determined that the freezing point was -0.49oC. Based on these experimental parameters, calculate the van't Hoff factor for BaCl2. Note that these are made-up numbers, and do not reflect a true experiment. Report your answer to two places after the decimal.

Answer 1

Answer:

The van't Hoff factor of BaCl2 is 1.80

Explanation:

Step 1: Data given

Mass of water = 23.651 grams

The freezing point of water = 0.2 °C

Mass of salt (BaCl2) = 1.014 grams

Molar mass BaCl2 = 208.23 g/mol

the freezing point of solution = -0.49 °C

Freezing point depression constant of water = 1.86 °C/m

Step 2: Calculate moles BaCl2

Moles BaCl2 = mass BaCl2 / molar mass BaCl2

Moles BaCl2 = 1.014 grams / 208.23 g/mol

Moles BaCl2 = 0.00487 moles

Step 3: Calculate molality

Molality = moles BaCl2 / mass H2O

Molality = 0.00487 moles / 0.023651 kg

Molality = 0.206 molal

Step 4: Calculate the van't Hoff factor

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 0.69 °C

⇒with i = the van't Hoff facor of BaCl2 = TO BE DETERMINED

⇒with Kf = the freezing point depression constant = 1.86 °C/m

⇒with m = the molality = 0.206 molal

0.69 °C = i * 1.86 °C/m * 0.206 m

i = 1.80

The van't Hoff factor of BaCl2 is 1.80