CE 112L Molar Mass Determination by Freezing Point Depression Pre- Lab Section:

Question

CE 112L Molar Mass Determination by Freezing Point Depression Pre- Lab Section: scientific and the answer to the correct units, the correct number of significant figures and/or the notation where necessary for full credit. Make sure you indicate the (s, l, or aq) of all reactants and products in the chemical equations that you 5 total points 1. pts) what is the molality of a solution that contains 1.50 g of urea omolar mass- 60.0 g/mol) in 200.0 2. (2 pts) Calculate the freezing point of a solution containing 5.85 g of Naci in 200.0 g of water? (Na: 22.99 g/mol: CI: 35.54 g/mol). 3. (2 pts) A solution containing 1.00 gof an unknown substance in 12.5 g of naphthalene (Kr 6.9°C/m) was found to freeze at 75.4°C. What is the molar mass of the unknown substance?

Explanation / Answer

Post one more question to get the remaining answers.

1) Molar mass of Urea = 60 gm/mol

Number of moles of Urea = 1.50/60 = 0.025 moles

molality = number of moles of solute/mass of solvent in Kg

=> 0.025/(200/1000)

=> 0.025 * 5

=> 0.125 m

2) Molar mass of NaCl = 22.99 + 35.54 = 58.53 gm/mol

Number of moles of NaCl = mass of NaCl/molar mass = 5.85/58.53 = 0.0999 moles

molality = number of moles of solute/per kg of solvent = 0.0999/(200/1000) = 0.4995 m

Change in freezing point = i (vont hoff factor) * Kf * m

=> 2 * 1.86 * 0.4995

=> 1.858

Freezing Point of water = 0 - 1.85 = -1.85 C

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