CE 112L Molar Mass Determination by Freezing Point Depression Pre-Lab Name Show

Question

CE 112L Molar Mass Determination by Freezing Point Depression Pre-Lab Name Show all work and the answer to the correct units, the correct number of significant figures and the notation where necessary for full Make sure you indicate the state (s, l, g, or aq) of all reactants and products in the chemical equations that you write. 5 total points 1. (1 pts) What is the molality of a solution that contains 1.50 g of urea (molar mass 600 g/mol) in 200.0 g of benzene? 2. (2 pts) Calculate the freezing point of a solution containing 5.85 200.0 of water? (Na: 22.99 g/mol; CI: g/mol). g ofNaci in g 3. (2 pts) A solution containing 1.00 g of 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

m = 1.5 g of urea

MW = 60.06

mol = 1.5/60 = 0.02497502497mol urea

m = 200 g B = 0.2 kg

m = mol/kg = 0.02497502497/0.2 = 0.124875 mol per kg

2)

m = 5.85 g of NACl

MW = 58

mol = 5.85/58 = 0.1

m = 200 g = 0.2 kg

molality = mol/kg = 0.1/0.2 = 0.5

then

dTf = -Kf*m*i = -1.86*0.5*2 = -1.86 °C

3)

m = 1 g of X

m = 12.5 g = 0.0125 kg

Kf = 6.9

Tnew = 75.4°C

dTf = 80.26 -75.4 = 4.86 °C

dT= K*m*i

4.86 = 6.9*m*1

m = 4.86/6.9 = 0.704347 molal

molal = mol /kg

mol = molal*kg = 0.704347*0.0125 = 0.0088043375 mol

MW = mass/mol = 1/0.0088043375 = 113.5

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