A solvent, S, having a density of 1.05 g/mL and K f = 3.90 ºC/m was uniformly co

Question

A solvent, S, having a density of 1.05 g/mL and Kf = 3.90 ºC/m was uniformly cooled. A graph of temperature-time readings showed a plateau (flat) region [S(l)/S(s)] at 16.6 ºC. A solution of 2.41 g   compound B was dissolved in  23.6 mL   of S. The solution was uniformly cooled and dropped in temperature until reaching  3.29 ºC  . It remained at that temperature for several minutes.

1) How many grams of S were used?

2)What is the freezing point of S in ºC?

3)What was the molality of compound B in the solution?

4)What is the molecular weight of compound B?

Explanation / Answer

1)

volume of S = 23.6 mL

density = 1.05 g/mL

mass of S = 23.6 x 1.05 = 24.78 g

mass of S used = 24.8 g

2)

freezing point = 16.6 oC

3)

delta Tf = Kf x m

16.6 - 3.29 = 3.90 x m

m = 3.41

molality of compound B = 3.41 m

4)

molality = moles / mass of solvent in kg

3.41 = moles / 0.0248

moles = 0.0846 mol

moles = mass / moalr mass

0.0846 = 2.41 / molar mass

molar mass = 28.5 g/mol

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