A solvent, S, having a density of 1.05 g/ml and Kf = 3.9 degrees C/m was uniform

Question

A solvent, S, having a density of 1.05 g/ml and Kf = 3.9 degrees C/m was uniformly cooled. A graph of temperature-time readings showed a plateau (flat) region [a S(1)/S(s)] slush at 6.65 degree C. A solution of 2.74 g compound b was dissolved in 15.5 ml of S. The solution was uniformly cooled and dropped in temperature until reaching 4.35 degrees C. It remained at that temperature for several minutes.

1. How many grams of S were used?

2. What was the molality of compound B in the solution?

3. What is the molecular weight of compound B?

Explanation / Answer

(1) Mass of S = volume x density of S

= 15.5 x 1.05

= 16.275 g = 16.3 g

(2) Freezing point depression DTf = 6.65 - 4.35 = 2.3 deg C

Molality = DTf/Kf

= 2.3/3.9

= 0.5897 m = 0.59 m

(3) Moles of B = molality x mass of S in kg

= 0.5897 x 0.016275 = 0.009598 mol

Molecular weight = mass/moles of B

= 2.74/0.009598

= 285 g/mol

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