Question

1. The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is cholesterol .

If 10.34 grams of cholesterol, C27H46O (386.6 g/mol), are dissolved in 157.9 grams of benzene ...

The molality of the solution is_______ m. The freezing point of the solution is________ °C.

2.The boiling point of ethanol CH₃CH₂OH is 78.50°C at 1 atmosphere.

A nonvolatile, nonelectrolyte that dissolves in ethanol is estrogen (estradiol).

How many grams of estrogen, C₁₈H₂₄O₂ (272.4 g/mol), must be dissolved in 293.0 grams of ethanol to raise the boiling point by 0.350 °C ?

_________g estrogen.

Answer 1

1)

Molality is given by number of moles of solute per kg of solvent.

Number of moles of cholesterol = amount of cholesterol (g)/ mw of cholesterol= 10.34 g/386.6 g/mole= 0.02675 moles

Molality of cholesterol solution= number of moles of cholesterol/amount of solvent in kg= 0.02675 mole/0.1579 kg= 0.1694 m

Depression in freezing point = Kf X m

Where Kf = molal freezing point depression (Kf for bezene= 5.12 centigrade/m) and m = molality of solution

Depression in freezing point = 5.12 centigrade/m X 0.1694 m= 0.87 centigrade

Therefore freezing point of solution = freezing point of benzene - depression in freezing point= 5.5 centigrade - 0.87 centigrade= 4.63 centigrade