Answer to Question #136865 in General Chemistry for Corriauna Woods

Question

The vapor pressure of water is 23.76 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in water is urea.

Calculate the vapor pressure of the solution at 25 °C when 7.936 grams of urea, CH4N2O (60.10 g/mol), are dissolved in 186.1 grams of water.

water = H2O = 18.02 g/mol.

VP(solution) =
mm Hg

The vapor pressure of ethanol is 54.68 mm Hg at 25°C.

How many grams of testosterone, C19H28O2, a nonvolatile, nonelectrolyte (MW = 288.4 g/mol), must be added to 240.1 grams of ethanol to reduce the vapor pressure to 53.93 mm Hg ?

ethanol = CH3CH2OH = 46.07 g/mol.

g testosterone

Answer 1

2nd question.

We use the same formula for this problem.

Let w gram of testosterone must be added.

Mol of solute n2 = w\288.4

Mol of solvent ethnol n1 = 240.1 \ 46.07

= 5.21 mol

Relative lowering of vapour pressure

= ( P0 - P) \ P0

= (54.68 - 53.93) \53.93

=0.75\53.93 = 0.014

So now we have to find out x2 and equate to it. By simplifying the concept that here mole of solute is very less compair to mol of solvent then x2 = n2/n1

=( w/288.4)/5.21

= w/1502.6

This is also equal to 0.014

So w/1502.6 = 0.014

w = 1502.6 × 0.014 = 21.03 gram (ans)