Question

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is testosterone. Calculate the vapor pressure of the solution at 25 °C when 7.752 grams of testosterone, C19H28O2 (288.4 g/mol), are dissolved in 208.0 grams of diethyl ether. diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol.

Answer 1

Answer: The vapor pressure of solution is 459.17 mmHg

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For testosterone:

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

`\text{Moles of testosterone}=(7.752g)/(288.4g/mol)=0.027mol`

• For diethyl ether:

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

`\text{Moles of diethyl ether}=(208.0g)/(74.12g/mol)=2.81mol`

Mole fraction of a substance is calculated by using the equation:

`\chi_A=(n_A)/(n_A+n_B)`

`\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}`

`\chi_{\text{testosterone}}=(0.027)/(0.027+2.81)\\\\\chi_{\text{testosterone}}=0.0095`

The formula for relative lowering of vapor pressure will be:

`(p^o-p_s)/(p^o)=i* \chi_{\text{solute}}`

where,

`p^o` = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

`p^s` = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

`\chi_{\text{solute}}` = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

`(463.57-p^s)/(463.57)=1* 0.0095\\\\p^s=459.17mmHg`

Hence, the vapor pressure of solution is 459.17 mmHg