Question

The vapor pressure of ethanol is 54.68 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 239.0 grams of ethanol to reduce the vapor pressure to 54.11 mm Hg ?

Answer 1

Answer: The mass of estrogen that must be added is 2.83 grams

Step-by-step explanation:

The equation used to calculate relative lowering of vapor pressure follows:

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

where,

`(p^o-p_s)/(p^o)` = relative lowering in vapor pressure

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

`\chi_(solute)` = mole fraction of solute = ?

`p^o` = vapor pressure of pure ethanol = 54.68 mmHg

`p_s` = vapor pressure of solution = 54.11 mmHg

Putting values in above equation, we get:

`(54.68-54.11)/(54.68)=1* \chi_{\text{estrogen}}\\\\\chi_{\text{estrogen}}=0.0104`

This means that 0.0104 moles of estrogen are present in the solution

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of estrogen = 0.0104 moles

Molar mass of estrogen = 272.4 g/mol

Putting values in above equation, we get:

`0.0104mol=\frac{\text{Mass of estrogen}}{272.4g/mol}\\\\\text{Mass of estrogen}=(0.0104mol* 272.4g/mol)=2.83g`

Hence, the mass of estrogen that must be added is 2.83 grams