The vapor pressure of ethanol is 54.68 mm Hg at 25°C. How many grams of estrogen

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 157.2 grams of ethanol to reduce the vapor pressure to 53.06 mm Hg ? ethanol = CH3CH2OH = 46.07

g/mol. g estrogen

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform , CHCl3, is 173.11 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 7.695 grams of the compound were dissolved in 186.2 grams of chloroform, the vapor pressure of the solution was 170.03 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound? chloroform = CHCl3 = 119.40 g/mol.

MW = g/mol

Explanation / Answer

(a)

According to Raoult's law for the case of non volatile compounds, we have :

P = PA0*XA

Here, PA0 is the vapor pressure of ethanol in pure form

XA is the mole fraction

Given : PA0 = 54.68 mm Hg

P = 53.06 mm Hg

Thus, we have : XA = PA/PA0 = 53.06/54.68 = 0.97

Let the mass of estrogen added be x grams.

Thus, moles added = mass/MW = x/272.4 moles

MW of ethanol = 46.07 grams

Thus, moles of ethanol = 1 mole

Mole fraction XA = moles of ethanol / Total moles = 1/(1 + x/272.4 ) = 0.97

Solving we get :

x = 8.424 g

(b)

Moles of chloform = mass/MW = 186.2/119.4 = 1.56 moles

Given :

PA0 = 173.11 mm Hg

P = 170.03 mm Hg

Applying the same formula as above, we get :

XA = 170.03/173.11 = 0.982

Assume the moles of compound present as x

Thus, we have :

XA = 0.982 = 1.56/(1.56 + x)

Solving we get : x = 0.0286 moles

Thus, MW of compound = mass/moles = 7.695/0.0286 = 269.05 g

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