Raoult’s law and vapor pressure (8282)

, por F_y_Q

|

Calculate the vapor pressure at 20\ ^oC of a solution containing 150 grams of glucose dissolved in 140 grams of ethyl alcohol. The vapor pressure of ethyl alcohol at 20\ ^oC is 43 mm Hg.

P.-S.

Glucose is \ce{C6H12O6} and has a molecular mass of 180 g/mol. Ethyl alcohol is \ce{CH3CH2OH} and has a molecular mass of 46 g/mol. Using this data, calculate the moles of each substance and determine the mole fraction of alcohol in the mixture.

150\ \cancel{g}\ \ce{C6H12O6}\cdot \frac{1\ \text{mol}}{180\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{0.833\ mol\ \ce{C6H12O6}}}

140\ \cancel{g}\ \ce{C2H6O}\cdot \frac{1\ \text{mol}}{46\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{3.043\ mol\ \ce{C2H6O}}}

The mole fraction of alcohol is:

x_{\ce{C2H6O}} = \frac{n_{\ce{C2H6O}}}{n_{\ce{C2H6O}} + n_{\ce{C6H12O6}}} = \frac{3.043\ \cancel{\text{mol}}}{(0.833 + 3.043)\ \cancel{\text{mol}}} = \color[RGB]{0,112,192}{\bf 0.785}

Now calculate the vapor pressure using Raoult’s law:

P_{\ce{C2H6O}} = x_{\ce{C2H6O}}\cdot P_T = 0.785\cdot 43\ \text{mm Hg} = \fbox{\color[RGB]{192,0,0}{\textbf{33.75\ \ce{mm\ Hg}}}}