When 8.05 g of an unknown compound X was dissolved in 100. g of benzene (C6H6), the vapor pressure of the benzene decreased from 100.0 torr to 94.8 torr at 26 degrees C. What is the mole fraction and molar mass of X?

I got #"114.51 g/mol"#.

The reduction of vapor pressure is a colligative property. From Raoult's law, we have:

#P_j = chi_jP_j^"*"#,

where #P_j# is the partial pressure above the solution from component #j#, #P_j^"*"# is the pure vapor pressure of #j# by itself, and #chi_j# is the mol fraction of #j# in the solution.

(We consider #bbj# to be the solvent and #bbi# to be the solute.)

But mol fractions are always #<= 1#. Therefore, for a two-component solution, #chi_j = 1 - chi_i#. Thus:

#P_j = (1 - chi_i) P_j^"*"#

Since #chi_j <=1#, it follows that #1 - chi_i <= 1# as well, or that #chi_i >= 0#. Therefore, adding any solute into a solvent decreases its vapor pressure.

This reduction is calculated as follows. Define #DeltaP_j = P_j - P_j^"*"# to get:

#color(green)(DeltaP_j) = chi_jP_j^"*" - P_j^"*"#

#= (chi_j - 1)P_j^"*"#

#= (1 - chi_i - 1)P_j^"*"#

#= color(green)(-chi_iP_j^"*")#

Since #P_j^"*"# and #chi_i# are positive, the reduction in vapor pressure is of course negatively-signed. From this equation then, we can determine the mol fraction of what was dissolved in benzene.

Plug #P_j = "94.8 torr"# and #P_j^"*" = "100.00 torr"# in to get:

#(94.8 - 100.00) "torr" = -chi_i ("100.00 torr")#

#=> color(blue)(chi_i = 0.052)#

Now that we have the mol fraction of #X#, we can find the molar mass of #X# simply by solving for the mols of #X#, and dividing the given mass by the mols of #X#.

#0.052 = (n_X)/(n_X + n_"Ben")#

#=> 0.052n_X + 0.052n_"Ben" = n_X#

#=> (0.052 - 1)n_X = -0.052n_"Ben"#

#=>n_X = 0.0549n_"Ben"#

But we know the molar mass of benzene; it's #6xx12.011 + 6xx1.0079 = "78.1134 g/mol"#. So, we get the mols of benzene by its molar mass:

#"100.0 g Benzene" xx "1 mol"/"78.1134 g" = "1.2802 mols Benzene solvent"#

Therefore, the mols of #X# are:

#n_X = 0.0549("1.2802 mols") = "0.0703 mols X"#

This gives a molar mass for #X# as:

#color(blue)(M_(m,X)) = "8.05 g X"/"0.0703 mols X" = color(blue)("114.51 g/mol")#