# Question #ab304

##### 1 Answer

I got

My best guess is that the gram atomic mass is supposed to be

This seems to be combining the idea of vapor pressures via Raoult's law and utilizing the mol fraction, but the question is conflicting... not sure why it asks for the molecular formula if it gives a gram atomic mass.

Define the solute as

#P_Y^"*" = "854 torr"#

#P_Y = "848.9 torr"# where

#"*"# indicates for the pure solvent.

From Raoult's law for ideal solutions, we have:

#P_Y = chi_(Y(l))P_Y^"*"#

#= (1 - chi_(X(l)))P_Y^"*"# ,where

#chi_(a(l)) = (n_(a(l)))/(n_(a(l)) + n_(b(l)))# is the mol fraction of#a# in the solution phase.

Since we know the drop in vapor pressure of the solvent

#chi_(X(l)) = 1 - P_Y/P_Y^"*"#

#= 1 - ("848.9 torr")/("854 torr")#

#= 0.005972#

We can obtain the mols of

#chi_(X(l)) = 0.005972 = n_(X(l))/(n_(X(l)) + n_(Y(l)))#

We know the molar mass of the solvent, so we can get the mols of solvent:

#100 cancel("g CS"_2) xx "1 mol"/(76.141 cancel("g CS"_2)) = "1.313 mols CS"_2# #"in solution"#

Therefore, we have:

#0.005972 = n_(X(l))/(n_(X(l)) + "1.313 mols")#

#=> 0.005972n_(X(l)) + 0.005972(1.313) = n_(X(l))#

#=> 0.9940n_(X(l)) = 0.007843#

#=> n_(X(l)) = "0.007890 mols"#

So, the **molecular mass** should be:

#"2.0 g"/"0.007890 mols" = color(blue)("253.47 g/mol")# .