What is the molar mass for a solute if #"12.0 g"# of it is dissolved in #"108 g"# of water, resulting in a vapor pressure that is #10%# lower?

The solute apparently is water? This problem makes no physical sense, because water does not change the composition of water.

You know that the solute does not vaporize, so Raoult's law applies. It is because it is nonvolatile that the vapor pressure of the solution is lowered relative to the pure solvent. Otherwise, the vapor pressure would be suppressed by less.

#P_j = chi_(j(l))P_j^"*"#

where:

• #P_j# is the vapor pressure of the solvent in solution.
• #P_j^"*"# is that of the pure solvent.
• #chi_(j(l))# is the mol fraction of solvent #j# in the liquid phase. We also have solute #i#.

In this case, you have enough info to find the molality. What we actually want to solve is within the mol fraction of the solvent.

#chi_(j(l)) = n_j/(n_i + n_j) = (108 cancel("g H"_2"O") xx "1 mol"/(18.015 cancel("g H"_2"O")))/("12 g solute" xx "1 mol"/(x " g solute") + 108 cancel("g solute") xx "1 mol"/(18.015 cancel("g H"_2"O")))#

#= ("5.995 mols H"_2"O")/(12/x "mols solute" + "5.995 mols H"_2"O")#

Next, the vapor pressure was lowered by #10%#, so:

#P_j = 0.9P_j^"*"#

This gives a mol fraction of:

#chi_(j(l)) = (P_j)/(P_j^"*") = (0.9P_j^"*")/(P_j^"*") = 0.9#

And so, the molecular mass of the solute can be solved for. We omit units for this until the end.

#0.9 = (5.995)/(12/x + 5.995)#

#6.661 = 12/x + 5.995#

#0.666 = 12/x#

#color(blue)(x = "18.015 g/mol")#

... well, apparently, APPARENTLY, someone had the bright idea of dissolving water in water and calling this a problem. This makes no physical sense, because the solute IS the solvent.