Question #d3727

The problem provides you with the mass of the solute and asks for its molar mass, which is another way of saying that you need to determine the number of moles of solute present in that solution.

Now, you know that pure water has a vapor pressure of #"148.361 torr"# at #60.0^@"C"#.

In order to be able to compare this vapor pressure with the solution's vapor pressure at the same temperature, you need to convert either atm to torr or vice versa by using the conversion factor

#"1 atm " = " 760 torr"#

So, your solution has a vapor pressure of

#0.1876color(red)(cancel(color(black)("atm"))) * "760 torr"/(1color(red)(cancel(color(black)("atm")))) = "142.576 torr"#

This is consistent with the fact that a solute has been dissolved in pure water, since the vapor pressure of the solution is always lower than the vapor pressure of the pure solvent.

You're dealing with a non-volatile solute, which tells you that the vapor pressure of the solution will depends on two things

• the vapor pressure of the pure solvent at that temperature, #P_"solvent"^@#
• the mole fraction of the solvent in the solution, #chi_"solvent"#

Mathematically, this is written as

#color(blue)(P_"sol" = chi_"solvent" xx P_"solvent"^@)#

Use water's molar mass to determine how many moles you have in that sample

#90.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.996 moles H"_2"O"#

Let's assume that #n# represents the number of moles of solute present in solution. The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution

#chi_"water" = (4.996color(red)(cancel(color(black)("moles"))))/((4.996 + n)color(red)(cancel(color(black)("moles")))) = 4.996/(4.996 + n)#

Plug this into the equation for the vapor pressure of the solution to get

#P_"sol" = 4.996/(4.996 + n) * P_"solvent"^@#

This will be equivalent to

#4.996/(4.996 + n) = (142.576 color(red)(cancel(color(black)("torr"))))/(143.316color(red)(cancel(color(black)("torr"))))#

#4.996/(4.996 + n) = 0.99484#

Rearrange to solve for #n#

#0.99484 * (4.996 + n) = 4.996#

#n = (4.996 - 0.99484 * 4.996)/0.99484 = 0.02591#

So, your solution contains #0.02591# moles of solute in #"10.6 g"# of solute. Since molar mass is defined as the mass of one mole of a substance, you can say that

#1color(red)(cancel(color(black)("mole solute"))) * "10.6 g"/(0.02591color(red)(cancel(color(black)("moles solute")))) = "409.12 g"#

Therefore, the solute's molar mass will be

#M_M = color(green)("409 g mol"^(-1)) -># rounded to three sig figs