# What is the molality of an aqueous solution whose new boiling point is #102.5^@ "C"# due to adding a nonvolatile solute?

##### 1 Answer

You know that you have to relate molality with boiling point, so this has to do with** colligative properties**. I had written an explanation on why:

http://socratic.org/questions/how-do-colligative-properties-affect-freezing-point

The above link explains why **adding solute** lowers the freezing point and **raises the boiling point**---the discussed diagram depicts the boiling point as the intersection between the liquid and gas lines, and *the intersection moves to the right* when the purity of the solvent decreases. i.e. *when you add a solute*.

The equation you should use is (*Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-3*):

#\mathbf(DeltaT_"vap" = T_"vap" - T_"vap"^"*" = K_b*m)#

#\mathbf(K_b = (M_"solvent")/("1000 g/kg")(R(T_"vap"^"*")^2)/(DeltabarH_"vap"))# where:

#T_"vap"^"*"# is the boiling point in#"K"# of thepure solution(no solute)#T_"vap"# is the boiling point in#"K"# ofthe solution(with solute)#M_"solvent"# is themolar massof the solvent in#"g/mol"# .#R# is theuniversal gas constant:#"8.314472 J/mol"cdot"K"# #DeltabarH_"vap"# is themolar enthalpy of vaporization, in#"kJ/mol"# (hence, "molar").#K_b# is known as theboiling point elevation constant.#m# is themolality, which is#"mols solute"/"kg solvent"# .

You should get to know the first equation, but generally, you may be given

Since you weren't given **water**, specifically:

#M_"solvent" = "18.015 g/mol"#

#T_"vap" = 102.5^@ "C" = "375.65 K"#

#T_"vap"^"*" = 100.0^@ "C" = "373.15 K"#

#DeltabarH_"vap" = "40.7 kJ/mol"#

I would solve for

#K_b = ("18.015" cancel"g""/mol")/("1000" cancel"g""/kg") (("8.314472" cancel"J""/"cancel"mol"cdotcancel"K")("373.15 K")^(cancel(2)))/("40700" cancel"J/mol")#

#= color(green)("0.5124 K"*"kg/mol")#

That was the hard part. Now, just solve for ** smaller** than the effect of freezing point depression though, so I would expect a concentration higher than

*The same amount of solute will raise the boiling point less than it will lower the freezing point.*

#m = (T_"vap" - T_"vap"^"*")/K_b#

#= (2.50 cancel"K")/("0.5124" cancel"K"*"kg/mol")#

#=# #color(blue)("4.879 mol particles/kg water")#