Answer:
103.76° C
Explanation:
Let's use Raoult's Law to determine the ration of the mole fraction of the solvent by substituting the given pressure values:
[tex]P_{soln}=X_{H_2O}P^o_{H_2O}\\\\X_{H_2O}=\frac{P_{soln}}{P^o_{H_2O}}\\\\X_{H_2O}=21.0 \ torr/23.78 \ torr\\X_{H_2O}=\frac{^nH_2O}{^nH_2O+n_{solute}}\\\\nH_2O=21.0 \ mol\\\\^nH_2O+n_{solute}=23.78\ mol\\\\n_{solute}=23.78 \ mol-21.0\ mol\\\\=2.78\ mol[/tex]
We then determine the molality of the solution:
[tex]m=\frac{2.78\ mol \ solute}{21 \ mol \ sol}\times \frac{1 \ mol \ solv}{18.0152g \ solv}\times \frac{1000g \ solv}{1kg solv}\\\\=7.3483m[/tex]
#Compute the boiling point elevation as:
[tex]\bbigtriangleup T_b=K_b\times m=(0.512\textdegree C/m)(7.3483m)\\\\=3.76\textdegree C[/tex]#Add to boiling point of solution:
[tex]BP=100.0\textdegree C+\bigtriangleup T_b\\=100.0\textdegree C+3.76\textdegree C\\\\=103.76\textdegree C[/tex]
Hence the boiling point of the aqueous solution is 103.76° C
