Question

What is the boiling point of an aqueous solution that has a vapor pressure of 21.0 torr at 25 ∘C? (P∘H2O=23.78 torr; Kb= 0.512 ∘C/m).

Answer 1

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:

$P_{soln}=X_{H_2O}P^o_{H_2O}\\\\X_{H_2O}=\frac{P_{soln}}{P^o_{H_2O}}\\\\X_{H_2O}=21.0 \ torr/23.78 \ torr\\X_{H_2O}=\frac{^nH_2O}{^nH_2O+n_{solute}}\\\\nH_2O=21.0 \ mol\\\\^nH_2O+n_{solute}=23.78\ mol\\\\n_{solute}=23.78 \ mol-21.0\ mol\\\\=2.78\ mol$

We then determine the molality of the solution:

$m=\frac{2.78\ mol \ solute}{21 \ mol \ sol}\times \frac{1 \ mol \ solv}{18.0152g \ solv}\times \frac{1000g \ solv}{1kg solv}\\\\=7.3483m$

#Compute the boiling point elevation as:

$\bbigtriangleup T_b=K_b\times m=(0.512\textdegree C/m)(7.3483m)\\\\=3.76\textdegree C$#Add to boiling point of solution:

$BP=100.0\textdegree C+\bigtriangleup T_b\\=100.0\textdegree C+3.76\textdegree C\\\\=103.76\textdegree C$

Hence the boiling point of the aqueous solution is 103.76° C