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If pure water boils at 99.8 degrees Celsius, what is the expected elevated boiling point of a solution with 2.50 g of [tex]CaCl_2[/tex] in 50.0 mL (i.e., 50.0 g) of [tex]H_2O[/tex]? For [tex]CaCl_2[/tex], [tex]i = 3[/tex].

Answer :

INFORMATION:

We know that:

- pure water boils at 99.8 degrees celcius

And we must calculate the expected elevated boiling point of a solution of 2.50g of CaCl2, in 50.0mL (i.e., 50.0g) of H2O

STEP BY STEP EXPLANATION:

To calculate it, we need to use that:

Boiling point of solution = boiling point of pure solvent + boiling point elevation (ΔTb)

The elevation in boiling point (ΔTb) is proportional to the concentration of the solute in the solution. It can be calculated via the following equation.

[tex]ΔTb=i\times k_b\times m[/tex]

Where,

- i is the Van’t Hoff factor

- Kb is the ebullioscopic constant

- m is the molality of the solute

From given information, we know that:

- i = 3

Now, the ebullioscopic constant (Kb) is often expressed in terms of °C * kg * mol^-1. The value of Kb for water is 0.512.

So, kb = 0.512 °C * kg * mol^-1

Then, we must calculate the molality

[tex]\begin{gathered} Molality=\frac{\text{ moles of solute}}{\text{ kg of solvent}} \\ Molality=\frac{\frac{2.5g}{110.98\frac{g}{mol}}}{0.05kg}=0.45\frac{mol}{kg} \end{gathered}[/tex]

So, m = 0.45 mol/kg

Replacing the values in the formula for ΔTb

[tex]\begin{gathered} ΔT_b=3\times0.512\frac{\degree C\cdot kg}{mol}\times0.45\frac{mol}{kg} \\ ΔT_b=0.69\degree C \end{gathered}[/tex]

Finally, the expected elevated boiling point of the solution would be

[tex]\text{ Boling point of solution}=99.8\degree C+0.69\degree C=100.49\degree C[/tex]

ANSWER:

The expected elevated boiling point of the solution is 100.49 °C

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Rewritten by : Jeany

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