Question

What is the boiling point of an aqueous solution that freezes at -2.05 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your answer using 2 decimal places!!!!

Answer 1

The boiling point of the aqueous solution is 99.44 °C.

Step-by-step explanation:

The boiling point of an aqueous solution can be determined using the formula:

Tbp = Tbpsolvent + (Kbp * m)

where:

Tbp is the boiling point of the solution

Tbpsolvent is the boiling point of the pure solvent

Kbp is the ebullioscopic constant for the solvent

m is the molality of the solution

In this case, we are given that the freezing point of the solution is -2.05 degrees C. We can use the formula for freezing point depression to find the molality:

AT = Kf * m

Rearranging the formula, we can solve for m:

m = AT / Kf

Substituting the given values:

m = (-2.05 °C) / (1.86 °C kg.mol-¹)

Rounding to 2 decimal places:

m = -1.10 mol/kg

Now, we can use the formula for boiling point elevation to find the boiling point:

Tbp = Tbpsolvent + (Kbp * m)

Substituting the given values:

Tbp = 100.00 °C + (0.512 °C/m * -1.10 mol/kg)

Calculating:

Tbp = 100.00 °C - 0.56 °C = 99.44 °C

Therefore, the boiling point of the aqueous solution is 99.44 °C.

Answer 2

To solve this problem we will apply the concepts of Boiling Point elevation and Freezing Point Depression. The mathematical expression that allows us to find the temperature range in which these phenomena occur is given by

Boiling Point Elevation

`\Delta T_b = K_b m`

Here,

`K_b` = Constant ( Different for each solvent)

m = Molality

Freezing Point Depression

`\Delta T_f = K_f m`

Here,

`K_f =` Constant ( Different for each solvent)

m = Molality

According to the statement we have that

`\Delta T_f = T_0 -T_f = 0-(-2.05)`

`\Delta T_f = 2.05\°C`

From the two previous relation we can find the ratio between them, therefore

`(\Delta T_b)/(\Delta T_f) = (K_b)/(K_f)`

`(\Delta T_b)/(\Delta T_f) = (0.512)/(1.86)`

`(\Delta T_b)/(\Delta T_f) = 0.275`

We already know the change in the freezing point, then

`\Delta T_b = 0.275 (\Delta T_f)`

`\Delta T_b = 0.275 (2.05)`

`\Delta T_b = 0.5643\°C`

The temperature difference in the boiling point is 100°C (Aqueous solution), therefore

`T_b -100 = 0.5643`

`T_b = 100.56\°C`

Therefore the boiling point of an aqueous solution is
`100.56\°C`