Question

What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your answer to 2 decimal places.

Answer 1

The freezing point of an aqueous solution can be calculated using the freezing point depression equation ΔTf = Kf * m. In this case, the solution boils at 101°C, indicating the boiling point elevation constant (Kb) is provided. Assuming complete dissociation, we can calculate the freezing point depression to be -0.512°C.

Step-by-step explanation:

The freezing point depression can be calculated by using the equation:

ΔTf = Kf * m

Where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution.

In this case, since we have the boiling point elevation constant (Kb), we need to use the equation:

ΔTf = Kb * m

Given that Kb is 0.512 K/m and assuming complete dissociation, a 1.0 m aqueous solution of a solute will contain 1.0 mol of particles per kilogram of water. Therefore, the freezing point depression (ΔTf) will be:

ΔTf = 0.512 K/m * 1.0 m = 0.512 K

Since the freezing point of pure water is 0°C, the freezing point of the aqueous solution will be:

0°C - 0.512 K = -0.512°C

Answer 2

-3.63 degree Celsius

Step-by-step explanation:

We are given that

Boiling point of solution=
`T_b=101^(\circ)` C

Boiling point water=100 degree Celsius

`K_f=1.86K/m`

`K_b=0.512 K/m`

`\Delta T_b=T-T_0`

Where
`T`=Boiling point of solution

`T_0=`Boiling point of pure solvent

`\Delta T_b=101-100=1^(\circ)`C

`\Delta T_b=k_bm`

Using the formula

`1=0.512* m`

Molality,
`m=(1)/(0.512)` m

`\Delta T_f=k_fm`

Using the formula

`\Delta T_f=(1)/(0.512)* 1.86`

`\Delta T_f=3.63 C`

We know that

`\Delta T_f=T_0-T_1`

Where
`T_0` =Freezing point of solvent

`T_1=` Freezing point of solution

Using the formula

`3.63=0-T_1`

Freezing point of water=0 degree Celsius

`T_1=0-3.63=-3.63 C`

Hence, the freezing point of solution=-3.63 degree Celsius