Question

When 130. g of urea are dissolved in 950.g of a certain mystery liquid , the freezing point of the solution is 2.90 C less than the freezing point of pure X. Calculate the mass of potassium bromide that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor = 1.9 for potassium bromide in X. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.

Answer 1

Answer:

The mass of potassium bromide that must be dissolved in 950g of the mystery liquid is 135.66g.

Explanation:

To solve this problem is necessary to use one of the Colligative properties, the Freezing-point depression. This property establishes that when a non-volatile solute is added to a liquid solvent the freezing point of the solution will be lower than the pure solvent. Also, the equation given by this property is

ΔT = $i.k_{c}.m$

where ΔT is the freezing-point depression, i is the van't Hoff factor, kc is the  cryoscopic constant and m is the molality.

First, it is necessary to calculate the cryoscopic constant of the solvent (the mystery liquid) with the molality of the urea solution and the freezing-point depression.

The molarity of the solution is $\frac{130g}{60g/mol x950g}x1000 = 2.28m$

Thus, the cryoscopic constant of the solvent is

$k_{c} =\frac{ΔT}{m} = \frac{2.9C}{2.28mol/g} = 1.27C.m^-1$

Second, it is necessary to calculate the molality of the potassium bromide solution with the same cryoscopic constant of the solvent and the correction of the van't Hoff factor

$m=\frac{ΔT}{k_{c}xi} = \frac{2.9C}{1.27C.m^-1x1.9}} =  1.2m$

The mass of potassium bromide that must be dissolved in 950g of the mystery liquid is

$\frac{950gx1.2moles}{1000}x119g/mol = 135.66g$

Answer 2

Answer:

We need 135.5 grams of potassium bromide

Explanation:

Step 1: Data given

Mass of urea = 130 grams

Mass of liquid = 950 grams = 0.950 kg

The freezing point of the solution is 2.90 C less than the freezing point of pure X

The van't Hoff factor = 1.9 for potassium bromide in X

Step 2: Calculate moles urea

Moles urea = mass urea / molar mass urea

Moles urea = 130 grams / 60.06 g/mol

Moles urea = 2.16 moles

Step 3: Calculate molality

Molality = moles urea / mass liquid X

Molality = 2.16 moles / 0.950 kg

Molality = 2.27 molal

Step 4: Calculate Kf

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 2.90 °C

⇒with i = the van't Hoff factor of urea = 1

⇒with Kf = the freezing point depression constant = TO BE DETERMINED

⇒with m = the molality = 2.27 molal

2.90 °C =  1 * Kf * 2.27 molal

Kf = 2.90 / 2.27

Kf = 1.28 °C/m

Step 5: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 2.90 °C

⇒with i = the van't Hoff factor of potassium bromide = 1.9

⇒with Kf = the freezing point depression constant = 1.28 °C/m

⇒with m = the molality = TO BE DETERMINED

2.90 °C = 1.9 * 1.28 °C/m * m

m = 1.19 molal

Step 6: Calculate moles potassium bromide

molality = moles KBr / mass X

1.19 molal = moles KBr / 0.950 kg

moles KBr = 1.19 * 0.950 kg

Moles KBr = 1.13 moles KBr

Step 7: Calculate mass KBr

Mass KBr = moles KBr * molar mass KBr

Mass KBr = 1.13 moles * 119.0 g/mol

Mass KBr = 135.5 grams

We need 135.5 grams of potassium bromide