Respuesta :
Answer:
A. 256
Explanation:
In a solution where a liquid is the sovent, we'll use the van't Hoff factor, which is the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to 1.
ΔTemp.f = i * Kf * b
where,
ΔTemp.f = the freezing-point depression;
i = the van't Hoff factor
Kf = the cryoscopic constant of the solvent;
b = the molality of the solution.
So the freezing-point depression by definition is the difference between the the freezing point of the pure solvent and the freesing point of the solution.
Mathematically,
ΔTemp.f = Temp.f° - Temp.f
where,
Temp.f° = the freezing point of the pure solvent.
Temp.f = the freezin point of the solution.
Freezing point of pure water = 0°C
ΔTemp.f = 0 - (-1.32)
= 1.32°C
i = 1,
Kf = 1.86 °Ckg/mol
Solving for the molality, b = ΔTemp.f/( i * Kf)
= 1.32/(1*1.86)
= 0.71 mol/kg
Converting from mol/kg to mol/g,
0.71 mol/kg * 1kg/1000g
= 0.00071 mol/g.
Mass of solvent = 110g
Number of moles = mass * molality
= 0.00071 * 110
= 0.078 mol.
To calculate molar mass,
Molar mass (g/mol) = mass/number of moles
Mass of solute (liquid) = 20g
Molar mass = 20/0.078
= 256.2 g/mol
A solution containing 20.0 g of an unknown liquid (molar mass 256 g/mol) and 110.0 g water has a freezing point of -1.32 °C.
What is the freezing point depression?
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent.
- Step 1: Calculate the molality of the solution.
We will use the following expression for non-electrolytes.
ΔT = Kf × b
b = ΔT/Kf = 1.32 °C/(1.86 °C/m) = 0.710 m
where,
- ΔT is the freezing point depression.
- Kf is the cryoscopic constant.
- b is the molality.
- Step 2. Calculate the molar mass of the unknown liquid (solute).
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
molar mass solute = mass solute / b × kg solvent
molar mass solute = 20.0 g / (0.710 mol/kg) × 0.1100 kg = 256 g/mol
A solution containing 20.0 g of an unknown liquid (molar mass 256 g/mol) and 110.0 g water has a freezing point of -1.32 °C.
Learn more about colligative properties here: https://brainly.com/question/23480761
