Respuesta :
Answer:
The mass we used from the solute is 175 g
Explanation:
ΔT = Kf . m . i → Formula for colligative property of freezing point depression
ΔT = T° freezing of pure solvent - T° freezing of solution
0°C - (-1.3°C) = 1.3 °C
Kf if the cryoscopic constant of solvent. In this case, for water and the value is 1.86 °C/m
m = molality → moles of solute / 1 kg of solvent
Let's replace → 1.3°C = 1.86°C . m
m = 1.3°C / 1.86 m → 0.731 mol/kg
0.731 mol/kg is the molality, moles of solute in 1kg of solvent. But in this solution we do not have 1 kg of solvent, we have 4.79 kg. So:
0.731 mol/kg . 4.79 kg = 3.501 moles
This is the moles of the solute we used, so let's convert them to mass:
3.501 mol . 50 g / 1 mol = 175 g
Note: Freezing point depression formula is completed with the Van't Hoff factor (i); this number corresponds to the ions dissolved in the solution.
In this case the solute is non volatile and nonionizing, so i = 1
ΔT = Kf . m . i
The mass of a nonvolatile, nonionizing compound is 175 g.
Freezing point depression:
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. It is given by:
ΔT =kf *i*m
ΔT = T° freezing of pure solvent - T° freezing of solution
0°C - (-1.3°C) = 1.3 °C
kf = constant of solvent. In this case, for water and the value is 1.86 °C/m
→ Molality:
It is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.
m = molality → moles of solute / 1 kg of solvent
Let's replace → 1.3°C = 1.86°C . m
m = 1.3°C / 1.86 m → 0.731 mol/kg
→ Calculation for number of moles:
0.731 mol/kg . 4.79 kg = 3.501 moles
This is the moles of the solute we used, so let's convert them to mass:
3.501 mol . 50 g / 1 mol = 175 g
It offers the on the colligative properties of solutions.
In this case the solute is non volatile and nonionizing, so i = 1
ΔT = Kf . m . i
Thus, the mass of a nonvolatile, nonionizing compound is 175 g.
Find more information about Molality here:
brainly.com/question/1370684
