Respuesta :
Answer:
molr mass (mm):
mm = 342.43 g/mol
Explanation:
freezing point (ΔTf):
- ΔTf = - Kf*mb
∴ ΔTf = - 1.3°C
∴ a: water (solvent)
∴ b: solute
∴ mb: molality of the solute [=] mol/Kg
∴ Kf: cryoscopic constant
⇒ Kf water = 1.86 °C.Kg/mol
∴ wb = 35.9 g
∴ wa = 150.0 g = 0.150 Kg
∴ molar mass solute (mm) [=] g/mol
⇒ mb = - ΔTf/Kf
⇒ mb = - ( - 1.3°C)/(1.86 °CKg/mol)
⇒ mb = 0.6989 mol/Kg
∴ moles solute (nb):
⇒ nb = (0.6989 mol/Kg)(0.150 Kg) = 0.1048 mol
molar mass:
⇒ mm = (35.9 g)/(0.1048 mol) = 342.43 g/mol
The freezing point depression is -1.3 °C for a solution of 35.9 g of a 3.3 × 10² g/mol solute in 150.0 g of water.
What is the freezing point depression?
Freezing point depression (ΔT) is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent.
The freezing points of solutions are all lower than that of the pure solvent and are directly proportional to the molality of the solute.
ΔT = i × Kf × m
m = ΔT / i × Kf
m = (1.3 °C) / 1 × (1.86 °C/m) = 0.70 m
where,
- i is the van 't Hoff factor (1 for nonelectrolytes).
- Kf is the cryoscopic constant (1.86 °C/m for water).
- m is the molality of the solution.
Since the 0.70 m solution has 150.0 g (0.1500 kg) of water, we will use the definition of molality to calculate the moles of solute.
m = moles of solute / kg of solvent
moles of solute = m × kg of solvent = 0.70 mol/kg × 0.1500 kg = 0.11 mol
0.11 moles of solute have a mass of 35.9 g. The molar mass of the solute is:
M = 35.9 g/0.11 mol = 3.3 × 10² g/mol
The freezing point depression is -1.3 °C for a solution of 35.9 g of a 3.3 × 10² g/mol solute in 150.0 g of water.
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