Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, freezing point depression formula is:
ΔT = kf×m×i
Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = 0.253m
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
Answer:
THE FREEZING POINT IS -1.41 °C
Explanation:
Using the formula of change in freezing point:
ΔTf = i Kf m
i = 3 (1 Ca, 2 Br)
i is the number of the individual elements in the molecules
Kf of water = 1.86 °C/m
mass of CaBr2 = 189.9 g
Calculate the Molar mass of CaBr2:
Molar mass = ( 40 + 80*2) = 200 g/mol
Calculatee the molarity:
molarity = 189.9 g * 1 mole / 200 g/mol / 3.75 kg of water
molarity = 0.2532 M
So therefore, the change in freezing point is:
ΔTf = 1 Kf * M
ΔTf = 3 * 1.86 * 0.2532
ΔTf = 1.41 °C
The freezing point = old freezing point - change in freezing point
The freezing point = 0 - 1.41 °C = - 1.41 °C
The freezing point therefore is -1.41 °C
