Question

What is the freezing point (in degrees Celcius) of 3.75 kg of water if it contains 189.9 g of C a B r 2?

Answer 1

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

Answer 2

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,

The freezing point of the solution is -1.4°C