Question

Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?

a) -2.53 °C
b) -4.12 °C
c) -1.37 °C
d) -3.68 °C

Answer 1

The freezing point of the solution of 9.04 g of I2 in 75.5 g of benzene is approximately 5.18 °C.

Step-by-step explanation:

Using the formula for freezing point depression, we can calculate the freezing point of the solution:

ΔTf = Kf x m

Where:

• ΔTf is the change in freezing point
• Kf is the freezing point depression constant for the solvent (benzene) = 5.12 °C/m
• m is the molality of the solution

To find the molality:

molality (m) = moles of solute / mass of solvent (in kg)

First, convert the mass of I2 and benzene into moles:

moles of I2 = mass of I2 / molar mass of I2

moles of benzene = mass of benzene / molar mass of benzene

Then, calculate the molality:

molality (m) = moles of I2 / mass of benzene (in kg)

With the molality calculated, substitute the values into the freezing point depression formula:

ΔTf = (5.12 °C/m) x molality

Solving for ΔTf, we find that the freezing point is depressed by 0.3202 °C. Subtracting this value from the freezing point of pure benzene (5.5 °C), we can calculate the freezing point of the solution:

Freezing point = 5.5 °C - 0.3202 °C = 5.18 °C

Therefore, the freezing point of the solution of 9.04 g of I2 in 75.5 g of benzene is approximately 5.18 °C.