Question

A 0.99 m aqueous solution of an ionic compound with the formula mx has a freezing point of -2.6 ∘c . calculate the van't hoff factor (i) for mx at this concentration.

Answer 1

To find the van't Hoff factor for the ionic compound MX, you apply the freezing point depression formula with the known constants and solve for i. The calculation indicates that MX disassociates into 1.4 particles on average in solution, which is less than the ideal factor due to incomplete dissociation and ionic interactions.

Step-by-step explanation:

To calculate the van't Hoff factor (i) for the ionic compound MX, you use the formula for freezing point depression:

ΔTf = Kf × m × i

Where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant, m is the molality of the solution, and i is the van't Hoff factor.

First, the freezing point depression (ΔTf) of the solution is given as -2.6°C. Since the freezing point of pure water is 0°C, the depression is 2.6°C. The molal freezing-point depression constant (Kf) for water is -1.86°C/m. The molality (m) of the solution is given as 0.99 m (which means 0.99 moles of solute per kilogram of solvent).

Plugging these values into the formula:

2.6°C = (1.86°C/m) × (0.99 m) × i

We can now solve for i:

i = 2.6°C / (1.86°C/m × 0.99 m)

After performing the calculation:

i ≈ 1.4

The calculated van't Hoff factor indicates that MX disassociates into 1.4 particles on average in solution. This is less than the ideal van't Hoff factor because not all ionic compounds dissociate completely in solution, leading to a lower observed van't Hoff factor compared to the theoretical one, which assumes complete dissociation. Ionic interactions and ion pairing in the solution can cause such deviations from ideality.

Answer 2

Formula for the depression in freezing point is:

`\Delta T_f = i* k_f* m` -(1)

where
`\Delta T_f` is depression in freezing point,

`i` is Van't Hoff factor,

`k_f` is molal freezing point depression constant, and

`m` is molality of the solution.

Molality of the solution,
`m = 0.99 m` (given)

Molal freezing point depression constant of water,
`k_f = 1.86^(o)C/m`

Depression in freezing point of solution,
`\Delta T_f = T_(water solvent) - T_(solution)`

`\Delta T_f = {0^(o)C} - ({-2.6^(o)C}) = 2.6^(o)C`

Substituting the values in equation (1):

`\2.6^(o)C = i* 1.86^(o)C/m* 0.99 m`

`i = (2.6^(o)C)/(1.86^(o)C/m* 0.99 m) = 1.412`

Hence, the Van't Hoff factor (i) for
`HX` is 1.412.