A 0.98 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.
depression= i*.98*1.86=2.6
i=2.6/(.98*1.86)=1.43
Well, let's break it down! The Van't Hoff factor (i) represents the number of particles that a solute dissociates into when it dissolves in a solvent. This factor is important because it affects colligative properties such as freezing point depression.
In this case, we have a 0.98 m aqueous solution of an ionic compound MX, and its freezing point is -2.6∘C. To calculate the Van't Hoff factor (i), we can use the formula:
∆Tf = Kf * i * m,
where ∆Tf is the freezing point depression, Kf is the cryoscopic constant, i is the Van't Hoff factor, and m is the molality of the solution.
Given that the molality is 0.98 m and the freezing point depression is -2.6∘C, we can rearrange the formula to solve for i:
i = ∆Tf / (Kf * m).
Unfortunately, I can't provide you with the exact value of Kf for MX without knowing the specific compound. However, if you have this value, you can plug it into the formula along with the molality to calculate the Van't Hoff factor (i). May I help you with anything else?
To calculate the van't Hoff factor (i) for MX at this concentration, we need to use the formula:
ΔTf = -i * Kf * m
Where:
ΔTf = Freezing point depression
Kf = Freezing point constant for the solvent (water)
m = molality of the solution
i = van't Hoff factor
Given:
ΔTf = -2.6°C
Kf = freezing point constant for water = 1.86 °C/m
m = 0.98 m
We need to solve for i.
Rearranging the formula, we have:
i = -ΔTf / (Kf * m)
Now, substitute the given values into the equation:
i = -(-2.6 °C) / (1.86 °C/m * 0.98 m)
Simplifying:
i = 2.6 / (1.86 * 0.98)
i ≈ 1.4
Therefore, the van't Hoff factor (i) for MX at this concentration is approximately 1.4.
To calculate the van't Hoff factor (i) for MX at this concentration, we need to use the equation:
ΔTf = - i * Kf * m
Where:
ΔTf is the freezing point depression
i is the van't Hoff factor
Kf is the freezing point depression constant (dependent on the solvent)
m is the molality of the solution
In this case, we are given:
ΔTf = -2.6 °C
m = 0.98 m (molality)
To find the van't Hoff factor (i), we need to find the freezing point depression constant for the solvent (water). For water, the freezing point depression constant (Kf) is approximately 1.86 °C/m.
Now, we can rearrange the equation and solve for i:
i = -ΔTf / (Kf * m)
Let's substitute the given values into the equation:
i = -(-2.6 °C) / (1.86 °C/m * 0.98 m)
Calculating the equation:
i = 2.6 °C / (1.82 °C*m)
Simplifying:
i ≈ 1.39
Therefore, the van't Hoff factor (i) for MX at this concentration is approximately 1.39.