The Van't hoff factor for CaCL2 is 2.71

  • Thread starter Jef123
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In summary: Therefore, using the equation ΔTf = kfmi, where ΔTf = -1.14°C, kf = 1.86°C/molal, and i = 2.71, we can solve for m, the molal concentration. m = ΔTf / (kf * i) = (-1.14°C) / (1.86°C/molal * 2.71) = -0.232 molal To find the mass %, we need to know the mass of CaCl2 in the solution, which we can calculate using the molal concentration and the molar mass of CaCl2 (110.98 g/mol). mass of CaCl2 = m * molar mass
  • #1
Jef123
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1. The van't hoff factor for CaCL2 is 2.71. What is its mass % in aqueous solution that has Tf = -1.14°C

The Attempt at a Solution



So I think I am suppose to use the change in freezing point equation: ΔTf= kfmi. Where kf = is the molal freezing point depression constant, i = van't hoff factor, m = molal concentration.

Other than that, I am kind of lost. It seems like there is not enough variables given. Any suggestions?
 
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  • #2
Jef123 said:
1. The van't hoff factor for CaCL2 is 2.71. What is its mass % in aqueous solution that has Tf = -1.14°C

The Attempt at a Solution



So I think I am suppose to use the change in freezing point equation: ΔTf= kfmi. Where kf = is the molal freezing point depression constant, i = van't hoff factor, m = molal concentration.

Other than that, I am kind of lost. It seems like there is not enough variables given. Any suggestions?

Congratulations for giving full information on the problem, which often we have to ask after. Often actually setting out the information gives the answer.

In this case you give a relation between four quantities, and three of the quantities are given. So the first step is to get the fourth from that.

From that result you have to get the mass %. Be explicit about what your definition of this is.

(I suppose this is the point of the exercise which is somewhat artificial IMO.
Apart from the fact I don't know how well regarded and used Van't Hoff's i is these days.)
 
  • #3
Water's freezing point depression constant is 1.86 degrees Celsius/molal.
 

1. What is the Van't Hoff factor for CaCl2?

The Van't Hoff factor for CaCl2 is 2.71.

2. How is the Van't Hoff factor calculated?

The Van't Hoff factor is calculated by dividing the number of particles formed when a substance dissolves in water by the number of moles of the substance dissolved.

3. Why is the Van't Hoff factor important?

The Van't Hoff factor is important because it helps determine the extent of dissociation of a substance into ions when it dissolves in water. This is crucial for understanding the properties and behavior of solutions.

4. How does the Van't Hoff factor for CaCl2 compare to other compounds?

The Van't Hoff factor for CaCl2 is higher than most other compounds because it dissociates into three ions (Ca2+ and 2 Cl-) when it dissolves in water.

5. Can the Van't Hoff factor change?

Yes, the Van't Hoff factor can change depending on the conditions of the solution, such as temperature, pressure, and concentration. It may also vary for different solvents or when the substance is in a different physical state.

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