I have to find the calculation of Van't Hoff factor "i" for NaCl and sucrose. The directions say to use the equation
delta Tf = kf X m X i
and to use Kf for water = -1.86C/m
So I have the molality of NaCl and sucrose as well as delta Tf so for NaCl I plugged the numbers into the equation.
2.5C = -1.86C/m X 0.6697mol/kg X "i"
2.5 = -1.2456 X i
i=-2
Did I do this correctly? Thank you for your help!!
If you have the right m, yes.
Is it ok that "i" is negative?
No, i is not negative, ever.
Forget the - sign for 1.86 unless you are going to be consistent and place a - sign for 2.5. Wasn't that 2.5 C BELOW the normal freezing point. That's why I don't worry about the signs. You KNOW m is positive you KNOW Kf is negative, you KNOW delta T is a depression. Usually you go through all this to calculate m and you KNOW molality is +.
Ok and for sucrose I did
-4.5C = -1.86C/m X 0.1004 mol X i
-4.5 = -0.1867i
i =24
Can i be a big number like this?
Nevermind, I made a mistake im sorry. Disregard the question.
I don't believe 24. i should be 1 for sucrose.
No, you made a mistake in your calculation. Let's go through the correct calculation step by step.
First, let's plug in the given values into the equation:
ΔTf = kf * m * i
Given: ΔTf = 2.5°C, kf for water = -1.86°C/m, and m for NaCl = 0.6697 mol/kg.
Now, substitute the known values into the equation:
2.5°C = (-1.86°C/m) * (0.6697 mol/kg) * i
Next, divide both sides of the equation by (-1.86°C/m) and by (0.6697 mol/kg):
2.5°C / (-1.86°C/m) / (0.6697 mol/kg) = i
Simplifying the right side:
i ≈ -2.025
So the correct calculation of the Van't Hoff factor (i) for NaCl is approximately -2, not -2 as you stated.
For the calculation of the Van't Hoff factor for sucrose, you would need to use the given values for the molality of sucrose (m) and the change in freezing point (ΔTf), then follow the same steps to calculate the value of i.