Question about freezing point and solution?

Question

Predict the freezing point (in Celsius) of a solution of 5.0 g phenol in 100 g water. The Kf of phenol is 7.40 degrees Celsius (kg)/mole. I am suppose to use1 this equation: M=(Kf * g) / (ΔTf * G), but I need to solve for Tf instead of M. I really need some help on how to get started with this problem. If anyone could help, your response would greatly be appreciated. Thanks.

Answer 1

I do not understand your formula. What is G? Anyway, you need to calculate the MOLALITY, not molarity of the solution. Molality with an L, is moles per kg of solvent. The molar mass of phenol is 94.11 or use the periodic table. From the given, the molality is [5g/94.11]/0.1 kg or m = 0.531 molal

a good formula is delta T = m k(f) where kf is the freezing constant of the solvent. Since your solvent is water, it is the one in greater amount, then use the water constant of -1.86

or delta T = m (-1.86) Please note that even if you are given the kf of phenol, phenol is not the solvent, so you do not need it.

Also, delta T meaning the change in freezing point of the solution. Now, it is very important to remember that the ideal behavior of solutions based on the colligative properties is that a solute (phenol) will DEPRESS the freezing point, compared to pure water. Or, a solution of phenol will have a lower freezing point. In contrast, a solute will make the boing point higher, or it will ELEVATE the boiling point. You need to remember this correlation.

Thus, you expect the freezing point of the solution of phenol to be lower than 0 C, lower than zero degrees celsius.

Thus Tf - 0 C = -1.86 x 0.531 or - .988C, which means the freezing point of has depressed or lowered.

Answer 2

There is no formula. As you add more and more solute, the bp goes up. For the mp the more particles you have will decrease it more. So NaCl dissociated into 2 particles so decreases it more than sugar that does not dissociate. CaCl2 decreases it more than NaCl