At what temperature would a 1.05m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 C/m.

deltaTf=1.9*-1.86*1.05

what mass due to waters of crystallization is present in a 3.38-g sample of copper (II) sulfate

ukh

To determine the freezing point of a solution, you need to use the equation:

ΔTf = Kf * i * m

Where:
- ΔTf is the change in freezing point of the solvent (water)
- Kf is the molal freezing point depression constant for the solvent (water)
- i is the van't Hoff factor, which represents the number of particles the solute dissociates into when dissolved in the solvent
- m is the molality of the solution, which is the moles of solute per kilograms of solvent

In this case, you are given the van't Hoff factor (i = 1.9) and the molal freezing point depression constant for water (Kf = 1.86 °C/m). However, you still need to determine the molality of the NaCl solution.

To calculate the molality, you need to know the number of moles of NaCl and the mass of water in the solution. In this case, you are given that the concentration of NaCl is 1.05 m. This means that there are 1.05 moles of NaCl in 1 kilogram of water.

Now, you can substitute the values into the equation to find the change in freezing point (ΔTf):

ΔTf = (1.86 °C/m) * (1.9) * (1.05 m)

Calculating this expression gives:

ΔTf = 3.7274 °C

Finally, to find the freezing point of the solution, you need to subtract the change in freezing point from the normal freezing point of water (0 °C):

Freezing point of the solution = 0 °C - 3.7274 °C = -3.73 °C

Therefore, the 1.05m NaCl solution would freeze at approximately -3.73 °C.