What is the new freezing point of water if 25g of NaCl is dissolved in 200 g of water?Water has a Kf of 1.860C/molal and a freezing point of 00C. Group of answer choices-7.960C7.960C-3.980C-0.4650C

To solve this problem, we need to use the formula for freezing point depression:

ΔTf = Kf * m

where ΔTf is the change in freezing point, Kf is the cryoscopic constant of the solvent (water in this case), and m is the molality of the solution.

First, we need to calculate the molality of the solution. Molality is defined as the number of moles of solute (NaCl in this case) divided by the mass of the solvent (water in this case) in kilograms.

The molar mass of NaCl is approximately 58.44 g/mol. So, the number of moles of NaCl is 25 g / 58.44 g/mol = 0.428 mol.

The mass of water is 200 g, which is 0.2 kg.

So, the molality of the solution is 0.428 mol / 0.2 kg = 2.14 mol/kg.

Now we can calculate the change in freezing point:

ΔTf = Kf * m = 1.86 °C/molal * 2.14 molal = 3.98 °C.

Since the freezing point of water is lowered by the presence of the solute, the new freezing point of the water is 0 °C - 3.98 °C = -3.98 °C.

So, the new freezing point of the water is -3.98 °C.