At what temperature (in °C) will the NaCl stop melting ice when there is 7.5 mol of NaCl in 6.95 kg of H₂O?
I don't understand the question. Is there more to the problem? Is this a subpart of a larger question?
...no. That was all we were given.
OK. Perhaps this is what they want.
dT = i*Kf*m
m is given as m = mols NaCl/kg solvent
Kf is 1.86 C/m for water
i is 2
Substitute and you get delta T. Subtract that from zero C to get the temperature at which water will freeze under those conditions.
To find the temperature at which NaCl stops melting ice, we need to determine the freezing point depression caused by the presence of 7.5 mol of NaCl in 6.95 kg of water. The freezing point depression formula is:
ΔT = Kf * molality
where ΔT is the change in freezing point, Kf is the cryoscopic constant, and molality is the concentration of the solute in moles per kilogram of solvent.
First, let's calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
Given:
moles of NaCl = 7.5 mol
mass of H2O = 6.95 kg
m = 7.5 mol / 6.95 kg
Next, we need to know the cryoscopic constant for water (Kf). The cryoscopic constant for water is approximately 1.86 °C/m.
Now we can calculate the change in freezing point (ΔT):
ΔT = Kf * m
Substituting the values:
ΔT = 1.86 °C/m * (7.5 mol / 6.95 kg)
Finally, to find the temperature at which NaCl stops melting ice, we subtract ΔT from the freezing point of water, which is 0 °C:
Freezing point = 0 °C - ΔT
Now you can calculate the freezing point by plugging in the values and performing the calculations.