What is the freezing point of the water that is derived from a 6.5 inch snowfall on a roadway that was treated with rocks out at a rate of 0.175 kg/m²? the density of the freshly fallen snow was determined to be 0.17 g/cm³?

You can calculate the depression of the frezzing point of water, based on that it is a colligative property that may be expressed as:

ΔTf = Kf * m

Kf is cryoscopic constant of water = 1.86 °C / m

You can find m as m = moles of solute / kg of solvent.

Take a base of 1 m^2 of snow, to determine the amount of solute an water.

1 m^2 * 0.175 kg/m^2 = 0.175 kg of salt rocks.

You need to know the kind of salt to determine the number of moles.

If the salt is CaCl2, the molar mass is 40 g/mol + 2*35.5 g/mol = 111 g/mol

Then the number of moles is:[ mass / molar mass] =

 (0.175kg * 1000g/kg) / 111 g/mol = 1.5766 moles.

The amount of sbiw in 1 m^2 * 6.5 inches of snow is:

1m^2 * (100cm/m)^2 * 6.5 inch *(2.54cm/inch) = 165,100 cm^3

You multiply the volume times the density to find the mass of snow: 165,100 cm^3 * 0.17 g/cm^3 = 28067 g = 28.067 kg

 molality, m = number of moles of solute / kg of water = 1.5766 / 28.067 = 0.056 m

And ΔTf = kf*m = 1.86°C/m * 0.056m = 0.10°C.

Then, the new freezing point of water is 0° - 0.10°C = - 0.1°C

- 0.1 °C