How many pounds of road salt per 1000gal of water to lower freeze point to 0 deg F?

Molal freezing point constant –1.86 °C /molal

since F changes by 1.8 degrees every time Celsius changes by 1 degree:

A Molal freezing point constant of –1.858 °C/ molal, is equal to –3.344 °F/ molal

so let's find the molality that drops the freezing point by 32 degrees to O F:

dT = Kf (molal)

-30 F = (-3.344 F / molal) (molality)

molality = 8.971 molal total

whoops, since NaCl releases two ions per molecule, you get an 8.971 molal solution by making a 4.486 molar NaCl solution.

which is good to catch, since 5.8 Molar NaCl is about saturated

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1000 gal @ 3.785 l/gal = 3785 litres water

3785 litres @ density of water of 1 kg / litre = 3785 kg water

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the molality required is 4.486 molal, which has 4.486 moles of salt per kilogram of solvent:

3785 kg H2O @ 4.486 moles salt / kg H2O = 16,978 moles of salt

16,978 moles of salt @ 58.44 g / mol = 992204 grams of salt

992.2 kg salt @ 2.2046 pounds / kg =2187 pounds of salt

your answer is 2187 pounds of salt

I am going to leave the rounding off to sig figs up to you, I am sure that they meant to have more sig figs that the "1000" gallons has, but there is no decimal

thanks for asking me to recheck this