Freezing pointIt's January in Nashville and your local weather station just informed you that the temperature outside is -8.77 degrees Celsius. You estimate the amount of ice on your driveway to be...
Freezing pointIt's January in Nashville and your local weather station just informed you that the temperature outside is -8.77 degrees Celsius. You estimate the amount of ice on your driveway to be approximately 30kg. What is the minimum amount of sodium chloride in kilograms (kg) needed to lower the freezing point of the water so that the ice melts?
The goal is to lower the freezing point of water to -8.77 so that the ice will thaw. For this we need to use the equation
deltaT = i Kf m
where deltaT is the change in temperature, i is the van't Hoff factor, Kf is the freezing point depression constant of water and m is the molality of the water.
We know deltaT = 0 - (-8.77) = 8.77 and the value of i will be 2 because each NaCl molecules breaks into two ions. For colligative properties, such as freezing point depression, we are concerned with how many particles there are rather than the concentration of the substance (i.e. concentration of ions vs. concentrtion of NaCl).
For Kf, we will use the freezing point depression constant of water, 1.86 K kg /mol. m is the molality of the solution which is found from the moles of solute over the kg of solvent (i.e. water).
We can fill in the equation
8.77 = 2 (1.86) (mol NaCl / 30 kg H2O)
and then solve for the moles of NaCl to find
70.7 moles NaCl
Which we need to convert to grams and then kilograms which we can do using the molar mass and a conversion factor.
70.7 mol (58.5 g/mol) (1 kg / 1000 g)
= 4.14 kg NaCl