How many grams of table salt would have to dissolved ?

Question

How many grams of table salt would have to dissolved in 135.0 g of water to lower the freezing point by 1.70 degrees celsius?


kf = 1.86 K-kg/mol for water


im lost, any explanation on this?

Answer 1

The freezing point depression of a solution can be found as:
ΔT = molality * k_f
Where molality is the concentration of the solute and k_f is the freezing point depression constant for the solvent.

You are given k_f in the question. You are told that the freezing point is lowered in the solution by 1.70 °C. All you need to calculate now is the concentration.

Rearranging the above equation to solve for the concentration,
Molality = ΔT / k_f
Which you should be able to easily plug in your known values and solve.

But we were asked to find the mass of NaCl needed, not just its concentration. We know that,
Molality = moles of solute / kilograms of solvent
We are given the mass of the solvent (in terms of gram, which we will need to concert to kilograms). We already know what the concentration of the solution should be. All we need to do now if find the number of moles of NaCl needed to achieve this concentration.
Rearranging to solve for the number of moles of solute,
Moles = Molality * kg of solvent
Which you can now plug in your known values to solve for.
But wait, Sodium Chloride is a strong electrolyte, meaning that in solution it will break up into Na+ and Cl- ions. The concentration (molality) we just found was for the concentration of the ions in solution, not NaCl, and we just calculated the moles of ions in solution. For every 1 mole of NaCl, there are 2 moles of ions. So we must remember to divide the number of moles just found by the number of ions per mole of solute for a strong electrolyte.
Moles of solute = moles of ions / 2
Moles of solute = Molality * kg of solvent / 2

Once we know the number of moles of solute needed, the last step (converting to grams of solute) is easy…simply multiply by the molar mass of Sodium Chloride and you will get the number of grams of NaCl needed to lower the freezing point of the water by 1.70 degrees.