Question

Salt, NaCl, has a solubility of 35.7g per 100g of water at 0 degrees C. What is the lowest possible melting point for ice that could be obtained considering the solubility of NaCl in water? The Kf for H2O is 1.86 C/m

Answer 1

`-22.7^@"C"`

Explanation:

Your tool of choice here will be the equation that expresses the freezing-point depression of a solution, which looks like this

`color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "`, where

`DeltaT_f` - the freezing-point depression;
`i` - the van't Hoff factor
`K_f` - the cryoscopic constant of the solvent;
`b` - the molality of the solution.

In your case, the cryoscopic constant of water is said to be

`K_f = 1.86^@"C kg mol"^(-1)`

So, the idea here is that you can only dissolve `"35.7 g"` of sodium chloride, `"NaCl"`, in water at `0^@"C"`.

The thing to remember about sodium chloride is that the compound is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations, `"Na"^(+)`, and chloride anions, `"Cl"^(-)`

`"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"^(-)`

This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute that are dissolved and the number of moles of solute particles produced in solution, will be equal to

`i = 2`

That is the case because one mole of sodium chloride will dissociate completely to form one mole of sodium cations and one mole of chloride anions, hence two moles of particles.

Now, let's assume that you're working with `"100 g"` of water at `0^@"C"`. Use sodium chloride's molar mass to determine how many moles of sodium chloride you'd get in that `"35.7-g"` sample

`35.7 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.443 color(red)(cancel(color(black)("g")))) = "0.61085 moles NaCl"`

The molality of a solution tells you how many moles of solute you get per kilogram of solvent

`color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))`

In your case, you will have

`b = "0.61085 moles"/(100 * 10^(-3)"kg") = "6.1085 mol kg"^(-1)`

The freezing-point depression of this solution would be

`DeltaT_f = 2 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 6.1085 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))`

`DeltaT_f = 22.72^@"C"`
The freezing-point depression is defined as

`color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "`, where

`T_f^@` - the freezing point of the pure solvent
`T_"f sol"` - the freezing point of the solution

The freezing point of the solution will thus be

`T_"f sol" = T_f^@ - DeltaT_f`

`T_"f sol" = 0^@"C" - 22.72^@"C" = color(green)(|bar(ul(color(white)(a/a)-22.7^@"C"color(white)(a/a)|)))`

I'll leave the answer rounded to three sig figs.