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
1 Answer
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
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
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,
#"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
#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
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.