What is the molality of #NaCl# if the freezing point of a #NaCl# solution is -5.58 C°?
In this case, the only information you have is that a given sodium chloride solution has a freezing point of
As you know, the freezing point of a solution is a colligative property, meaning that is depends exclusively on the concentration of solute particles.
Since sodium chloride is an electrolyte, it will dissociate completely in aqueous solution to form sodium cations and chloride anions
#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#
Notice that one mole of sodium chloride produces one mole of sodium cations and one mole of chloride anions in solution, i.e. two moles of particles.
This mans that sodium chloride has a van't Hoff factor,
Now, you can compare the freezing point of the solution with that of pure water, which is equal to
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "#, where
In your case, the freezing-point depression will be equal to
#DeltaT_f = 0^@"C" - (-5.58^@"C") = 5.58^@"C"#
The freezing-point depression is also related to the molality of the solution
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "#, where
The cryoscopic constant of water is equal to
Rearrange the above equation to solve for
#DeltaT_f = i * K_f * b implies b = (DeltaT_f)/(i * K_f)#
Plug in your values to get
#b = (5.58color(red)(cancel(color(black)(""^@"C"))))/(color(red)(2) * 1.86color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)"1.50 mol kg"^(-1)color(white)(a/a)|)))#
The answer is rounded to three sig figs.