Question #9afee

You know that the mole fraction of sodium hydroxide is equal to 1/5. This means that the mole fraction of water will be

#chi_"water" = 1 - chi_(NaOH)#

#chi_"water" = 1 - 1/5 = 4/5#

This means that, for evey mole of sodium hydroxide, your solution contains 4 times more moles of water. Let #x# be the number of moles of sodium hydroxide.

The number of moles of water will thus be

#n_"water" = 4 * x = 4x#

The equation for freezing point depression looks like this

#DeltaT_"f" = i * K_f * b#, where

#K_f# - the cryoscopic constant - depends on the solvent;
#b# - the molality of the solution;
#i# - the van't Hoff factor - the number of ions per individual molecule of solute.
#DeltaT_"f"# - the freezing point depression - defined as
#T_"f"^@ - T_"f sol"#.

The molality of the solution is defined as the number of moles of solute, in your case sodium hydroxide, divided by the mass of the solvent, in your case water, expressed in kilograms!

You can determine the mass of water by using its molar mass

#4xcancel("moles") * "18.02 g"/(1cancel("mole")) = 4x * "18.02 g"#

#b = (x " moles")/(underbrace(4x * "18.02 g")_(color(blue)("mass of water in grams")) * 10^(-3))#

#b = (cancel(x)"moles")/(4cancel(x)cancel("moles") * 18.02"g"/cancel("mol") * 10^(-3)) = 1/(4 * 18.02) * 1000 = "13.9 molal"#

Since sodium hydroxide is a strong electrolyte, it will dissociate completely in aquesous solution to give sodium cations, #Na^(+)#, and hydroxide anions, #OH^(-)#.

#NaOH_((aq)) -> Na_((aq))^(+) + OH_((aq))^(-)#

This means that 1 mole of sodium hydroxide produces 1 mole of #Na^(+)# and 1 mole of #OH^(-)#, so the van't Hoff factor will be equal to 2.

Therefore, the freezing point depression will be

#DeltaT_"f" = 2 * 1.86^@"C"/cancel("molal") * 13.9cancel("molal")#

#DeltaT_"f" = 51.7^@"C"#

Rounded to two sig figs, despite the fact that you only gave one sig fig for the mole fraction of sodium hydroxide, the answer will be

#DeltaT_"f" = color(green)(52^@"C")#