Question #0c4c5

1 Answer
Jun 13, 2016

Answer:

#T_"f sol" = -1.16^@"C"#

Explanation:

When dealing with the freezing point of a solution, your first goal is to determine its molality by

  • determining the number of moles of solute present in the solution
  • determining the mass of the solvent in kilograms

Once you know the molality of the solution, you can find the freezing-point depression by using the equation

#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)#

Glucose is a non-electrolyte, i.e. it remains undissociated in aqueous solution, which means that its van't Hoff factor will be #i=1#.

So, you know that your solution contains #"53.4 g"# of glucose. To determine how many moles of glucose it contains, use the compound's molar mass

#53.4 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))) = "0.2964 moles glucose"#

The solution contains #"475 g"# of water, which is the solvent. Convert this to kilograms

#475color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.475 kg"#

The molaity of the solution, which is defined as

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

will thus be

#b = "0.2964 moles"/"0.475 kg" = "0.624 mol kg"^(-1)#

Now plug in your values and solve for the freezing-point depression

#DeltaT_f = 1 * 1.86^@"C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 0.624 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 1.161^@"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

This means that the freezing point of the solution will be

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

#T_"f sol" = 0^@"C" - 1.161^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-1.16^@"C")color(white)(a/a)|)))#

The answer is rounded to three sig figs.