Question

Question #0c4c5

Answer 1

`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.