Question

Question #e695c

Answer 1

`T_"f" = -1.28^@"C"`

Explanation:

I'm not entirely sure why the problem provides the density of the water, since that is not needed to find the boiling point and the freezing point of the solution.

I can only assume that your solution contains `"57.8 g"` glucose dissolved in `"465 mL"` of water. Since the density of water is to be taken as `"1.00 g/mL"`, this will not make any difference for the final result.

So, the first thing would be to determine the mass of water

`465color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "465 g"`

The equation for freezing-point depression looks like this

`DeltaT_f = i * K_f * b" "`, where

`DeltaT_f` - the freezing-point depression;
`i` - the van't Hoff factor, equal to `1` for non-electrolytes;
`K_f` - the cryoscopic constant of the solvent;
`b` - the molality of the solution.

In your case, glucose is a non-electrolyte, which means that it does not dissociate to form ions in solution. As a result, you have `i=1`.

In order to calculate the molality of the solution, you need to know two things

• the number of moles of solute, in your case sucrose
• the mass of the solvent, in your case water, expressed in kilograms

Use glucose's molar mass to determine how many moles you have in that sample

`57.8color(red)(cancel(color(black)("g"))) * " 1 mole glucose"/(180.156color(red)(cancel(color(black)("mL")))) = "0.3208 moles glucose"`

The molality of the solution will thus be - do not forget to convert the mass of the water from grams to kilograms

`color(blue)(b = n_"solute"/m_"solvent")`

`b = "0.3028 moles"/(465 * 10^(-3)"kg") = "0.6899 moles/kg" = "0.6899 molal"`

You can find the values of water's cryoscopic and ebullioscopic constants here

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

So, plug in your values and determine the freezing-point depression

`DeltaT_"f" = 1 * 1.86^@"C" color(red)(cancel(color(black)("kg")))/color(red)(cancel(color(black)("mol"))) * 0.6899color(red)(cancel(color(black)("moles")))/color(red)(cancel(color(black)("kg"))) = 1.283^@"C"`

The freezing-point depression is defined as

`DeltaT_"f" = T_"f"^@ - T_"f"" "`, where

`T_"f"^@` - the freezing point of the pure solvent
`T_"f"` - the freezing point of the solution

This means that the freezing point of the solution will be

`T_"f" = T_"f"^@ - DeltaT_"f"`

`T_"f" = 0^@"C" - 1.283^@"C" = color(green)(-1.28^@"C")`

I'll leave the calculation of the solution's boiling point to you as practice. The equation for boiling-point elevation is

`DeltaT_b = i * K_b * b" "`, where

`DeltaT_b` - the boiling-point elevation;
`i` - the van't Hoff factor
`K_b` - the ebullioscopic constant of the solvent;
`b` - the molality of the solution.

Plug in your values and solve for `DeltaT_b`, then use

`DeltaT_"b" = T_"b" - T_"b"^@`