# Question #e695c

##### 1 Answer

#### Answer:

#### 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

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

*van't Hoff factor*, equal to

*cryoscopic constant* of the solvent;

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

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

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

**pure solvent**

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

*van't Hoff factor*

Plug in your values and solve for

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