# How do you calculate the freezing point and the boiling point of the solution formed when 0.150 g of glycerol (C_3H_8O_3) is added to 20.0 g of water? K_(er) (H_2O) = 1.86°C, K_b (H_2O) = 0.52 C?

Jul 7, 2017

Freezing point: $- {0.152}^{\text{o""C}}$

Boiling Point: ${100.042}^{\text{o""C}}$

#### Explanation:

We're asked to find the new freezing point and boiling point of a solution when glycerol is added to water, with known masses and constants.

Freezing Point

The formula for the new freezing point of the solution (which is always lower!) is

$\Delta {T}_{f} = i m {K}_{f}$

where

• $\Delta {T}_{f}$ is the change in freezing point temperature,

• $i$ is the van't Hoff factor, which is essentially the number of dissolved particles per unit of the solute. Since glycerol is a molecule, it is present solely as whole molecules, so the van't Hoff factor is $1$ in his case

• $m$ is the molality of the solution, which is the number of moles of solute per kilogram of solvent:

$\text{molality" = "mol solute"/"kg solvent}$

We need to find the moles of solute present, by using the given mass of glycerol and its molar mass (calculated to be $92.09$ $\text{g/mol}$):

0.150cancel("g C"_3"H"_8"O"_3)((1color(white)(l)"mol C"_3"H"_8"O"_3)/(92.09cancel("g C"_3"H"_8"O"_3))) = 0.00163 ${\text{mol C"_3"H"_8"O}}_{3}$

The kilograms of the solvent (water) is

20.0cancel("g H"_2"O")((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = 0.0200 $\text{kg H"_2"O}$

The molality is thus

"molality" = (0.00163color(white)(l)"mol C"_3"H"_8"O"_3)/(0.0200color(white)(l)"kg H"_2"O") = color(red)(0.0814m

• ${K}_{f}$ is the molal freezing point constant for the solvent (water), given as $1.86$ $\text{^"o""C/} m$.

Plugging in known values, we have

DeltaT_f = (1)(color(red)(0.0814)cancel(color(red)(m)))(1.86(""^"o""C")/(cancel(m))) = 0.152^"o""C"

The new freezing point is thus value subtracted from the normal freezing point of water, ${0}^{\text{o""C}}$:

$\text{freezing point" = 0^"o""C" - 0.152^"o""C" = color(blue)(-0.152^"o""C}$

Boiling Point

Opposite to the freezing point, the boiling point always increases.

The new boiling point is found essentially the same way as the freezing point, except replacing the freezing point quantities with boiling point quantities:

$\Delta {T}_{b} = i m {K}_{b}$

We're given that the molal boiling point constant for the solvent (water) is $0.512$ $\text{^"o""C/} m$. Plugging in known values, we have

DeltaT_b = (1)(color(red)(0.0814)cancel(color(red)(m)))(0.512(""^"o""C")/(cancel(m))) = 0.0417^"o""C"

The new boiling point is this value added to the normal boiling point of water, ${100}^{\text{o""C}}$:

$\text{boiling point" = 100^"o""C" + 0.0417^"o""C" = color(green)(100.042^"o""C}$

which I'll round off to three decimal places, as I did with the freezing point.