Question

How do you calculate the freezing point and the boiling point of the solution formed when 0.150 g of glycerol $({C_3}{H_8}{O_3})$ is added to 20.0 g of water? ${K_{er}}({H_2}O) = 1.86^\circ C$, ${K_b}({H_2}O) = 0.52^\circ C$ ?

Answer 1

The freezing point is defined as the temperature at which the liquid substance starts to freeze and the boiling point is defined as the point at which the liquid substance starts to boil. To calculate the change in freezing point and change in boiling point the main terms used are molality, Van’t Hoff factor, cryoscopic constant and molal boiling point constant.

Complete step by step answer: Given,
Mass of glycerol is 0.150 g.
Mass of water is 20.0 g.
${K_f}({H_2}O)$of water is $1.86^\circ C$
${K_b}({H_2}O)$of water is $0.52^\circ C$
As we know that the change in freezing point temperature is calculated by the formula as shown below.
$\Delta {T_f} = i \times m \times {K_f}$
Where,
$\Delta {T_f}$ is the change in freezing point temperature.
i is the Van’t Hoff factor
m is the molality
${K_f}$ is the cryoscopic constant.
The molecular weight of glycerol is 92.09 g/mol.
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molar mass
To calculate the number of moles of glycerol, substitute the value in the above equation.
$\Rightarrow n = \dfrac{{0.150}}{{92.09g/mol}}$
$\Rightarrow n = 0.00163mol$
The formula to calculate the molality is shown below.
$m = \dfrac{n}{M}$
Where,
m is the molality
M is the mass of solvent in Kg.
n is the number of moles
To calculate the molality, substitute the values in the above expression.
$\Rightarrow m = \dfrac{{0.00163}}{{0.0200Kg}}$
$\Rightarrow m = 0.0814m$
To calculate the change in freezing point temperature, substitute the value in the formula.
$\Rightarrow \Delta {T_f} = (1) \times (0.0814) \times (1.86)$
$\Rightarrow \Delta {T_f} = 0.152^\circ C$
Thus, the freezing point is calculated by subtracting the new freezing point with the freezing point of water.
$\Rightarrow {T_f} = 0^\circ C - 0.152^\circ C$
$\Rightarrow {T_f} = - 0.152^\circ C$
The formula to calculate the change in boiling point is shown below.
$\Delta {T_b} = i \times m \times {K_b}$
$\Delta {T_b}$ is the change in boiling point temperature.
i is the Van’t Hoff factor
m is the molality
${K_b}$ is the molal boiling point constant.
To calculate the change in boiling point, substitute the values in the above equation.
$\Rightarrow \Delta {T_b} = 1 \times 0.0814 \times 0.512$
$\Rightarrow \Delta {T_b} = 0.0417^\circ C$
The boiling point temperature is calculated by adding the boiling point of water with change in boiling point of water.
$\Rightarrow {T_b} = 100 + 0.0417$
$\Rightarrow {T_b} = 100.042^\circ C$

Note:
Make sure to convert the value of mass given in grams into kilograms as the molality is calculated in terms of mole per kilogram. 1 gram is equal to $\dfrac{1}{{1000}}kg$. The Van’t Hoff factor of glycerol is taken as 1 because it is a nonelectrolyte.