Question

When 0.500 g of vitamin k is dissolved in 10.0 g of camphor (kf = 40.0°c/m), the freezing point of the solution is 4.43°c lower than that of pure camphor. assuming vitamin k is a nonelectrolyte in camphor, calculate its molar mass. 55.4 g/mol 451 g/mol 0.451 g/mol 3.54 × 104 g/mol?

Answer 1

The solution for this problem is:
1. number of mole / kg of vitamin K = (4.43 °C) / (40.0 °C/m) = 0.11075 m *molality freezing point depression constant for camphor is 37.7 C Kg/mo
2. (0.11075 mol/kg) x (0.0100 kg) = 0.0011075 mol 
3. (0.500 g) / (0.0011075 mol) = 451 g/mol

Answer 2

The molar mass of vitamin K is $\boxed{451.46{\text{ g/mol}}}$.

Further Explanation:

Colligative properties depend only on concentration of solute particles, not on their identities. Four colligative properties are mentioned below.

• Relative lowering of vapor pressure
• Elevation in boiling point
• Depression in freezing point
• Osmotic pressure

The expression for freezing point depression is,

$\Delta {{\text{T}}_{\text{f}}} = {{\text{k}}_{\text{f}}}{\text{m}}$                                                                                  …… (1)

Here,

$\Delta {{\text{T}}_{\text{f}}}$ is depression in freezing point.

${{\text{k}}_{\text{f}}}$ is molal freezing point constant.

m is molality of solution.

Rearrange equation (1) to calculate m.

 ${\text{m}} = \dfrac{{\Delta {{\text{T}}_{\text{f}}}}}{{{{\text{k}}_{\text{f}}}}}$                                                        …… (2)

Substitute $4.43{\text{ }}^\circ {\text{C}}$ for $\Delta {{\text{T}}_{\text{f}}}$ and $40.0{\text{ }}^\circ {\text{C/m}}$ for ${{\text{k}}_{\text{f}}}$ in equation (2).

$\begin{aligned}{\text{m}} &= \frac{{4.43{\text{ }}^\circ {\text{C}}}}{{40.0{\text{ }}^\circ {\text{C/m}}}} \\ &= 0.11075{\text{ m}} \\\end{aligned}$  

The formula to calculate molality of solution is as follows:

${\text{Molality of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}$                                     …… (3)

Rearrange equation (3) for moles of solute.

${\text{Moles of solute}} = \left( {{\text{Molality of solution}}} \right)\left( {{\text{Mass of solvent}}} \right)$                        …… (4)

Substitute 0.11075 m for molality of solution and 10 g for mass of solvent in equation (4) to calculate moles of vitamin K.

$\begin{aligned}{\text{Moles of Vitamin K}} &= \left( {{\text{0}}{\text{.11075 m}}} \right)\left( {{\text{10}}{\text{.0 g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ kg}}}}{{1{\text{ g}}}}} \right) \\&= 0.0011075{\text{ mol}} \\\end{aligned}$  

The molar mass of vitamin K can be calculated from its moles as follows:

 $\begin{aligned} {\text{Molar mass of Vitamin K}} &= \frac{{0.500{\text{ g}}}}{{0.0011075{\text{ mol}}}} \\&= 451.46{\text{ g/mol}} \\\end{aligned}$

Learn more:

1. Choose the solvent that would produce the greatest boiling point elevation: brainly.com/question/8600416
2. What is the molarity of the stock solution? brainly.com/question/2814870

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration

Keywords: colligative properties, freezing point, vitamin K, 451.46 g/mol, 0.500 g, 0.11075 m, 0.0011075 mol, depression in freezing point, molality, relative lowering of vapor pressure, osmotic pressure, elevation in boiling point.