Respuesta :
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
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
The molar mass of vitamin K is [tex]\boxed{451.46{\text{ g/mol}}}[/tex].
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,
[tex]\Delta {{\text{T}}_{\text{f}}} = {{\text{k}}_{\text{f}}}{\text{m}}[/tex] …… (1)
Here,
[tex]\Delta {{\text{T}}_{\text{f}}}[/tex] is depression in freezing point.
[tex]{{\text{k}}_{\text{f}}}[/tex] is molal freezing point constant.
m is molality of solution.
Rearrange equation (1) to calculate m.
[tex]{\text{m}} = \dfrac{{\Delta {{\text{T}}_{\text{f}}}}}{{{{\text{k}}_{\text{f}}}}}[/tex] …… (2)
Substitute [tex]4.43{\text{ }}^\circ {\text{C}}[/tex] for [tex]\Delta {{\text{T}}_{\text{f}}}[/tex] and [tex]40.0{\text{ }}^\circ {\text{C/m}}[/tex] for [tex]{{\text{k}}_{\text{f}}}[/tex] in equation (2).
[tex]\begin{aligned}{\text{m}} &= \frac{{4.43{\text{ }}^\circ {\text{C}}}}{{40.0{\text{ }}^\circ {\text{C/m}}}} \\ &= 0.11075{\text{ m}} \\\end{aligned}[/tex]
The formula to calculate molality of solution is as follows:
[tex]{\text{Molality of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}[/tex] …… (3)
Rearrange equation (3) for moles of solute.
[tex]{\text{Moles of solute}} = \left( {{\text{Molality of solution}}} \right)\left( {{\text{Mass of solvent}}} \right)[/tex] …… (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.
[tex]\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}[/tex]
The molar mass of vitamin K can be calculated from its moles as follows:
[tex]\begin{aligned} {\text{Molar mass of Vitamin K}} &= \frac{{0.500{\text{ g}}}}{{0.0011075{\text{ mol}}}} \\&= 451.46{\text{ g/mol}} \\\end{aligned}[/tex]
Learn more:
- Choose the solvent that would produce the greatest boiling point elevation: https://brainly.com/question/8600416
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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.
