Purchase Solution

Molecular Weight by Freezing Point Depression

Not what you're looking for?

Ask Custom Question

For unknown sample #1, there is 2 g of the unknown sample and 10 mL of water. The freezing point was -1.7 C so Td=1.7.

In unknown sample #2, there is 2g of the sample and 10 mL of water. Freezing point = -1.8. So Td = 1.8.

How do I obtain the molecular weights for these two unknown samples?

Purchase this Solution
Solution Summary

The molar mass of unknown samples is calculated from the freezing point data.

Solution Preview

You have to use molality equations

Molality = moles of solute / kg of solvent

Td = Kf x molality

Hence, Td = Kf ...

Purchase this Solution
Free BrainMass Quizzes
General Chemistry - Classification of Matter

This test will assess your knowledge on the classification of matter which includes elements, compounds and mixtures.

Match Elements with their Symbols

Elements are provided: choose the matching one- or two-letter symbol for each element.

Thermochemistry

The quiz helps in revising basic concepts about thermochemistry.

Functional groups in Organic Chemistry

You will be tested on the names of functional groups in Organic Chemistry. It is very important to know the functional groups to understand Organic reactions.

Organic Chemistry Naming: Alkanes

This is a quiz which is designed to assist students with learning the nomenclature used to identify organic compounds. This quiz focuses on the organic compounds called Alkanes.

View More Free Quizzes
Related BrainMass Solutions

  • Molecular Weight by Freezing Point Depression of Solution

    112029 Molecular Weight by Freezing Point Depression of Solution A 0.573g sample of a nonvolatile solute is added to 15.6g of napthalene and the mixture warmed until melted.

  • Percent by mass of palmatic acid in Commercial steric acid

    freezing point depression Δ T f = K f m Δ T f = freezing point depression Kf= molal freezing point depression constant m= molality of the solution=concentration in moles of solute per kilogram of solvent For benzene the normal

  • Freezing Point Depression of Water by Sucrose

    539150 Freezing Point Depression of Water by Sucrose We need to calculate the molar mass of sucrose from the freezing point depression of water using a MeasureNet temp probe to get the cooling curves of water and sucrose.

  • Colligative Properties

    point depression delta T f = K f m delta T f = freezing point depression Kf= molal freezing point depression constant m= molality of the solution For benzene the normal freezing point is 5.5 degrees Celsius The solution

  • Freezing point of water containing sucrose

    56446 Freezing point of water containing sucrose Calculate the freezing point of 250 ml of water containing 7.5 g of sucrose. For water, Kf=1.86 K kg/mol. The molecular weight of sucrose is 342.3 g/mol.

  • Freezing Point of a Solution of an Unknown Substance

    Freezing points are used to determine the molecular weight of solutes. A colligative property (freezing point depression) is employed in this calculation. All work is shown and explained.

  • Solution and Solvent Problems

    The freezing point depression constant of this solvent is 29.8 degrees C/m, while the freezing point is -22.3 degrees C. 4. A 15.9 %, by mass, solution of propanol, CH3CH2CH2OH, is found to have a density of 1.14 g/mL in water.

  • freezing point of naphthalene and benzene mixture

    Molal Freezing Point Depression and Molal Boiling Point Elevation Constants: Solvent Normal Freezing K(fr) Normal Boiling K(br) Point,C C m^-1 Point, C C m^-1 Acetic acid

  • Boiling Point Elevation of Chloroform

    for chloroform is 61.70 °C Therefore boiling point of the solution= 64.83 °C =61.7+3.13 Answer: boiling point of a solution= 64.83 °C The question on freezing point depression would be solved in the same way The solution calculates

View More Related Solutions