2) What is the molality of solution that contains 50 g ethanol (C,H,OH) in 325 g water?The molality of a solution is equal to the moles of solute divided by the mass of solvent inkilograms. The answer is 3.338 mol/kg|3) What will be the freezing point change for the solution in question 2?

Answered: 2) What is the molality of solution…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

2) What is the molality of solution that contains 50 g ethanol (C,H,OH) in 325 g water?
The molality of a solution is equal to the moles of solute divided by the mass of solvent in
kilograms. The answer is 3.338 mol/kg|
3) What will be the freezing point change for the solution in question 2?

Expert Solution

Step 1

$Mass of enthanol = 50 g : mass of solvent = 325 g Molecular mass of ethanol = 46.07 g / mol i . e, 46.07 g / mol no . of moles ethanol = \frac{50}{46.07} Molarity, m = \frac{no . of moles}{mass of solvent (g)} \times 1000 = \frac{50}{46.07 \times 325} \times 1000 i . e, \frac{50 \times 1000}{46.07 \times \times 325} = 3.339 g / mol i . e, 3.339 g / mol ∆ T_{F} = K_{F} m K_{F} for water = 1.86 ∆ T_{F} = 1.86 \times 3.339 ∆ T_{F} = 6.21 ∆ T_{F} = T °_{F} - T_{F} where T °_{F} = freezing point of H_{2} O (pure) 6.21 = 0 - T_{F} T_{F} = freezing point of solution . T_{F} = - 6.21 ° C freezing point change by 6.21 ° C for solution .$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions