   Chapter 2, Problem 14P

Chapter
Section
Textbook Problem

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.Determination of the Molecular Weight of a Solute by Freezing Point Depression A 100-g amount of a solute was dissolved in 1000 g of water. The freezing point of this solution was measured accurately and determined to be — 1.12°C. What is the molecular weight of the solute?

Interpretation Introduction

To determine:

The molecular weight of the solute.

Introduction:

When a solution is prepared, it contains properties different from the solvent which used to prepare the solution. Depression of the freezing point is such altered property. This alteration is due to the number of particles present in the solution, which is also known as colligative properties. Colligative properties are used to determine the molecular weight of a soluble solid.

Explanation

Given information:

Freezing point = 1.120C

Solute amount = 100 g

Water amount = 1000 g

Formula used:

Tf=Kf×b

TF=Freezing point

KF=Freezing point constant

b=Molality

Calculation:

The molality of solution is defined as the number of moles of solute in 1 kg of the solvent.

Thus,

b=nM(kg)

Here, number of moles of solute can be calculated as follows:

n=mM.M

Here, m is mass and M

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