CHEMISTRY-ALEK 360 ACCES 1 SEMESTER ONL
CHEMISTRY-ALEK 360 ACCES 1 SEMESTER ONL
12th Edition
ISBN: 9781259292422
Author: Chang
Publisher: MCG
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Chapter 12, Problem 12.135QP

(a) Derive the equation relating the molality (m) of a solution to its molarity (M),

m = M d M M 1000

where d is the density of the solution (g/mL) and ℳ is the molar mass of the solute (g/mol). (Hint: Start by expressing the solvent in kilograms in terms of the difference between the mass of the solution and the mass of the solute.) (b) Show that, for dilute aqueous solutions, m is approximately equal to M.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The given equation relating molality (m) of the solution to its molarity (M) has to be derived.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

Molality (m) =Numberofmolesofsolute1kgofsolvent

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

Molarity (M) =Numberofmolesofsolute1literofsolution

Molarity is estimation of moles in the total volume of the solution while molality is estimation of moles in relationship with solvent in the solution

Explanation of Solution

Given data:

m =Md-M.M'1000

Where,

d- Density of the solution (g/mL)

M’- Molar mass of the solute (g/mol)

m- Molality of the solution

M- Molarity of the solution

Relate molality (m) to molarity (M):

The equations relating the molality (m) of the solution to its molarity (M) are as follows,

kg solvent =[mass of the solution (g)- mass of the solute (g)]×1kg1000g(1)

If we assume 1L of the solution then we can determine the mass of the solution from its volume and density and the mass of the solute from molar mass and molarity.

d =massVolumemass of the solution = d(gmL)×1000mL

mole =massMolar massmolar mass of solute = M(molL)×1L× molar mass(gmol)

Substituting these expressions into equation (1) we get,

kg solvent =(d)(1000) - M × molar mass1000kg solvent = d-M×molar mass1000......(2)

From molality definition we know that,

kg solvent =moles of the solute (n)m......(3)

We assume that 1L of the solution, we also know that mol solute (n) = Molarity (M), then the equation (3) becomes,

kg solvent =Mm

Substitute this expression into equation (2) we get,

Mm= d-M × molar massm

Take inverse of both sides of the equations,

mM = 1d-M× molar mass1000m =Md-M×molar mass1000

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

For dilute aqueous solutions, m is approximately equal to M has to be shown.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

Molality (m) =Numberofmolesofsolute1kgofsolvent

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

Molarity (M) =Numberofmolesofsolute1literofsolution

Molarity is estimation of moles in the total volume of the solution while molality is estimation of moles in relationship with solvent in the solution

Explanation of Solution

m is approximately equal to M For dilute aqueous solutions is shown as follows,

For dilute aqueous solution the density is approximately 1g/mL, because water’ density is approximately 1g/mL.

In dilute solutions, d >>M×molar mass1000

For example, 0.010M NaCl solution

M×molar mass1000=(0.010mol/L)(58.44g/mol)1000= 5.8×10-4g/L<<1

with d >> M×molar mass1000

The derived equation reduces to

mMd

Because

d1g/mL.mM

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Chapter 12 Solutions

CHEMISTRY-ALEK 360 ACCES 1 SEMESTER ONL

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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY