**Home** / **g ml to g liter** / mL () 1000 L . 1060 () gmol - Loyola University Chicago

"A" students work

SOLUTIONS

(without solutions manual) solvent

~10 problems/night.

solute

Dr. Alanah Fitch

Flanner Hall 402

508-3119

afitch@luc.edu

Office Hours Th & F 2-3:30 pm

Define some measurement scales and units G3: Science Is referential

Module #13 molessolute

Solution Properties 1. Molarity M =

Lsolution

Defining types of concentrations 2. mole fraction nA A

For mixtures nA + nB .

3. molality molality = m = molessolute

kg solvent

Method for Conversion from Molarity to molality Example Density of an aqueous solution of ammonium sulfate is 1.06g/mL and the

molarity is 0.886. What is the molality? ( NH4 )2

SO4

1. Assume 1 L volume

2. Calculate moles of solute 1. Assume 1 L volume Atoms amu total

molessolute 2. Calculate moles of solute

Lsolution (1L) = molessolute molessolute 2 N 14.01 28.02

Lsolution (1L) = molessolute = 0.886 8( H) 8(1.008) 8.064

S 32.07 32.07

4(O) 4(16.00) 64

3. Calculate g of solution from d, V 3. Calculate g of solution from d, V total 133.098

103 mL

d solution L (1Lsolution ) = gsolution

g 1000mL

1.06 mL L ( Lsolution ) = 1060gsolution

4. Calculate g of solute from M

M (1Lsolution ) gsolute 4. Calculate g of solute from M

molsolute = gsolute mol

5. Subtract to get g of solvent 0.886 L (1Lsolution ) 133.098gsolute

molsolute = 117.9248gsolute

5. Subtract to get g of solvent

gsolution - gsolute = gsolvent 1060gsolution - 117.9248gsolute = 942.1gsoluent

6. Calculate g solute/g solvent

molality = m = molessolute 6. Calculate g solute/g solvent

kgsolvent

molality = m = molessolute = 0.886molessolute = 0.941

kgsolvent 0.9421kgsolvent

1

Method for Conversion from molality to Molarity Example Density of an aqueous solution of KOH is 1.43g/mL and the molality is

1. Assume 1 kg solvent 14.2. What is the molarity?

2. Calculate moles of solute 1. Assume 1 kg solvent

molessolute 2. Calculate moles of solute

kgsolvent 1kgsolvent = molessolute

14.2molesKOH

3. Calculate g of solute from MM 1kgwater [1kgwater] = 14.2molesKOH

(molessolute ) gsolute

molessolute = gsolute 3. Calculate g of solute from MM

4. Sum masses to get total mass of solution [14.2molesKOH] 39.10 + 16.00 + 1.08gKOH

moleKOH = 797.756gKOH

1000gsolvent + gsolute = gsolution 4. Sum masses to get total mass of solution

5. Calculate V of solution from density gsolution = gH2O + gKOH = 1000g + 797.756g = 1797.756gsolution

(gsolution ) mLsolution L

gsolution 103 mL = ( Lsolution ) 5. Calculate V of solution from density

6. Calculate moles/V (1797.756gsolution ) 1mLKOHsolution

M=

1.43gKOHsolution = 1,257.17mL

molessolute

Lsolution 6. Calculate moles/V

moles =

14.2moles,KOH = 11.296

M = Lsolution 1.257 LKOHsolution

Methods of measurement are related but not equivalent "A" students work

16 50 (without solutions manual)

Density

14 Molality 45 ~10 problems/night.

Weight %

40

12

35

10 30

Dr. Alanah Fitch

Flanner Hall 402

8 25 508-3119

afitch@luc.edu

6 20

15 Office Hours Th & F 2-3:30 pm

4

10

25

Module #13

Solution Properties

00

0 2 4 6 8 10 12 14 16

Molarity Water/Salt Solutions 1

2

Molaity or (densityx10)

weight percent

SOLUTIONS A mixture is usually plotted as variation in mole fractions

Any two components A and B

nnnAAA +++ nnnBBB == nnBB ++ nnBB = 1

AAA +++ BBB === nnnAAA +++ nnnBBB ... nnnAAA +++ nnnBBB ... nnAA ++ nnBB ..

nB B nA A

Mixed in any mole fraction nA + nB .

nA + nB.

1

0.8

0.6

0.4

0.2

0

B A 0 0.2 0.4 0.6 0.8 1

Solvent is usually Solute is usually Pure B moles of A Pure A

Considered the "dissolver" Considered the "dissolved"

mixture

Solvation Diagrammed as an Example of Hess's Law

kJ

2- H = - 91.2 mol

2+ + SO4,aq

SOLUTIONS MgSO4,s H2O Mgaq

solvent Any two components A and B

solute

Mixed in any mole fraction

-1284.9kJ

mole

-1376.1kJ

mole

water

kJ

2- H = - 91.2 mol

2+ + SO4,aq

MgSO4,s H2O Mgaq

3

mole fraction

How to convert from ML to liter?Common conversions from mL to liters:100 mL = 0.1 L250 mL = 0.25 L500 mL = 0.5 L750 mL = 0.75 L1000 mL = 1 L1250 mL = 1.25 L1500 mL = 1.5 L1750 mL = 1.75 L2000 mL = 2 L

Title: Microsoft PowerPoint - 13 Solution Properties.ppt

Author: NTUser

Creator: PScript5.dll Version 5.2

Producer: Acrobat Distiller 6.0 (Windows)

CreationDate: Mon Feb 5 08:55:26 2007

ModDate: Mon Feb 5 08:55:26 2007

Tagged: no

Form: none

Pages: 20

Encrypted: no

Page size: 612 x 792 pts (letter) (rotated 90 degrees)

File size: 1557544 bytes

Optimized: yes

PDF version: 1.4