Download solutions and molarity for votech

Solutions and Molarity
Solutions
► The
solute is the substance dissolved in the
solution
► The solvent is the substance in which the
solute is dissolved
► Aqueous Solution is a solution in which the
solvent is water
►
►
Heterogeneous Mixture:
 A mixture in which the composition is not uniform
throughout
 Sand and water
 A salad
 Oil and water
Homogeneous Mixture:





A mixture in which the composition is uniform throughout
May be solids, liquids and/or gases
Also called a solution
Salt and water
Ice tea mix and water
Solutions
► What
is an ionic substance?
 A bond between a metal and a non-metal
 The metal gives up one or more electrons to
the non-metal and becomes positively
charged
 The non-metal accepts one or more electrons
and becomes negatively charged.
 The two ions are held together by
electrostatic attraction
 [Na+][Cl-]
Solutions
► Ionic
substances readily dissociate (come
apart) in water.
► The positive and negative ions become
attracted to individual water molecules
► Salt dissolving
► Ionic equation for salt dissolving in water:
NaCl -> Na+ + Cl-
Solutions
►
►
►
►
An electrolyte is a substance, when dissolved in water,
conducts electricity
A strong electrolyte is a solute that produces many ions in
water (sodium chloride)
A weak electrolyte is a solute that produces only a few ions
in water
A non-electrolyte does not dissociate into ions in water and
does not conduct electricity (sucrose)
Conversions
► 1000
ml = 1 L (liter)
► To convert from ml to L, divide by 1000
► To convert from L to ml, multiply by 1000
► 350 ml = ? L
► 350 ml = .350 L
► 1.7 L = ? ml
► 1.7 L = 1700 ml
Molarity
► Molarity
is an expression of solution
concentration
► Molarity = moles of solute/liter of solution
► Steps for calculating molarity
 Convert volume from ml to L (if necessary)
 Determine the molarity by dividing the number
of moles of solute by the solution volume in
liters
► Volumetric
flasks
Molarity
► 100.5
ml/1000 = .1005 L
► Molar mass of glucose = 180.16 g/mol
► 5.10 g/180.16 g/mol = .03 mol glucose
► .03 mol/.1005 L = .29 mols per liter
► .29 M
Molarity
► Practice
Problem #1: .15 M
► Practice Problem #2: .13 M
► Practice Problem #3: .008 M
Preparation of Solutions
► Convert
ml to L (if necessary)
► Determine molar mass of solute
► Set up a proportion
► Solve to get number of moles of solute
needed
► Multiply that number by the molar mass of
the solute.
Preparation of Solutions
► Example
1:
► 0.1 M means .1 mol/1 Liter





Convert to L -> 100.0 ml/1000 = 0.10 L
Molar mass NaOH = 40 g/mol
Proportion -> .1mol/1 L * x mol/.1 L
X = .01 mol of NaOH
.01 * 40.0 g/mol = 0.4 g
Preparation of Solutions Alternate
► Convert
ml to L (if necessary)
► Determine molar mass of solute
► Multiply all the numbers:
 Liters x mols/liter x molar mass
► Example
1:
 0.10 L x .1 mol/1 Liter x 40.0 g/mol = 0.4 g
Preparation of Solutions
► Practice




11.1
80.0
11.1
30.0
g
g
g
g
Problems
CaCl2
NaOH
CaCl2
NaOH
Dilutions
Solutions you work with in the lab (dropper bottles) are
made by diluting premixed concentrated stock solutions.
► By adding additional solvent to concentrated solutions
decreases the ratio of solute to solvent particles.
► Use the following relationship to calculate the volume of
solvent you may need
 M1V1 = M2V2
 M1V1 represent the molarity and volume of the stock
solution
 M2V2 represent the molarity and volume of the dilute
solution
 Remember to convert ml to L !
►
Dilutions
► Example




problem
2.00 M * V1 = 0.300 M * 0.50 L
2.00 M * V1 = 0.150
Solve for V1 = 0.150/2.00
V1 = .075 L or 75.0 ml
Dilutions
► Practice
Problems
 V1 = 0.125 L or 125 ml
 V1 = .005 L or 5.0 ml
 M2 = .7 M