Moles of solute
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moles of solute. liters of solution. moles of solute. m =. mass of solvent (kg). M =. Concentration Units Continued. Molarity (M). Molality (m). 12.3. moles of solute. M = molarity =. How many moles of KI are required to make 500. mL of a 2.80 M KI solution?.

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Concentration Units Continued

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Concentration units continued

moles of solute

liters of solution

moles of solute

m =

mass of solvent (kg)

M =

Concentration Units Continued

Molarity(M)

Molality(m)

12.3


Concentration units continued

moles of solute

M = molarity =

How many moles of KI are required to make

500. mL of a 2.80 M KI solution?

liters of solution

Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.


Concentration units continued

mass of solute

Mass %=

What is the mass percentage of 2 moles of NaOH

dissolved in 358 ml of water?

mass of solution

Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Another way this can be expressed is in mass percentage.

X 100


Concentration units continued

moles of solute

m =molality =

What is the molality of a solution that has contains

83.05g of KI dissolved in 500ml of water?

kg of solvent

Solution Stoichiometry

The concentration of a solution can also be expressed as the moles of solute per kg of solvent.


Concentration units continued

moles of solute

m=

mass of solvent (kg)

What is the molality of a solution made from 155g of sodium chloride and 1500g of water?

155g of NaCl = 2.65 mol NaCl

1500g of water = 1.5kg of water

2.65 moles NaCl

____________________

= 1.8m

1.5kg water


Change in boiling point

Change in Boiling Point

Common Applications of Boiling Point Elevation


Concentration units continued

12.6


Concentration units continued

0

DTf = T f – Tf

moles of solute

m=

mass of solvent (kg)

=

3.202 kg solvent

1 mol

62.01 g

478 g x

0

Tf = T f – DTf

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

DTf = Kfm

Kf water = 1.86 0C/m

= 2.41 m

DTf = Kfm

= 1.86 0C/m x 2.41 m = 4.48 0C

= 0.00 0C – 4.48 0C = -4.48 0C

12.6


Concentration units continued

actual number of particles in soln after dissociation

van’t Hoff factor (i) =

number of formula units initially dissolved in soln

Colligative Properties of Electrolyte Solutions

0.1 m NaCl solution

0.1 m Na+ ions & 0.1 m Cl- ions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

0.1 m NaCl solution

0.2 m ions in solution

i should be

1

nonelectrolytes

2

NaCl

CaCl2

3

12.7


Change in freezing point

Change in Freezing Point

  • Which chemical below would be besttode-ice a frozen street and why?

  • sand, SiO2

  • Rock salt, NaCl

  • Ice Melt, CaCl2


Freezing point depression

Freezing Point Depression

At what temperature will a 5.4 molal solution of NaCl freeze?

Solution

∆TFP = Kf • m • i

∆TFP = (1.86 oC/molal) • 5.4 m • 2

∆TFP = 20.oC

FP = 0 – 20. = -20.oC


Triple point diagram

Triple point diagram

solid

liquid

gas


Phase diagram

Phase diagram


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