Concentration cells.
Concentration Cells are electrochemical cells that have two equivalent half-cells of the same material differing only in concentrations. One can calculate the potential developed by such cells using the Nernst Equation. A concentration cell produces a voltage in the process of reaching equilibrium, which will occur when the concentration in both cells are equal.
Concentration cell methods of chemical analysis compare a solution of known concentration with an unknown, determining the concentration of the unknown via the Nernst Equation.
Saturday, February 2, 2008
IIT JEE Revision Ch 9. SOLUTIONS - Core Points
Jee Syllabus
Solutions:
Raoult's law;
Molecular weight determination from lowering of vapor pressure,
Molecular weight determination from elevation of boiling point
Molecular weight determination from depression of freezing point.
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Raolt's Law
In the case of a solution of two liquids, A and B, the total vapor pressure Ptot(P total) above the solution is equal to the sum of the vapor pressures of the two components, PA and PB and
PA = PA° * Am
PB = PB° * Bm
Where
PA° = vapour pressure created by 1 mol of liquid A
Am = mole fraction of liquid A in the solution
PB° = vapour pressure created by 1 mol of liquid A
Bm = mole fraction of liquid A in the solution
The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure.
If a small amount of non-volatile solute is added to the the solvent, the vapour pressure of the solution becomes less than that of the pure solvent.
Some properties of the solution depend only on the number of solute particles but on the nature of the solute. These are called colligative properties or democratic properties.
The four important ones are:
i) Relative lowering in vapour pressure
ii) elevation in boiling point
iii) depression in freezing point
iv) osmotic pressure
Molecular weight determination from lowering of vapor pressure
Molar mass of a solute can be found from the property of lowering of vapor pressure of a solution.
Mb = (Wb*Ma)/[Wa*(Pa°-Pa)/Pa°]
Wb = weight of solute particles, Wa= weight of solvent
(Pa°-Pa)/Pa° = decrease in vapour pressure of solution
Ma = Molar mass of solvent
Molecular weight determination from elevation of boiling point
Mb = [Kb*Wb*1000]/[ΔTb*Wa]
ΔTb = increase in boiling point of the solution after adding the solute
Kb = molal elevation constant or ebulloscopic constant
= the elevation in boiling point for 1 molal solution, i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of the solvent.
Molecular weight determination from depression of freezing point.
when a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.
The depression in freezing temperature is proportional to the molal concentration of the solution.
ΔTf α m Or ΔTf = Kf*m
Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
Mb = [Kf*Wb*1000]/[ΔTf * Wa]
Solutions:
Raoult's law;
Molecular weight determination from lowering of vapor pressure,
Molecular weight determination from elevation of boiling point
Molecular weight determination from depression of freezing point.
-----------------
Raolt's Law
In the case of a solution of two liquids, A and B, the total vapor pressure Ptot(P total) above the solution is equal to the sum of the vapor pressures of the two components, PA and PB and
PA = PA° * Am
PB = PB° * Bm
Where
PA° = vapour pressure created by 1 mol of liquid A
Am = mole fraction of liquid A in the solution
PB° = vapour pressure created by 1 mol of liquid A
Bm = mole fraction of liquid A in the solution
The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure.
If a small amount of non-volatile solute is added to the the solvent, the vapour pressure of the solution becomes less than that of the pure solvent.
Some properties of the solution depend only on the number of solute particles but on the nature of the solute. These are called colligative properties or democratic properties.
The four important ones are:
i) Relative lowering in vapour pressure
ii) elevation in boiling point
iii) depression in freezing point
iv) osmotic pressure
Molecular weight determination from lowering of vapor pressure
Molar mass of a solute can be found from the property of lowering of vapor pressure of a solution.
Mb = (Wb*Ma)/[Wa*(Pa°-Pa)/Pa°]
Wb = weight of solute particles, Wa= weight of solvent
(Pa°-Pa)/Pa° = decrease in vapour pressure of solution
Ma = Molar mass of solvent
Molecular weight determination from elevation of boiling point
Mb = [Kb*Wb*1000]/[ΔTb*Wa]
ΔTb = increase in boiling point of the solution after adding the solute
Kb = molal elevation constant or ebulloscopic constant
= the elevation in boiling point for 1 molal solution, i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of the solvent.
Molecular weight determination from depression of freezing point.
when a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.
The depression in freezing temperature is proportional to the molal concentration of the solution.
ΔTf α m Or ΔTf = Kf*m
Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
Mb = [Kf*Wb*1000]/[ΔTf * Wa]
IIt JEE Revision - Solutions - Introduction
a solution is a homogeneous mixture of two or more substances whose composition can be varied within certain limits.
In a solution, the component which is in excess is called solvent.
The component, that has lesser quantity is called solute.
In a solution particles are of molecular size (about 1000 pm) and the different components cannot be separated by any of the physical methods such as filtration, settling etc.
The concentration of a solution may be defined as the amount of solute present in the given quantity of the solution.
volume percent
Vol% = 100*volume of solute/(volume of solute + volume of solvent)
mass percent
mass % = 100*(mass of solute)/(mass of solute + mass of solvent)
parts per million
ppm = 10^6*(mass of solute)/(mass of solute + mass of solvent)
molality
m = number of moles of solute/kilograms of solvent
molar concentration or moles per liter or Molarity
M = number of moles of solute/liters solution
Mole fraction
Xy (y is subscript) = number of moles of y in mixture/totals moles in mixture
Normality
It is the number of gram equivalents of the solute dissolved per litre of the solution. It is denoted by N.
Normality (N) =
[Number of gram equivalents of solute]/[Volume of solution in litres]
Units of normality are gm equivalent per litre.
Formality
It is the number of formula masses of the solute dissolved per litre of the solution. It is represented by F.
Formality = [Number of formula masses of solute]/[Volume of the solution in litre]
Formality is used to express the concentrations of ionic substances like NaCl, CuSO4 etc. in solutions. They do not exist in solutions as discrete molecules. In these solutions, the sum of the atomic masses of various atoms constituting the formula of the compound (ionic) is called gram formula mass instead of molar mass.
Solubility of Gases in Liquids
Gases dissolve in liquids to form homogeneous solutions. The solubility of different gases in the same solvent varies. Gases which react with the solvent will be most soluble. The solubility of a gas decreases with temperature and increases with with increase of pressure over the solution at a given temperature.
Henry's Law
The mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure of the gas in equilibrium with the solution.
m is proportional to p
where m = mass of the gas dissolved in a unit volume of the solvent
p = pressure of the gas in equilibrium.
If pressure is more, more mass of gas is dissolved
In a solution, the component which is in excess is called solvent.
The component, that has lesser quantity is called solute.
In a solution particles are of molecular size (about 1000 pm) and the different components cannot be separated by any of the physical methods such as filtration, settling etc.
The concentration of a solution may be defined as the amount of solute present in the given quantity of the solution.
volume percent
Vol% = 100*volume of solute/(volume of solute + volume of solvent)
mass percent
mass % = 100*(mass of solute)/(mass of solute + mass of solvent)
parts per million
ppm = 10^6*(mass of solute)/(mass of solute + mass of solvent)
molality
m = number of moles of solute/kilograms of solvent
molar concentration or moles per liter or Molarity
M = number of moles of solute/liters solution
Mole fraction
Xy (y is subscript) = number of moles of y in mixture/totals moles in mixture
Normality
It is the number of gram equivalents of the solute dissolved per litre of the solution. It is denoted by N.
Normality (N) =
[Number of gram equivalents of solute]/[Volume of solution in litres]
Units of normality are gm equivalent per litre.
Formality
It is the number of formula masses of the solute dissolved per litre of the solution. It is represented by F.
Formality = [Number of formula masses of solute]/[Volume of the solution in litre]
Formality is used to express the concentrations of ionic substances like NaCl, CuSO4 etc. in solutions. They do not exist in solutions as discrete molecules. In these solutions, the sum of the atomic masses of various atoms constituting the formula of the compound (ionic) is called gram formula mass instead of molar mass.
Solubility of Gases in Liquids
Gases dissolve in liquids to form homogeneous solutions. The solubility of different gases in the same solvent varies. Gases which react with the solvent will be most soluble. The solubility of a gas decreases with temperature and increases with with increase of pressure over the solution at a given temperature.
Henry's Law
The mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure of the gas in equilibrium with the solution.
m is proportional to p
where m = mass of the gas dissolved in a unit volume of the solvent
p = pressure of the gas in equilibrium.
If pressure is more, more mass of gas is dissolved
JEE Revision - Raolt's Law
Raolt's Law
In the case of a solution of two liquids, A and B, the total vapor pressure Ptot(P total) above the solution is equal to the sum of the vapor pressures of the two components, PA and PB and
PA = PA° * Am
PB = PB° * Bm
Where
PA° = vapour pressure created by 1 mol of liquid A
Am = mole fraction of liquid A in the solution
PB° = vapour pressure created by 1 mol of liquid A
Bm = mole fraction of liquid A in the solution
In the case of a solution of two liquids, A and B, the total vapor pressure Ptot(P total) above the solution is equal to the sum of the vapor pressures of the two components, PA and PB and
PA = PA° * Am
PB = PB° * Bm
Where
PA° = vapour pressure created by 1 mol of liquid A
Am = mole fraction of liquid A in the solution
PB° = vapour pressure created by 1 mol of liquid A
Bm = mole fraction of liquid A in the solution
JEE Revision Solutions Mol. weight - Decrease Vap Pressure
Molecular weight determination from lowering of vapor pressure
Molar mass of a solute can be found from the property of lowering of vapor pressure of a solution.
Mb = (Wb*Ma)/[Wa*(Pa°-Pa)/Pa°]
Wb = weight of solute particles, Wa= weight of solvent
(Pa°-Pa)/Pa° = decrease in vapour pressure of solution
Ma = Molar mass of solvent
Molar mass of a solute can be found from the property of lowering of vapor pressure of a solution.
Mb = (Wb*Ma)/[Wa*(Pa°-Pa)/Pa°]
Wb = weight of solute particles, Wa= weight of solvent
(Pa°-Pa)/Pa° = decrease in vapour pressure of solution
Ma = Molar mass of solvent
IIT Revision Solutions
Molecular weight determination from elevation of boiling point
Molecular weight determination from elevation of boiling point
Mb = [Kb*Wb*1000]/[ΔTb*Wa]
ΔTb = increase in boiling point of the solution after adding the solute
Kb = molal elevation constant or ebulloscopic constant
= the elevation in boiling point for 1 molal solution, i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of the solvent.
Example to be added
Molecular weight determination from elevation of boiling point
Mb = [Kb*Wb*1000]/[ΔTb*Wa]
ΔTb = increase in boiling point of the solution after adding the solute
Kb = molal elevation constant or ebulloscopic constant
= the elevation in boiling point for 1 molal solution, i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of the solvent.
Example to be added
IIT JEE Revision Molecular weight - freezing point.
Molecular weight determination from depression of freezing point.
The freezing point is the temperature at which the solid and liquid states the substance have the same vapour pressure.
When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.
The depression in freezing temperature is proportional to the molal concentration of the solution (m).
ΔTf α m Or ΔTf = Kf*m
ΔTf = depression in freezing point.
Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
When m =1; ΔTf = Kf
Depression in freezing point is a colligatvie property as it is directly proportional to the molar concentration of the solute.
To find the molar mass of an unknown substance (nonvolatile compound), a known mass of it is dissolved in a known mass of a solvent and depression in its freezing point (ΔTf)is measured.
weight of solute be Wb g
weight of the solvent be Wa g
Molar mass of the solute be Mb
Molality of the solution, m = Wb*1000/Mb*Wa
Substitute the value of m in ΔTf = Kf*m = Kf*Wb*1000/Mb*Wa
From the above equation Mb can be calculated.
Mb = Kf*Wb*1000/Wa*ΔTf
Example:
Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg molˉ¹, calculate the molar mass of the compound. (IIT 1992)
The formula of Mb is available above.
weight of solute be Wb g = 0.643 g
weight of the solvent be Wa g = 50*0.879 = 43.95 g
Change in freezing point = 5.51 - 5.03 = 0.48°C
Mb = (5.12 * 0.643 * 1000)/(43.95*0.48)
Mb = [Kf*Wb*1000]/[ΔTf * Wa]
The freezing point is the temperature at which the solid and liquid states the substance have the same vapour pressure.
When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.
The depression in freezing temperature is proportional to the molal concentration of the solution (m).
ΔTf α m Or ΔTf = Kf*m
ΔTf = depression in freezing point.
Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
When m =1; ΔTf = Kf
Depression in freezing point is a colligatvie property as it is directly proportional to the molar concentration of the solute.
To find the molar mass of an unknown substance (nonvolatile compound), a known mass of it is dissolved in a known mass of a solvent and depression in its freezing point (ΔTf)is measured.
weight of solute be Wb g
weight of the solvent be Wa g
Molar mass of the solute be Mb
Molality of the solution, m = Wb*1000/Mb*Wa
Substitute the value of m in ΔTf = Kf*m = Kf*Wb*1000/Mb*Wa
From the above equation Mb can be calculated.
Mb = Kf*Wb*1000/Wa*ΔTf
Example:
Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg molˉ¹, calculate the molar mass of the compound. (IIT 1992)
The formula of Mb is available above.
weight of solute be Wb g = 0.643 g
weight of the solvent be Wa g = 50*0.879 = 43.95 g
Change in freezing point = 5.51 - 5.03 = 0.48°C
Mb = (5.12 * 0.643 * 1000)/(43.95*0.48)
Mb = [Kf*Wb*1000]/[ΔTf * Wa]
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