Importance of Core Revision Points: Core Revision Points are important because if you remember them strongly, many more points related to them will come out of your memory and help you to answer question and problems. Read them many times and make sure you remember them very strongly.
Sections in the Chapter
3.1 Types of solutions
3.2 Methods for expressing the concentration of a solution: units of solution
3.3 Solubilioty of gases and solids in liquids
3.4 Vapour pressure of solutions
3.5 Ideal and Nonideal solutions
3.6 Colligative properties
3.7 Relative lowering of vapour pressure
3.8 Elevation in boiling point
3.9 Depression in freezing point
3.10 Osmosis and osmotic pressure
3.11 Electrolytic solutions – Abnormal molar masses
Sections in the Chapter
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.
3.2 Methods for expressing the concentration of a solution: units of solution
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
When a liquid is placed in a vessel and is covered with jar, from the liquid evaporation takes place and the vapour of the liquid or molecules of the liquid in gap form fill the available space. As the evaporation takes place over a period of time, the number of gaseous molecules goes up. As evaporation is taking place some molecules in the gaseous phase collide with the surface of the liquid and become liquid molecules. Thus both evaporation and condensation take place simultaneously. But initially there is more evaporation and less condensation. At the some stage, rate of evaporation equals rate of condensation and equilibrium is established between gas and liquid phases. The pressure exerted by the vapours at the equilibrium stage is called vapour pressure.
Definition
The pressure exerted by the vapours above the liquid surface (in a closed vessel) in equilibrium with the liquid at a given temperature is called vapour pressure.
Vapour pressure changes from liquid to liquid. It depends on intermolecular forces. if the forces in a liquid are weak, there is more gas formation and hence more vapour pressure.
A higher temperature there is more gas formation and hence for the same liquid vapour pressures increase with temperature.
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
3.5 Ideal and Nonideal solutions
3.6 Colligative properties
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
3.8 Elevation in boiling 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]
3.10 Osmosis and osmotic pressure
3.11 Electrolytic solutions – Abnormal molar masses
Solute: the substance dissolved; the substance present in a solution in the lesser amount.
Solvent: the dissolving medium; the substance in a solution in the greater amount.
Solution: a homogeneous mixture of two or more substances.
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
Solutions Part 1
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3. Solutions - JEE - CBSE Class XII - Revision Questions
Sections in the Chapter
3.1 Types of solutions
3.2 Methods for expressing the concentration of a solution: units of solution
3.3 Solubilioty of gases and solids in liquids
3.4 Vapour pressure of solutions
3.5 Ideal and Nonideal solutions
3.6 Colligative properties
3.7 Relative lowering of vapour pressure
3.8 Elevation in boiling point
3.9 Depression in freezing point
3.10 Osmosis and osmotic pressure
3.11 Electrolytic solutions – Abnormal molar masses
Sections in the Chapter
3.1 Types of solutions
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.
3.2 Methods for expressing the concentration of a solution: units of solution
3.3 Solubilioty of gases and solids in liquids
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
3.4 Vapour pressure of solutions
When a liquid is placed in a vessel and is covered with jar, from the liquid evaporation takes place and the vapour of the liquid or molecules of the liquid in gap form fill the available space. As the evaporation takes place over a period of time, the number of gaseous molecules goes up. As evaporation is taking place some molecules in the gaseous phase collide with the surface of the liquid and become liquid molecules. Thus both evaporation and condensation take place simultaneously. But initially there is more evaporation and less condensation. At the some stage, rate of evaporation equals rate of condensation and equilibrium is established between gas and liquid phases. The pressure exerted by the vapours at the equilibrium stage is called vapour pressure.
Definition
The pressure exerted by the vapours above the liquid surface (in a closed vessel) in equilibrium with the liquid at a given temperature is called vapour pressure.
Vapour pressure changes from liquid to liquid. It depends on intermolecular forces. if the forces in a liquid are weak, there is more gas formation and hence more vapour pressure.
A higher temperature there is more gas formation and hence for the same liquid vapour pressures increase with temperature.
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
3.5 Ideal and Nonideal solutions
3.6 Colligative properties
3.7 Relative lowering of vapour 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
3.8 Elevation in boiling point
3.9 Depression in 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]
3.10 Osmosis and osmotic pressure
3.11 Electrolytic solutions – Abnormal molar masses
Glossary
Solute: the substance dissolved; the substance present in a solution in the lesser amount.
Solvent: the dissolving medium; the substance in a solution in the greater amount.
Solution: a homogeneous mixture of two or more substances.
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
Solutions Part 1
_________________
_________________
ExamFearVideos
35 Videos Playlist
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Revision Questions
3. Solutions - JEE - CBSE Class XII - Revision Questions
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