Solutions and Their Types

  • A solution is a homogeneous mixture of two or more than two or more components. A solution has a single phase.
  • The component that is present in the largest quantity is known as a solvent. Solvent determines the physical state in which solution exists. One or more components present in the solution other than solvent are called solutes. If solution consists of two chemical components, the solution is called binary solution. If it contains three or four chemical components it is called ternary or quaternary solution.
  • For example when sugar is dissolved in water, then a sugar solution in water is formed. In this case, water is the solvent, sugar is solute and their mixture is a solution. the state of solvent is liquid, hence the state of the resultant solution is also liquid.
  • If the composition and properties are uniform throughout the mixture, the solution is called homogeneous solution.
  • It is to be noted that the constituent particles of solution cannot be separated by filtration, settling or centrifugal action.


Classification of Solutions on the Basis of Size of Solute Particles:

  • Depending upon the size of solute particles, the solutions are broadly classified into three types.

Coarse Solution:

  • When the size of constituent particles is relatively bigger, the the coarse solution is obtained.
  • e.g. a mixture of salt and sugar.

Colloidal Solution:

  • When the size of the particles dispersed in the solvent is in the range of 10-9 m to 10-6 m., then the colloidal solution is formed.
  • The stability of colloidal solution is due to the same nature, of the charge carried by the colloidal particles.
  • e.g. Ferric hydroxide sol

True Solution:

  • Aa homogeneous mixture of two or more substances, the composition of which is not fixed and may be varied within the certain limit is called a true solution.
  • The size of the particles dispersed in the solvent is less than 10-6 m.
  • e.g. Solution of common salt in water.

Classification of Solutions on the Basis of Phases of Solute and Solvent:

  • Depending upon the phases of solute and solvent the solutions are broadly classified into three types.

Gaseous Solution

Sr.No.

Solute Phase

Solvent Phase

Name of Solution

Examples
1 Solid Gas Solid in gas

Iodine in air, moth balls in air

2 Liquid Gas Liquid in gas

Water vapours in air, chloroform in nitrogen

3 Gas Gas Gas in gas

Mixture of non-reacting gases (O2+ N2), air, natural gas, oxygen + Acetylene mixture

Liquid Solutions

4

Solid Liquid Solid in liquid

Sugar in water, salt in water, amalgams

5

Liquid Liquid Liquid in liquid

Ethanol in water, antifreeze in water

6

Gas Liquid Gas in liquid

Carbon dioxide in water, Oxygen in fresh river water

Solid Solutions

7 Solid Solid

Solid in solid

Alloys like brass, bronze
8 Liquid Solid

Liquid in solid

Amalgam of mercury with metal
9 Gas Solid

Gas in solid

Hydrogen gas on palladium or platinum


  • Sometimes solutions of liquid in gas and solid in the gas are not considered as solutions because the mixture may not be homogeneous.

Concept of Concentration of Solution

  • A Concentration of a solution is the measure of the composition of a solution. For a given solution, the amount of solute dissolved in a unit volume of solution (or a unit volume of solvent) is called concentration of the solution.
  • It can be expressed either qualitatively or quantitatively. For example, qualitatively we can say that the solution is dilute (i.e., relatively very small quantity of solute) or it is concentrated (i.e., relatively very large quantity of solute). But in practice, it is not useful hence it is not used in chemistry. Quantitative description method gives an exact concentration of the solution and hence its concentration can be compared with the concentration of other solution. e.g. 20% HCl, 10% NaOH etc.

Methods of Expressing Concentration of the Solution Quantitatively:

Percentage by Mass or Mass Percentage (w/w):

  • The mass of solute in gram dissolved in a solvent to form 100 gram of solution is called percentage by mass.
  • The ratio of the mass of solute to the mass of solution is called mass fraction.
  • For example, if a solution is described by 10% glucose in water by mass, it means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution.

  • This method is used for solid in liquid solution. Concentration described by mass percentage is commonly used in industrial chemical applications. For example, commercial bleaching solution contains 3.62 mass percentage of sodium hypochlorite in water.

Example – 1:

  • 6 g of urea was dissolved in 500 g of water. Calcualte the percentage by mass of urea in solution.
  • Solution:
  • Given: Mass of solute = 6 g, Mass of Solvent = 500 g
  • To Find: Percentage by mass (W/W) = ?

Mass of Solution = Mass of solute + Mass of Solvent = 6 g + 500 g = 506 g

Percentage by mass (W/W) = (Mass of solute /Mass of solution) × 100

∴ Percentage by mass (W/W) = (6 g/506 g) × 100 = 1.186 % by mass

Ans: percentage by mass of urea in solution is 1.186



Example – 2:

  • 34.2 g of glucose is dissolved in 400 g of water. Calculate the percentage by mass of glucose solution.
  • Solution:
  • Given: Mass of solute = 34.2 g, Mass of Solvent = 400 g
  • To Find: Percentage by mass (W/W) = ?

Mass of Solution = Mass of solute + Mass of Solvent = 34.2 g + 400 g =434.2 g

Percentage by mass (W/W) = (Mass of solute /Mass of solution) × 100

∴ Percentage by mass (W/W) = (34.2 g/434.2 g) × 100 = 7.877 % by mass

Ans: percentage by mass of glucose in solution is 7.877

Example – 3:

  • A solution is prepared by dissolving a certain amount of solute in 500 g of water. The percentage by mass of a solute in a solution is 2.38. Calculate mass of the solute.
  • solution.
  • Solution:
  • Given: Mass of solute = x g, Mass of Solvent = 500 g, Percentage by mass (W/W) = 2.38
  • To Find: Mass of solute = x = ?

Mass of Solution = Mass of solute + Mass of Solvent = x g + 500 g =( x + 500) g

Percentage by mass (W/W) = (Mass of solute /Mass of solution) × 100

∴  2.38 = (x g/( x + 500) g) × 100

∴  2.38( x + 500) = 100 x

∴  2.38 x + 1190 = 100 x

∴ 100 x – 2.38 x = 1190

∴ 97.62 x = 1190

∴ x = 1190/97.62 = 12.19 g

Ans: Mass of solute is 12.19 g

Percentage by Volume (V/V):

  • For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL water such that the total volume of the solution is 100 mL.

Solutions

  • This method is used for liquid in liquid solution.


Example – 4:

  • 58 cm3 of ethyl alcohol was dissolved in 400 cm3 of water to form 454 cm3 of a  solution of ethyl alcohol. Calculate percentage by volume of ethyl alcohol in water.
  • Solution:
  • Given: Volume of solute = 58 cm3, Volume of Solution = 454 cm³
  • To Find: Percentage by volume (V/V) = ?

Percentage by volume (V/V) = (Volume of solute /Volume of solution) × 100

∴ Percentage by volume (V/V) = (58 cm³/454 cm³) × 100 = 12.78 % by volume

Ans: percentage by volume of ethyl alcohol in solution is 12.78

Example – 5:

  • 12.8 cm3 of benzene is dissolved in 16.8 cm3 of xylene. Calculate percentage by volume of benzene.
  • Solution:
  • Given: Volume of solute = 12.8 cm3, Volume of Solvent = 16.8 cm³
  • To Find: Percentage by volume (V/V) = ?

Volume of solution = Volume of solute + Volume of solvent

Volume of solution = 12.8 cm³ + 16.8 cm³ = 29.6  cm³

Percentage by volume (V/V) = (Volume of solute /Volume of solution) × 100

∴ Percentage by volume (V/V) = (12.8 cm³/29.6 cm³) × 100 = 43.24 % by volume

Ans: percentage by volume of benzene in solution is 43.24

Percentage by Mass by Volume (w/V):
  • It is the mass of solute dissolved in 100 mL of the solution.

  • This method is commonly used in medicine and pharmacy.

Parts per million:

  • When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm) and is defined as:

  • Example: A litre of seawater (which weighs 1030 g) contains about 6 × 10–3 g of dissolved oxygen (O2). Such a small concentration is also expressed as 5.8 g per 106 g (5.8 ppm) of seawater.
  • The concentration of pollutants in water or atmosphere is often expressed in terms of μ g mL–1 or ppm.

Strength or Concentration (Grams per litre):

  • It is defined as the amount of the solute in gram present in the one litre of the solution.


Mole Fraction:

  • The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all components of the solution.

  • The concept of mole fraction is very useful in relating some physical properties of solutions, such as vapour pressure with the concentration of the solution and quite useful in describing the calculations involving gas mixtures.
  • It is to be noted that the sum of mole fraction of solute and mole fraction of liquid is 1.
  • Mole fraction is independent of temperature

Example – 6:

  • 23 g of ethyl alcohol (molar mass 46 g mol-1) is dissolved in 54 g of water (molar mass 18 g mol-1). Calculate the mole fraction of ethyl alcohol and water in solution.
  • Solution:
  • Given: Mass of ethyl alcohol = 23 g, molar mass of ethyl alcohol = 46 g mol-1, mass of water = 54 g, molar mass of water = 18 g mol-1
  • To Find: Mole fraction = ?

Number of moles of solute = nsolute = mass of ethyl alcohol/ molar mass of alcohol

∴  nsolute = 23 g /46 g mol-1  = 0.5 mol

Number of moles of solvent = nsolvent = mass of water/ molar mass of water

∴  nsolvent = 54 g /18 g mol-1  = 3.0 mol

Mole fraction of solute (ethyl alcohol) =  (nsolute )/(nsolute + nsolvent)

∴ Mole fraction of solute (ethyl alcohol) = 0.5/(0.5 + 3) = 0.5/3.5 = 0.1429

Mole fraction of solvent (water) =  (nsolvent )/(nsolute + nsolvent)

∴ Mole fraction of solute (water) = 3.0/(0.5 + 3) = 3.0/3.5 = 0.8571

Ans:  mole fraction of ethyl alcohol is 0.1429 and that of water is 0.8571

Example – 7:

  • 4.6 cm3 of methyl alcohol is dissolved in 25.2 g of water. Calculate a) percentage by mass of methyl alcohol b) mole fraction of methyl alcohol and water. Given density of methyl alcohol = 0.7952 g cm-3, and C = 12, H = 1 and O = 16.
  • Solution:
  • Given: Volume of methyl alcohol = 4.6 cm3, Mass of water = 25.2 g, density of methyl alcohol = 0.7952 g cm-3, and C = 12, H = 1 and O = 16.
  • To Find:  percentage by mass of methyl alcohol = ? Mole fraction = ?

Molecular mass of methyl alcohol (CH3OH) = 12 g  × 1 + 1 g  × 4 + 16 g × 1 = 32 g mol-1 

Molecular mass of water (H2O) = 1  g  × 2 + 16 g  × 1  = 18 g mol-1 

Mass of methyl alcohol = volume of methyl alcohol × its density

∴   Mass of methyl alcohol =4.6 cm3  ×0.7952 g cm-3  = 3.658 g

Mass of Solution = Mass of solute + Mass of Solvent = 3.658 g + 25.2 g = 28.858 g

Percentage by mass (W/W) = (Mass of solute /Mass of solution) × 100

∴ Percentage by mass (W/W) = (3.658 g/28.858 g) × 100 = 12.68 % by mass

Number of moles of solute = nsolute = mass of methyl alcohol/ molar mass of alcohol

∴  nsolute = 3.658 g /32 g mol-1  = 0.1143 mol

Number of moles of solvent = nsolvent = mass of water/ molar mass of water

∴  nsolvent = 25.2 g /18 g mol-1  = 1.4 mol

Mole fraction of solute (methyl alcohol) =  (nsolute )/(nsolute + nsolvent)

∴ Mole fraction of solute (methyl alcohol) = 0.1143/(0.1143 + 1.4) = 0.1143/1.5143 = 0.0755

Mole fraction of solvent (water) =  (nsolvent )/(nsolute + nsolvent)

∴ Mole fraction of solute (water) = 1.4/(0.1143 + 1.4) = 1.4/1.5143 = 0.9245

Ans:  percentage by mass of methyl alcohol is 12.68,

mole fraction of methyl alcohol is 0.0755 and that of water is 0.9245



Example – 8:

  • Calculate the mole fraction of HCl in a solution of HCl containing 24.8 % of HCl by mass. Given H = 1, Cl = 35.5
  • Solution:
  • Given: 24.8 % of HCl by mass, H = 1, Cl = 35.5
  • To Find:  percentage by mass of methyl alcohol = ? Mole fraction = ?

Molecular mass of HCl = 1 g  × 1 + 35.5 g  × 1 =  36.5 g mol-1 

Molecular mass of water (H2O) = 1  g  × 2 + 16 g  × 1  = 18 g mol-1 

Consider 100 g of HCl solution

Mass of HCl = 24.8 g and Mass of water = 100 -24.8 = 75.2 g

Number of moles of solute = nsolute = mass of HCl/ molar mass of HCl

∴  nsolute = 24.8 g /36.5 g mol-1  = 0.6795 mol

Number of moles of solvent = nsolvent = mass of water/ molar mass of water

∴  nsolvent = 75.2 g /18 g mol-1  = 4.178 mol

Mole fraction of solute (HCl) =  (nsolute )/(nsolute + nsolvent)

∴ Mole fraction of solute (HCl) = 0.6795/(0.6795 + 4.178) = 0.6795/4.8575 = 0.1399

Ans:  Mole fraction of HCl is 0.1399

Molarity (Molar Concentration):

  • Molarity (M) is defined as a number of moles of solute dissolved in one litre (or one cubic decimetre) of the solution. Unit of molarity is mol L-1 0r mol dm-3 or M.

  • Number of moles of a substance can be found using the formula

  • Molarity changes with temperature because volume changes with temperature.
  • Molarity can be expressed as Decimolar = M/10 (0.1 M), Semimolar = M/2 (0.5 M), Pentimolar = M/5 (0.2 M), Centimolar = M/100 (0.01 M) or milimolar = M/1000 (0.001 M).

Example – 9:

  • A solution of NaOH (molar mass 40 g mol-1) was prepared by dissolving 1.6 g of NaOH in 500 cm3 of water. Calculate the molarity of NaOH solution.
  • Solution:
  • Given: Mass of NaOH = 1.6 g, Molar mass of NaOH = 40 g mol-1,  Volume of water = 500 cm³ = 500 ×10-3 dm³ =  0.5 dm³
  • To Find:  Molarity of solution =?

Number of moles of solute (NaOH) = nsolute = mass of NaOH/ molar mass of NaOH

∴  nsolute = 1.6/40 g mol-1  = 0.04 mol

Molarity = Number of moles of solute/ volume of solvent in  dm³

∴  Molarity = 0.04/ 0.5 = 0.08 M or 0.08 mol dm-3.

Ans: Molarity of solution is 0.08 M or 0.08 mol dm-3



Molality:

  • Molality (m) is defined as a number of moles of solute expressed in kg dissolved in one kg of solvent, Molality has no unit.
  • Molality is a better way of expressing concentration than molarity because there is no term of volume of solvent is involved. The volume of solvent depends on the temperature of the solvent. Thus there is no effect of change of temperature on the molality.

  • Molality is related to solubility as

Example – 10:

  • 11.11 g of urea (NH2CONH2) was dissolved in 100 g of water. calculate the molality of the solution. Given N = 14, H = 1, C = 12, O = 16.
  • Solution:
  • Given: Mass of urea = 11.11 g, Mass of water = 100g = 0.1 kg, N = 14, H = 1, C = 12, O = 16
  • To Find:  Molarity of solution =?

Molar mass of urea (NH2CONH2) = 14 × 2 + 1 × 4 + 12 × 1  + 16 × 1 = 60 g mol-1

Number of moles of solute (urea) = nsolute = mass of urea/ molar mass of urea

∴  nsolute = 11.11/60 g mol-1  = 0.1852 mol

Molality = Number of moles of solute/ volume of solvent in  kg

∴  Molality = 0.1852/ 0.1 = 1.852 mol kg-1.

Ans: Molality of the solution is 1.852 mol kg-1.

Example – 11:

  • 34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality and mole fraction of sugar in the syrup. Given C = 12, H = 1 and O = 16.
  • Solution:
  • Given: Mass of sugar = 34.2 g, Mass of sugar syrupr = 214.2 g,  = 0.1 kg, C = 12, H = 1 and O = 16.
  • To Find:  Molality of solution =?, Mole fraction of sugar = ?

Molar mass of sugar (C12H22O11) = 12 × 12 + 1 × 22 + 16 × 11 = 342 g mol-1

Number of moles of solute (Sugar) = nsolute = mass of sugar/ molar mass of sugar

∴  nsolute = 34.2 g/342 g mol-1  = 0.1  mol

Mass of water = 214.2 – 34.2 = 180 g = 0.18 kg

Molality = Number of moles of solute/ volume of solvent in  kg

∴  Molality = 0.1 / 0.18 = 0.556 mol kg-1.

Molar mass of water (H2O) = 1 × 2 + 16 × 1 = 18 g mol-1

Number of moles of solvent (water) = nsolvent = mass of water/ molar mass of water

∴   nsolvent = 180 g/18 g mol-1  = 10  mol

Mole fraction of solute (sugar) =  (nsolute )/(nsolute + nsolvent)

∴ Mole fraction of solute (sugar) = 0.1/(0.1 + 10) = 0.1/10.1 = 0.0099

Ans: Molality of the solution is 0.556 mol kg-1 and mole fraction of sugsr = 0.0099



Example – 12:

  • Calculate the mole fraction of solute in its 2 molal aqueous solution.
  • Solution:
  • Given: Molality of solution = 2 molal.
  • To Find:  Mole fraction = ?

Molality of solution = 2 molal – 2 mol kg-1

Number of moles of solute = 2

Mass of solvent (water) = 1 kg = 1000 g

Molar mass of water = 18 g mol-1

Number of moles of solvent (water) = nsolvent = mass of water/ molar mass of water

∴   nsolvent = 1000 g/18 g mol-1  = 55.56  mol

Mole fraction of solute =  (nsolute )/(nsolute + nsolvent)

∴ Mole fraction of solute = 2/(2 + 55.56) = 2/57.56 = 0.0347

Ans: Mole fraction of solute = 0.0347

Example – 13:

  • Calculate molarity and molality of the sulphuric acid solution of density 1.198 g cm-3 containing 27 % by mass of sulphuric acid (molar mass 98 g mol-1).
  • Solution:
  • Given: Density of sulphuric acid = 1.198 g cm-3 ,27 % by mass of sulphuric acid, molar mass of sulphuric acid 98 g mol-1= 98 × 10-3 kg mol-1
  • To Find:  Molarity = ?, Molality = ?

Consider 100 g of sulphuric acid solution

Mass of sulphuric acid = 27 g

Mass of water = 100 – 27 = 73 g

Molality = Number of moles of solute/mass of solvent in  kg

∴  Molality = Mass of solute/ (Molar mass of solute × mass of solvent in  kg)

∴  Molality = 27 g/ (98 g mol-1 × 73 × 10-3 kg)  = 3.774 mol kg-1.

Density of solution = mass of solution / volume of solution

∴ Volume of solution  = mass of solution / Density of solution

∴ Volume of solution = 100 g  /1.198 g cm-3 = 83.47 cm³ = 83.47 × 10-3 dm³

Molarity = Number of moles of solute/ (Molar mass of solute × volume of solution in  dm³)

Molarity = Number of moles of solute/ (Molar mass of solute × volume of solution in  dm³)

∴  Molarity = 27 g/ (98 g mol-1 × 83.47 × 10-3 dm³)  = 3.301 mol dm-3.

Ans: Molarity of solution is 3.301 mol dm-3 and molality is 3.774 mol kg-1.



Example – 14:

  • Calculate the mole fraction, molality, and molarity of HNO3 in a solution containing 12.2 % HNO3. Given density of HNO3 as 1.038 g cm-3, H = 1, N = 14, O = 16.
  • Solution:
  • Given:12.2 % HNO3, Density of HNO3 as 1.038 g cm-3, H = 1, N = 14, O = 16.
  • To Find: Molarity = ? , molality of solution =?, Mole fraction of HNO= ?

Molar mass of sugar HNO = 1 × 1 + 14 × 1 + 16 × 3 = 63 g mol-1

Consider 100 g of HNO3 solution

Mass of sulphuric acid = 12.2 g

Mass of water = 100 – 12.2 = 87.8 g

Number of moles of solute ( HNO3) = nsolute = mass of HNO3/ molar mass of HNO3

∴  nsolute = 12.2 g/63 g mol-1  = 0.1937  mol

Molecular mass of solvent (H2O) = 18 g mol-1

Number of moles of solven (H2O) = nsolute = mass of H2O/ molar mass of H2O

∴  nsolvent = 87.8 g/18  g mol-1  = 4.878  mol

Mole fraction of solute (HNO3) =  (nsolute )/(nsolute + nsolvent)

∴ Mole fraction of solute (HNO3) =  0.1937/(0.1937 + 4.878) = 0.1937 / 5.072 = 0.0382

Molality of (HNO3) =Mass of solute/ (Molar mass of solute × mass of solvent in kg)

Molality of (HNO3) =12.2 g / (63 g × 87.8 k× 10-3 g ) = 2.205 mol kg-1

Density of solution = Mass of solution / volume of solution

Volume of solution = Mass of solution / Density of solution = 100 g/1.038 g cm-3  =  96.34 cm³ = 96.34 × 10-3 dm³

Molarity of (HNO3) =Mass of solute/ (Molar mass of solute × volume of solution in dm³)

Molarity of (HNO3) =12.2 g / (63 g × 96.34 × 10-3 dm³ ) = 2.01 mol dm-3

Example – 15:

  • Sulphuric acid is 95.8 % by mass. Calculate mole fraction and molarity of H2SO4 of density 1.91 g cm-3. Given H = 1, S = 32, O = 16
  • Solution:
  • Given: Density of sulphuric acid = 1.198 g cm-3 ,95.8 % by mass of sulphuric acid, density of sulphuric acid = 1.91 g cm-3.
  • To Find:  Mole fraction = ?, Molality = ?

Consider 100 g of sulphuric acid solution

Mass of sulphuric acid = 95.8 g

Mass of water = 100 – 95.8 = 4.2 g

Molar mass of sulphuric acid = 1 × 2 + 32 × 1 + 16 × 4 = 98 g mol-1

Molar mass of water = 18 g mol-1

Number of moles of solute (sulphuric acid) = mass of sulphuric acid / molar mass of sulphuric acid

∴ Number of moles of solute (sulphuric acid) = 95.8 g/98 g mol-1  = 0.9776

∴ Number of moles of solvent (water) = 4.2 g/18 g mol-1  = 0.2333

Mole fraction of solute (sulphuric acid) =  (nsolute )/(nsolute + nsolvent)

∴ Mole fraction of solute (sulphuric acid) = 0.9776/(0.9776 +0.2333) = 0.9776/1.2109 = 0.8073

Density  = mass  / volume

∴ Volume   = mass  / Density

∴ Volume of sulphuric acid = 95.8 g  /1.91 g cm-3 = 50.16 cm³ =50.16 × 10-3 dm³

∴ Volume of water = 4.2 g  /1 g cm-3 = 4.2 cm³ = 4.2 × 10-3 dm³

 

Molarity = Number of moles of solute/  volume of solution in  dm³

∴  Molarity = 0.9776 mol/52.36 × 10-3 dm³ = 3.301 mol dm-3.

Ans: Molarity of solution is 3.301 mol dm-3 and molality is 3.774 mol kg-1.

. (0.80730, 17.98 mol dm-3)



Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. If its density is 1.1 g cm-3, calculate the molarity of HCl solution and also calculate the mole fraction of HCl and H2O.  (11.45 mol dm-3, 0.232, 0.768)

An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm-3. Calculate molarity, molality and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16 (2.675 mol dm-3, 2.77 mol kg-1, 0.0476, 0.9523)

Battery acid 4.22 M aqueous H2SO4 solution, and has density 1.21 g cm-3. What is the molality of H2SO4. Given H = 1, S = 32, O = 16 (5.298 mol kg-1)

Calculate the mole fraction and molality of HNO3 in a solution containing 12.2 % HNO3. Given atomic masses H = 1, N = 14 and O = 16.

Calculate molarity and molality of 6.3 % solution of nitric acid having density 1.04 g cm-3. Given atomic masses H = 1, N = 14 and O = 16.

A solution of glucose in water is labelled as 10 % (w/w). Calculate a) molality and b) molarity of the solution. Given the density of the solution is 1.20 g mL-1  and molar mass of glucose is 180 g mol-1.

Sulphuric acid is 95.8 % by mass. Calculate the mole fraction and molarity of H2SO4. Density of H2SO4 solution is 1.91 g cm-3.

The density of 5.35 M H2SO4 solution is 1.22 g cm-3. What is molality of a solution?

10.0 g KCl is dissolved in 1000 g of water. If the density of the solution is 0.997 g cm-3, calculate a) molarity and b) molality of the solution. Atomic masses K = 39 g mol-1 , Cl = 35.5 g mol-1.

Normality:

  • Normality (N) is defined as gram-equivalent of solute dissolved in one litre (or one cubic decimetre) of the solution, Unit of molarity is N.
  • A solution having normality equal to unity is called normal solution.
  • Decinormal = N/10 (0.1 N), seminormal = N/2  (0.5 N)
  • Normality × equivalent mass = strength of solution in g/L.

Formality:

  • Formality is the number of formula mass in gram present per litre of a solution.
  • If the formula mass of solute is equal to its molar mass, then the formality is equal to molarity.
  • The formality of a solution depends on temperature. This concept is used in case of ionic substances.
  • The mole of the ionic compound is called formole and its molarity is called formality. Thus, the formality of a solution may be defined as a number of moles of ionic solute present in one litre of the solution.

The relation between mole fraction and molality:

  • The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all components of the solution.
  • Let us consider a binary solution components solvent (A) and solute (B)

Molarity of Dilution:

Molarity of Mixing:

Relation Between Molarity and Molality:

Density of a solution is in g/mL

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