Chemistry FBISE SSC-1 Keybook Chapter 6

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  1. Define solution, aqueous solution, solute and solvent?

Ans.     Solution:

A solution is a homogeneous mixture of substances that has uniform composition throughout.

Solute:

In a solution the substance that is present in lesser amount is called solute.

Solvent:

In a solution the substance that is present in larger amount is called solvent.

Aqueous solution:

A solution in which water is solvent is called an aqueous solution.

 

  1. What is supersaturated solution?

A solution that contains more of the solute than is contained in the saturated solution is called supersaturated.

 

  1. How we know whether a solution is saturated or supersaturated?

Ans.     A supersaturated solution is not stable in the presence of crystals of solute. If we add a crystal to its saturated solution, it will simply drop to the bottom. But if we add it to a supersaturated solution, crystallization will start.

 

The maximum amount of sodium acetate that dissolves in 100g of water at 0 oC is 119g and 170g at 100 oC.

(a)        If you add 170g of sodium acetate in 100g of water at 0oC, how much will dissolve?

Ans.     119g

 

(b)        Is the solution saturated, unsaturated or supersaturated?

Ans.     Saturated.

 

(c)        If the solution is heated to 100 oC, is the solution now saturated, unsaturated or supersaturated.

Ans.     Supersaturated.

 

(d)        If the solution is cooled back to 0 oC and no crystals appear. Is the solution saturated, unsaturated or supersaturated?

Ans.     Unsaturated.

 

  1. Why Ozone is important for us?

Ans.     Ozone is found in the upper atmosphere. It has an important biological function. It prevents most of sun’s ultraviolet solar radiation from reaching the earth surface. Long exposure to this radiation can cause cancer.

 

  1. Explain how gaseous solutions are important for aquatic animals?

Ans.     Gases when dissolved in liquid produce some important solutions. Water dissolve small amount of air giving a solution whose oxygen content (solute) is important for aquatic animals.

  1. How many types of solution. Also give examples.

Ans.     There are nine types of solution:

 

SoluteSolventState of resulting solutionExample
GasGasGasAir
GasLiquidLiquidSoda water
GasSolidSolidH2 absorbed on Ni, Pt, Pd
LiquidGasGasMist, fog, clouds
LiquidLiquidLiquidAlcohol in water
LiquidSolidSolidAmalgams
SolidGasGasCarbon particles in air
SolidLiquidLiquidSugar in water
SolidSolidSolidAlloys

 

What are the physical states of solute and solvent in each of the following solutions? Also identify the type of solution.

(a)        Deep sea divers use breathing mixture of helium and oxygen.

Ans.     Helium is solute and oxygen is solvent. It is a gas in gas solution.

 

(b)        Brass contains 80% copper and 20% zinc

Ans.     Zinc is solute and copper is solvent. It is a solid in solid solution.

 

(c)        Dental filling.

Ans.     Mercury is solute and silver is solvent. It is a liquid in solid solution.

 

(d)        Brine (Salt in water)

Ans.     Salt is solute and water is solvent. It is a solid in liquid solution.

 

(e)        Drinking water containing chlorine as disinfectant.

Ans.     Chlorine is solute and water is solvent. It is a gas in liquid solution.

 

(f)        Gem stones, Ruby contains Cr2O3 and Al2O3

Ans.     Cr2O3 is a solute and Al2O3 is a solvent. It is a solid in solid solution.

 

(g)        Conc. H2SO4, we use in the laboratory is 98% H2SO4 and contains only 2% H2O.

Ans.     H2O is a solute and H2SO4 is a solvent. It is a liquid in liquid solution.

 

  1. Define concentration.

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

 

  1. Define percentage?

Ans.     The mass or volume of solute dissolved in 100g or 100cm3 of solution.

 

  1. Write four ways to express percentage of solutions.

Ans.     There are four ways to express percentage of solutions:

  1. Percentage – Mass/Mass (m/m)
  2. Percentage – Mass/Volume (m/v)
  • Percentage – Volume/Mass (v/m)
  1. Percentage – Volume/volume (v/v)

 

  1. A saline solution is administered intravenously to a person suffering from severe dehydration. This is labeled as 0.85% m/v of NaCl. What does it mean?

Ans.     It means 0.85g NaCl is dissolved in sufficient water to make 100cm3 solution. The resulting solution will be 0.85% m/v.

 

  1. Define Molarity:

Ans.     Molarity is defined as the number of moles of solute dissolved per dm3 of solution.

M =

Its unit is mol.dm3

  1. If you read label on the bottle of concentrated H2SO4 you will notice 98% H2SO4 by mass and also 18M H2SO4. What does 18M stands for?

Ans.     This means there are 18 moles of H2SO4 in each dm3 of solution.

 

  1. Urea (NH2CONH2) is a white solid used as fertilizer and starting material for synthetic plastic. A solution contains 40g urea dissolved in 500cm3 of solution. Calculate the molarity of this solution.

Solution:

Mass of Urea   = 40g

Molar mass of urea (NH2CONH2) = 14+1×2+12+16+14+1×2

= 60g/mole

Moles of Urea =

= = 0.667 moles

Now

Molarity       =

=

                                                = 1.334 M

Example 6.1

            Potassium permanganate (KMnO4) is a dark blue black compound. When it dissolves in water, it forms a bright purple solution. It is used as disinfectant in water tanks. It is also known as pinky. A solution contains 0.05 moles of KMnO4 in 600 cm3 of solution. Calculate molarity of this solution.

 

Solution:

Moles of KMnO4         = 0.05

Volume of solution = 600cm3 =

= 0.083

Potassium chlorate (KClO3) is a white solid. It is used in making matches and dyes. Calculate the molarity of solution that contains.

(a)        1.5 moles of this compound dissolved in 250cm3 of solution.

            Solution:

Volume of solution = 250cm3 = dm3 = 0.25dm3

Moles of KClO3           =          1.5

Molarity           =          ?

Molarity           =

=

                                                =          6 M

 

(b)        75 g of this compound dissolved to produce 1.25dm3 of solution.

Mass of KClO3 = 75 g

Molar mass of KClO3  = 39+35.5+48

= 122.5g

Volume of solution      =          1.25 dm3

Molarity                       = ?

Moles of KClO3           =

=

= 0.588 moles

Now,

Molarity           =

=

= 0.47 M

(c)        What is the molarity of a 50cm3 sample of potassium chlorate solution that yields 0.25g residue after evaporation of the water.

Solution:

Mass of KClO3            = 0.25g

Molar mass of KClO3 = 39+35.5+48 = 122.5g

Volume of solution     = 50cm3 = dm3 = 0.05dm3

Molarity = ?

Moles of KClO3           =

=

= 2.04×10-3

 

Now

Molarity           =

=

= 0.04M

 

Example 6.2

A flask contains 0.25M NaOH solution. What mass of NaOH is present per dm3 of solution.

Solution:

Molarity           =          0.25M

Moles of NaOH           =          0.25 (Molarity means number of

moles per dm3 of solution)

Volume of solution =   1dm3

Molar mass of NaOH =            23+16+1 = 40g/mole

Moles of solute                        =

Mass of solute = Moles of solute x Molar mass

= 0.25 x 40

= 10g

 

Example 6.3

Potassium hydroxide (KOH) is used in the manufacture of shaving creams, paint and varnish. An analyst makes up a solution by dissolving 58g of KOH in one dm3 of solution. Calculate the molarity of this solution.

Solution:

Mass of KOH              = 5.6g

Molar mass of KOH    = 39+16+1 = 56g/ml

Moles of KOH             =

 

=

= 1 mole

 

  1. Sodium hydroxide solutions are used to neutralize acids and in the preparation of soaps and rayon. If you dissolve 25g of NaOH to make 1dm3 of solution, what is the molarity of this solution.

Solution:

Mass of solute             = 25g

Molar mass of NaOH   = 23+16+1 = 40g

Volume of solution      = 1dm3

Molarity                       = ?

Moles of solute            =

=

=          0.625 mole

Now

Molarity           =

=

=          0.625 M

  1. A solution of NaOH has concentration 1.2M. Calculate the mass of NaOH in g/dm3 in this solution.

Solution:

Molarity           = 1.2 M

Volume of solution= 1dm3

Molar mass of NaOH = 23+16+1 = 40g

Mass of solute = ?

Molarity           =

 

Moles of solute = Molarity x Volume of solution in dm3

=   1.2 x 1

= 1.2 mole

Moles of solute            =

Mass of solute  = Moles of solute x Molar mass

= 1.2 x 40

= 48g/dm3

 

 

  1. 3. A solution is prepared by dissolving 10g of hemoglobin in enough water to make up 1dm3 in volume. Calculate molarity of this solution. Molar mass of hemoglobin is 6.51×104 g/mole

Solution:

Molar mass of hemoglobin      = 6.51 x 104 g/mole

Mass of hemoglobin                = 10g

Volume of solution                  = 1dm3

Molarity = ?

Moles of solute            =

=

                                Moles of solute            = 1.54 x 10-4 mole

Now

Molarity           =

=

= 1.54 x10-4 M

 

  1. Prepare 0.2 M KMnO4 solution

Solution

To prepare 0.2M KMnO4 in 100cm3 volumetric flask. First we calculate mass of KMnO4.

Volume of solution = 100cm3 = dm3 = 01.dm3

Molarity = 0.2 M

Molar mass of KMnO4 = 39+55+16×4 = 158 g/mole

Molarity           =

Moles of solute = Molarity x Vol. of solution in dm3

= 0.2 x 0.1

Moles of solute= 0.02 mole

Now

Moles of solute=

Mass of solute = Moles of solute x Molar mass

= 0.02 x 158

= 3.16 g

Mass of KMnO4= 3.16 g

Follow the following steps

  1. Weigh out 3.16g of KMnO4 (0.02 moles)
  2. Add this solid into beaker; add some water to dissolve it.
  3. Transfer this solution to the 100cm3
  4. Keep adding water until the volume of solution rises to the etched line and mix the solution.

This is directed solution.

 

  1. How can you prepare 500 cm3 of 0.2 M KMnO4 solution.

            Solution:

Volume of solution      = 500cm3 = dm3 = 0.5 dm3

Molarity           = 0.2M

First we calculate mass of KMnO4

Molar mass of KMnO4 = 39+55+16×4 = 159 g/mole

Molarity           =

Moles of solute            = Molarity x Volume of sol. In dm3

= 0.2 x 0.5

Moles of solute = 0.1 moles

Now

Moles of solute            =

Mass of solute = Moles of solute x Molar mass

= 0.1 x 158

Mass of solute             = 15.8

Mass of KMnO4           = 15.8g

Follow the following steps to prepare 0.2M KMnO4 in 500 cm3 volumetric flask.

  • Weight out 15.8g of KMnO4 (0.1 moles)
  • Add this solid into a beaker, add some water to dissolve in it.
  • Transfer this solution to the 500cm3 volumetric flask and add more water.
  • Keep adding water until the volume of solution rises to the etched line and mix the solution.

This is desired solution.

 

  1. How can you prepare 25cm3 of 0.25M solution of CuSO4.5H2O (Blue vitriol)

Solution:

Volume of solution      =2.5 cm3 = dm3 = 0.025dm3

Molarity           = 0.25M

First we find mass of CuSO4.5H2O

Molar mass of CuSO4.5H2O= 63.5 + 32 + 16×4 + 5(2+16)

=63.5 + 32 + 64 + 90

= 249.5 g/mole

Molarity           =

Moles of solute            = Molarity x Volume of solution in dm3

= 0.25 x 0.25

Moles of solute            = 0.00625 mole

Now

Moles of solute            =

Mass of solute = Moles of solute x Molar mass

= 0.00625 x 249.5

= 1.56 g

Mass of            CuSO4.5H2O   = 1.56 g

Follow the following steps to prepare 0.25M CuSO4.5H2O in 25cm3 volumetric flask.

  1. Weight out 1.56g of CuSO4.5H2O (0.00625 moles).
  2. Add this solid into a beaker, add some water to dissolve it.
  3. Transfer this solution to the 25cm3 volumetric flask and add more water.
  4. Keep adding water until the volume of solution rises to the etched line and mix the solution.

 

Example 6.5

            Concentrated sulphuric acid is 18M H2SO4. How many cm3 of this acid is needed to produce 250cm3 of 0.1M H2SO4?

            Solution:

Given H2SO4                     Desired H2SO4

M1 V1       =               M2 V2

                                                V1        =               M2 V2

M1

=          0.1 x 250

18

=          1.39 cm3

Transfer 1.39cm3 of 18 M H2SO4 to a 25cm3 volumetric flask and a dilute it by adding water up to the mark and mix. Resulting solution is 0.1M H2SO4

 

  1. A stock solution of hydrochloric acid is 121.1M. How many cm3 of this solution should you use to prepare 500cm3 of 0.1M HCl?

Solution:

Given HCl                   Desired HCl

M1 V1             =             M2 V2

M1        = molarity of given HCl = 12.1 M

V1        = volume of HCl needed to dilute=?

M2        = molarity of required HCl solution = 0.1M

V2       = volume of required HCl = 500 cm3

M1 V1             =             M2 V2

V1                  =             M2 V2

M1

=

=

=             4.13 cm3

  1. Potassium dichromate (K2Cr2O7) is a red-orange compound. It is a strong oxidizing agent and is used in the estimation of iron content in ores. A stock solution is 2.5M K2Cr2O7. How many cm3 of this solution you need to dilute to make 500cm3 of 0.05M K2Cr2O7

Solution:

Given K2Cr2O7      =     Desired K2Cr2O7

M1 V1            =          M2 V2

M1        = molarity of given K2Cr2O7 = 2.5 M

V1        = volume of K2Cr2O7 needed to dilute=?

M2        = molarity of required K2Cr2O7 solution = 0.05M

V2       = volume of required K2Cr2O7 = 50 cm3

M1 V1  =          M2 V2

                                V1        =          M2 V2

M1

=

=

V1        =          1cm3

 

  1. Commercial acetic acid is 17.8 molar. How can you convert this into 0.1M acetic acid.

Solution:

Given Acetic Acid       Desired Acetic Acid

M1 V1       =    M2 V2

                M1        = Molarity of given Acetic Acid = 17.8 M

V1        = Volume of Acetic Acid needed to dilute=?

M2        = Molarity of required Acetic Acid solution = 0.1M

V2       = Volume of required Acetic Acid = 1000 cm3

M1 V1  =          M2 V2

                                V1        =          M2 V2

M1

=

=

V1        =          5.62cm3

We can convert 17.8 molar of commercial acetic acid into 0.1M acetic acid by adding water up to the mark and mix in a 1000 cm3 volumetric by adding 5.62cm3 of acetic acid.

 

  1. Define solubility

The amount of solute that dissolves in 100g of a solvent at a particular temperature is called its solubility.

 

  1. Why methanol and water are miscible?

Ans.     “Like dissolves like”. Water molecules are polar. Two H-atoms bonded to an O-atom are slightly positively charged and O-atom has a slightly negative charge. Water molecules form hydrogen bonds with one another. Methanol molecules are also polar and exhibit hydrogen bonding. Thus methanol and water are miscible.

 

  1. Q. How sodium chloride dissolves readily in water?

Ans.     Sodium Chloride is an ionic compound. The negative end of water molecules is attracted to sodium ions and the positive end of water molecules is attracted to chloride ions. These attractive forces are strong enough to overcome the attractions between ions in NaCl. Thus sodium chloride dissolves readily.

 

  1. Sodium Chloride and glucose both are soluble in water. But the solubility of NaCl is greater than glucose. Explain why?

Ans.     “Like dissolves like” Glucose is covalent compound. Whereas sodium chloride is polar ionic compound. Water is also polar. So solubility of NaCl is greater than glucose.

 

  1. In which liquid of each of the following pairs you would expect KCl, an ionic solid, to be more soluble.

(a)        H2O or CCl4

“Like dissolves like”. H2O is polar where CCl4 is non-polar. KCl is also polar ionic compound. So KCl is more soluble in H2O.

  • CH3OH or Benzene

“Like dissolve like” Benzene (C6H6) is non-polar whereas CH3OH is polar compound. KCl is also polar ionic compound. So KCl is more soluble in CH3OH

 

  1. Which of the following pairs of liquids are miscible?

(a)        Water and Benzene:

Ans.     “Like dissolve like” Benzene (C6H6) is non-polar and water is a polar compound. So water and benzene are not miscible.

 

(b)        Benzene and CCl4

“Like dissolve like” Benzene is non-polar and CCl4 is also a non-polar compound. So benzene and CCl4 are miscible.

 

(c)        An oil and Benzene

“Like dissolve like: Benzene is non-polar and oil is also non-polar. So oil and benzene are miscible.

 

  1. Explain the effect of temperature on solubility?

Ans.     The solubility of an ionic compound generally increases with the increase in temperature. However, solubility of some solids decreases with temperature.

 

  1. Why fish shows signs of stress on a hot day?

Ans.     Air is less soluble in hot water than in cold water, air comes out of water in the form of bubbles. This mean solubility of air in water decreases with increasing temperature. So the fish shows signs of stress on a hot day.

 

Use figure and answer the following questions:

  1. At what temperature the solubility of KNO3 and KBr is same?

Ans.     83°C

  1. What is the solubility of KBr at 50°C.

Ans.     80g

  1. Which is greater at 40 oC, the solubility of NaNO3 or solubility of KBr?

Ans.     NaNO3

 

  1. Identify from the graph the compound whose solubility is little affected with increase in temperature.

Ans.     NaCl

 

  1. Explain Solution, Colloids and Suspensions

Ans.     Solutions:

A solution is a homogeneous mixture in which the particles are individual molecules or ions distributed evenly throughout the surrounding fluid.

Colloids:

A heterogeneous mixture of tiny particles of a substance dispersed through a medium is called a colloid.

Suspensions:

Such a heterogeneous mixture containing particles large enough to b seen with the naked eye and clearly distinct from the surrounding fluid is called a suspension.

 

  1. Compare the properties of solution, suspension and colloids.

 

S.

No.

SolutionSuspensionsColloids
1.HomogeneousHeterogeneousHeterogeneous
2.Particles size vary from 0.1 to 1 nmParticles size is greater than 103nmParticles size vary from 1 to103 nm
3.Particles are invisible by naked eye, ordinary microscope as well as electron microscopeParticles are visible by naked eye.Particles are invisible by naked eye and in ordinary microscope but under electron microscope

 

Review Questions

Q.1      Encircle the correct answer

  • The maximum amount of sodium acetate that dissolves in 100g of water at 0oC is 119g and 170g at 100o If you place 170g of sodium acetate in 100g of water at 0°C. the resulting solution would be:
  1. unsaturated
  2. saturated
  3. supersaturated
  4. 1M
  • How many moles of sodium atoms are present in 2.3g Na?
  1. 1
  2. 5
  3. 1
  4. 15
  • What is the mass of 5 moles of hydrogen gas?
  1. 5g
  2. 04g
  3. 08g
  4. 008g
  • How many atoms are there in 28g of nitrogen gas?
  1. 2
  2. 1
  3. 022 x 1023
  4. 044 x 1023
  • How many atoms are there in 0.1 mole of carbon?
  1. 022 x 1023
  2. 022 x 1022
  3. 022 x 1024
  4. 022 x 1021
  • A solution of NaOH has a concentration of 4g/dm3. What is the mass of NaOH contained in 250cm of this solution?
  1. 40g
  2. 20g
  3. 1g
  4. 2g
  • Which of the following solution is more dilute?
  1. 1M
  2. 2M
  3. 1M
  4. 009M
  • A solution of NaOH contains 20g of this compound in 2dm3 of solution. What is the molarity of this solution?
  1. 2M
  2. 1M
  3. 25M
  4. 5M
  • Which quantity is the same for one mole of hydrogen gas and one mole of water?
  1. mass
  2. number of atoms
  3. number of molecules
  4. number of gram atomic mass
  • If one mole of Na contains x atoms of sodium, what is the number of moles contained in 46g of sodium?
  1. x
  2. 2
  3. 2x
  4. 5x

Answers:

i) Bii) Ciii) Civ) Cv) B
vi) Avii) Dviii) Cix) Cx) C

 

Q.2      Give short answers

  1. i) Differentiate between saturated and unsaturated solution?

Ans.     Saturated:

The solution which cannot dissolve more solute at a particular temperature is called a saturated solution.

Unsaturated:

A solution which can dissolve more of the solute at a given temperature is called an unsaturated solution.

 

  1. ii) Give example of a solid solution containing two solids.

Ans.     Example of solid in solid solution.

  1. Copper is added in gold
  2. Brass is an alloy of copper and zinc

 

iii)        Can you call colloid a solution?

Ans.     Yes, we can call colloid a solution. Colloids are heterogeneous mixture of tiny particles and scatter beam of light due to which they are not true solutions. They are called false or colloidal solutions.

 

  1. iv) Gasoline does not dissolve in water?

Ans.     Gasoline does not dissolve in water. Gasoline molecules are non-polar in nature. The attraction between a water molecule and gasoline molecule is very weak. So it is insoluble in water.

  1. v) Are gem stones solution?

Ans.     Yes, gemstones are solutions. Many naturally occurring gemstones are solid solutions. For example Ruby, Opal; in these solutions, a solid solute dissolves in a solid solvent. These solutions are solid in solid.

 

Q.3      A tiny crystal of a solid substance is added to an aqueous solution of the same substance. What would happen if the original solution was:

(a)        Supersaturated:

Crystallization will start.

(b)        Unsaturated:

Tiny crystals of solid substance will dissolve.

(c)        Saturated:

Tiny crystal of substance will settle down at the bottom.

 

Q.4      Explain why CH3OH is soluble in water but C6H6 is not?

Ans.     “Like dissolves like”. Water is polar solvent. Methanol (CH3OH) is polar due to which it is soluble in water. Benzene (C6H6) is non-polar so it is insoluble in water.

 

Q.5      How can you prepare 250 cm3 of 0.5M MgSO4 from a stock solution of 2.5M MgSO4?

Solution:

Given MgSO4               Desired MgSO4

M1 V1               =          M2 V2

M1        = Molarity of given MgSO4 = 2.5M

V1        = Volume of MgSO4 needed = ?

M2        = Molarity of required MgSO4 solution            = 0.5M

V2        = Volume of required MgSO4 solution = 250cm3

 

M1 V1               =          M2 V2

V1                    =          M2 V2

M1

                                                                =          0.5 x 250

2.5

=          125

2.5

=          50cm3

 

Q.6      Copy and complete the following table for aqueous solution of NaOH.

 Mass of soluteMoles of soluteVolume of solutionMolarity
1.20 g 500cm3 
2. 0.25 0.25
3.  200cm30.1

Solution:

  1. Molar mass of NaOH = 23 + 16 + 1 = 40 g/mole

Moles of solute                        = Mass of solute

Molar mass

= 20

40

Moles of solute                        = 0.5 mole

Molarity                       = Moles of solute

Volume of solution in dm3

Volume of solution      = 500cm3 = 500 dm3 = 0.5dm3

1000

Molarity                       = 0.5    = 1

0.5

  1. Moles of solute =

Mass of solute = Moles of solute x Molar mass

= 0.25 x 40

= 10

Molarity           =

Volume of solution in dm3 =

=

= 1 dm3

= 1 x 1000 cm3

= 1000 cm3

 

  1. Molarity =

Moles of solute = Volume of solution in dm3 x Molarity

Volume of solute = 200 cm3 = dm3 = 0.2 dm3

Moles of solute            = 0.2 x 0.1 = 0.02

Moles of solute   =

Mass of solute  = Moles of solute x molar mass

= 0.02 x 40

= 0.8g

 Mass of soluteMoles of soluteVolume of solutionMolarity
1.20 g0.5500cm31
2.10 g0.251000cm30.25
3.0.8 g0.02200cm30.1

 

Q.7      Give examples of the following solution:

(a)        a liquid solution of a liquid solvent and gaseous solute.

Ans.     Soda water, all soft drinks.

 

(b)        A solid solution of two solids.

Ans.     All metal alloys, like Brass.

 

Q.8      What is the molarity of a solution prepared by dissolving 1.25g of HCl gas into enough water to make 30cm3 of solution?

Solution:

Mass of solute = 1.25 g

Molar mass of HCl = 1 + 35.5 = 36.5g

Volume of solution = 30cm3 = = 0.03dm3

Moles of solute            =

=

Moles of solute            = 0.0342

Now

Molarity           =

=

= 1.14M

 

Q.9      Formalin is an aqueous solution of formaldehyde (HCHO), used as a preservative for biological specimens. A biologist wants to prepare 1dm3 of 11.5M formalin. What mass of formaldehyde he requires?

Solution:

Volume of solution      =          1 dm3

Molarity                       =          11.5M

Mass of Solute             =          ?

Molar mass of HCHO = 1+12+1+16 = 30g

Molarity           =

Moles of solute            = Molarity x Volume of Solution in dm3

= 11.5 x 1

= 11.5

Now,

Moles of solute            =

Mass of solute = Moles of solute x Molar mass

= 11.5 x 30

Mass of Solute = 345 g

 

Q.10    A solution of Ca(OH)2 is prepared by dissolving 5.2 mg of Ca(OH)2 to a total volume of 1000cm3. Calculate the molarity of this solution.

Solution:

Mass of solute =          5.2 mg = = 0.0052g

Molar mass of Ca(OH)2 = 40 +2 x16 +2×1= 74g

Volume of solution = 1000 cm3 = dm3 = 1 dm3

Molarity = ?

Moles of solute            =

=

Moles of solute            = 7.0 x 10-5

Now,

Molarity =

=

= 7.0 x 10-5

 

Q.11    Calculate the number of moles of solute present in 1.25cm3 of 0.5M H3PO4 solution.

Solution:

Molarity           = 0.5M

Volume of solution in dm3 = 1.25 cm3 = dm3

     = 0.00125dm3

Moles of solute = ?

Molarity           =

Moles of solute = Molarity x Vol of Solution in dm3

= 0.5 x 0.00125

Moles of solute = 6.3 x 10-4 moles

Q.12       Calculate the new molarity when 100cm3 of water is added to 100 cm3 of 0.5M HCl.

Solution:

M1        = new molarity             = ?

V1        = Volume of solution = 100+100=200cm3

M2        = Given molarity = 0.5M

V2        = Volume of HCl = 100 cm3

M1V1    =          M2V2

                                M1        =          M2V2

                                                                     V1

=

=

M1        =          0.25 M

 

Q.13    How are solutions useful for society?

Ans.     Most of the substances we need for our existence are solutions.

  1. The air we breathe is gaseous solution.
  2. Natural water is a liquid solution.
  • Commercial products are also liquid solution.

 

ThinkTank

  1. A 10.0g of solid solute is placed in 100g of water at 20oC and all of it dissolves. Then another 4.0g of the solute is added at 20oC and all of it dissolves.
  2. Is the first solution saturated, unsaturated or supersaturated?

Ans.       Unsaturated, because it can absorb more solute.

  1. Is it possible to tell from this information that the final solution is unsaturated or saturated?

Ans.       No. Because it may still dissolve more solute.

  1. What should you do to change:
  2. a saturated solution to an unsaturated solution.

Ans.       We can add more solvent to make it unsaturated.

  1. an unsaturated solution to a saturated solution.

Ans.        We can add more solute until it can’t dissolve anymore.

  1. Knowing the molarity of a solution is more meaningful than knowing whether a solution is dilute or concentrated. Explain.

Ans.    Knowing the molarity is more meaningful because it not only tells if solution is dilute or concentrated but also it tells the actual concentration.

  1. Design an experiment to determine the solubility of table sugar in water at room temperature.

Ans.   Prepare saturated solution of sugar in 100g of water. Take this solution in pre-weighted china dish. Place china dish on the burner and heat it slowly till water evaporates completely. Cool china dish and weigh it. Calculate the mass of sugar present in it. Solubility of sugar in 100g of water at room temperature is found to be 204g.

  1. Design an experiment to prepare 10% mass by volume solution of CuSO4.5H2O (Nelathota).

Ans.   If we dissolve 10g of CuSO4.%H2O in sufficientwater to make 100cm3 solution. The resulting solution will be 10% mass by volume.

  1. Which solution is more dilute, 50cm3 of 0.2M NaOH or 100cm3 of 0.1M NaOH?

Ans.    50cm3 of 0.2M NaOH:

Volume of solution in dm3 = = 0.05dm3

Molarity (M) = 0.2M

Number of moles=?

 

Number of moles = 0.2×0.05 = 0.01moles

100cm3 of 0.1M NaOH:

                Volume of solution in dm3 = = 0.1dm3

Molarity (M) = 0.1M

Number of moles=?

 

Number of moles = 0.1×0.1 = 0.01moles

Both have equal number of moles dissolved.

  1. Which solution is more concentrated 100cm3 of 0.1M HCL or 100cm3 of 0.1M NaOH?

Solution:

100cm3 of 0.1M HCl:

Volume of solution in dm3 = = 0.1dm3

 

Mass of HCl =

100cm3 of 0.1M NaOH:

Volume of solution in dm3 = = 0.1dm3

 

Mass of HCl =

Since mass of NaOH is greater than the mass of HCl in 100cm3 of solution therefore 0.1M solution is more concentrated.

  1. Benzene is a common organic solvent its use is now restricted because this can cause cancer. The recommended limit of exposure to benzene is 0.32mg per dm3 of air. Calculate the molarity of this solution.

Ans.    Mass of solute C6H6= 0.32mg = 0.32/1000g = 0.00032g

Molar mass of solute C6H6 = 6×12+1×6= 72+6 = 78g

Volume of solution = 1dm3

Molarity (M)=?

Molarity = 4×10-6M

  1. A patient in a hospital is often administered an intravenous (IV) drip containing an aqueous solution. this aqueous solution contains 0.85% (mass by volume) of sodium chloride or 5% (mass by volume) of glucose. Calculate the molarity of both these solutions.

Solution:

0.85% (mass by volume) of Sodium Chloride:

Mass of solute = 0.85g

Volume = 100cm3=100/1000dm3=0.1dm3

Molar Mass of NaCl = 23+35.5.5 =58.5g/mol

 

5% (mass by volume) of Glucose(C6H12O6):

Mass of solute = 5g

Volume = 100cm3=100/1000dm3=0.1dm3

Molar Mass of NaCl = 6×12+1×12+16×6= 180g/mol

 

  1. 100cm3 of NaOH solution was heated to complete dryness, 1.5g residue left behind. What was the molarity of the solution.

Ans. Mass of solute = 1.5g

Volume = 100cm3=100/1000dm3=0.1dm3

Molar Mass of NaOH = 23+16+1 = 40g/mol