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Download 9th Class Chemistry Notes (Unit # 6) for Federal Board (FBISE) & Punjab Boards

9th Class Chemistry Notes (Unit # 6) for FBISE Islamabad / Punjab Boards. Class 9 / IX / SSC / Matric. Download Complete Guide / Key Book as PDF. These notes are as per latest syllabus / course prescribed by Federal Board of Intermediate & Secondary Education (FBISE) Islamabad and all Boards of Punjab Province. Students of Sindh, KPK, Balochistan and Azad Kashmir can also utilize these notes as per their course / books.

9th Class Chemistry Notes (Unit # 6) for FBISE Islamabad / Punjab Boards

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9th Class Chemistry Notes (Unit # 6) for FBISE Islamabad / Punjab Boards

 

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REVIEW QUESTION

 

Differentiate between saturated and unsaturated solution?

Saturated solution:

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

Unsaturated solution:

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

 

Give example of a solid solution containing two solids.

  • Brass is an alloy of copper and zinc.
  • Steel is an alloy of iron containing small amounts of carbon and silicon.

 

Can you call collide a solution?

Yes, collides are heterogeneous solution. For example:

  1. Starch solution
  2. White of an egg.
  • Gelatin, glue, gums Milk, rubber, fog, dust in the air, jellies, paints, blood and starch in water.

 

Gasoline does not dissolve in water, why?

  • Gasoline and oils do not dissolve in water.
  • Gasoline and oil molecules are non-polar in nature.
  • The attraction between a water molecule and oil or gasoline molecule is very weak so these liquids are insoluble in water.

 

Are gem stones solutions?

  • Yes, many naturally occurring gemstones are solid solutions.
  • For example Ruby, Opal, in these solutions a solid solute dissolves in a solid solvent.
  • We call these solutions as solids in solids.

 

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) super saturated b) unsaturated c) saturated

  1. a) Crystallization will start.
  2. b) Tiny crystal of a solid substance will again dissolve
  3. c) Tiny crystal of a substance start settling down at the bottom.

 

Explain why CH3OH is soluble in water but C6H6 is not.

  • Methanol (CH3OH) is soluble in water due to hydrogen bonding.
  • The general principal is “Like dissolve like”.
  • As water is polar solvent and Benzene is non polar, therefore benzene will not dissolve in water.

How can prepare 250 cm3 of 0.5M MgSO4 from stock solution of 2.5M MgSO4?

M1     =      molarity of given MgSO4               =      2.5M

V1     =      volume of MgSO4 need to dilute    =      ?

M2     =      molarity of required MgSO4           =      0.5M

V2     =      volume of required MgSO4             =      250cm3

 

M1V1                          =                      M2V2

 

M2V2

V1                             =                      ——-

M1

 

0.5 x 250

V1                             =                      ———–

2.5

 

125

V1                             =                      —–          =      50 cm3

2.5

 

Give examples of the following:

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

Soda water, hydrochloric acid

  1. b) a solid solution of two solids
  • Brass is an alloy of copper and zinc.
  • Steel is an alloy of iron containing small amounts of carbon and silicon.

 

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

Mass of solute HCl              =              1.25g

Molar Mass of solute HCl      =              1 + 35.5            =      36.5g

Volume of Solution              =              30cm3

 

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

 

1.25                  1000

Molarity     =      ——–       x       ———-

36.5                  30

 

1250

Molarity     =      ——–       =      1.14M

1095

 

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

Volume of Solution              =              1 dm3                        =      1000cm3

Molarity                             =              11.5M

Molar Mass of solute (HCOH) =              1+12+1+16        =      30g

 

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

 

Mass of solute            1000

11.5                  =      —————-     x       ———-

30                             1000

 

Mass of solute

11.5                  =      —————-

30

 

Mass of solute    =      11.5 x 30           =      345g

 

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

Mass of solute Ca(OH)2                    =      5.2mg        =      5.2/1000    =      0.0052g

Molar mass of Ca(OH)2         =      40+2×16+2×1     =      40+32+2   =      74g

Molarity     =      ?

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

0.0052               1000

Molarity     =      ——–       x       ———-

74                     1000

 

0.0052

Molarity     =      ——–       =      0.00007     =      7 x 10-5M

74

 

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

Volume of solution in dm3    =      1.25/1000  =      0.00125 dm3

Molarity                             =      0.5M

Number of Moles                 =      ?

 

 

Number of moles

Molarity             =      —————————–

Volume of solution in dm3

 

Number of moles

0.5                   =      —————————–

0.00125

 

Number of moles =      0.5 x 0.00125     =      0.00063     =      6.3 x 10-4 moles

 

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

M1     =      new molarity                              =      ?

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

M2     =      molarity of HCL given                  =      0.5M

V2     =      volume of HCL                            =      100cm3

 

M1V1                          =              M2V2

 

M2V2

M1                             =              ——-

V1

 

0.5 x 100           0.5

M1                             =              ———–   =      —-   =      0.25M

200                   2

 

How are solutions useful for society? Give three examples.

  • We use many solution substances in our daily life such as air, soft drinks, juices, shampoo, petrol, natural gas, diesel, kerosene, cough syrup etc.
  • Most of the chemical reactions that take place in the bodies of living organisms occur in aqueous solutions.
  • Brass, steel, german silver are also solutions. These are widely used for making cooking utensils, surgical tools, cutlery, musical instruments and many other objects.

 

A 10g of solid solute is placed in 100 g of water at 20oC and all of it dissolves. Then another 4g of the solute is added at 20oC and all of it dissolves.

  1. a) Is the first solution saturated, unsaturated or supersaturated?
  2. b) Is it possible to tell from this information that the final solution is unsaturated or saturated?
  3. a)
  4. b) Because a solution which can dissolve more of the solute at a given temperature is called an unsaturated solution.

 

What should you do to change:

  1. a) a saturated solution to an unsaturated solution.
  2. b) an unsaturated solution to a saturated solution.
  3. a) i) Add more solvent.
  4. ii) Increase temperature.
  5. b) Keep adding solute until the solvent cannot dissolve any more at that given temperature.

 

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

Knowing the molarity is more meaningful because by knowing it you can not only know if it is diluted or concentrated, but also the actual concentration.

 

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

  • Prepare saturated solution of sugar in 100g of water.
  • Take this solution in a pre-weighed 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 204g.
  • Solubility of Sodium Chloride and Sucrose in Water from 0-100oC

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

If we dissolve 10g CuSO4.5H2O (Nelathota) in sufficient water to make 100 cm3 solution, the resulting solution will be 10% m/v.

 

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

Case 1:

Volume of solution in dm3    =      50/1000     =      0.05 dm3

Molarity                             =      0.2 M

Number of moles                 =      ?

Number of Moles

Molarity             =      —————————

Volume of solution in dm3

 

Number of Moles

0.2                    =      ——————

0.05

 

Number of Moles =      0.2 x 0.05  =      0.01 moles.

Case 2:

Number of Moles

Molarity             =      —————————

Volume of solution in dm3

 

Number of Moles

0.1                    =      ——————

0.1

 

Number of Moles =      0.1 x 0.1    =      0.01 moles.

 

Which solution is more concentrated 100cm3 of 0.1M HCl or 100cm3 of 0.1M NaOH.

Case 1:

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

Mass of HCl                                1000

0.1            =      ———————-      x       ——

36.5                                  100

 

Mass of HCl

0.1            =      ———————-      x       10

36.5

0.1 x 36.5

Mass of HCl                        =      ————          =      0.365g

10

 

Case 2:

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

Mass of NaOH                     1000

0.1            =      ———————-      x       ——

40                                     100

 

Mass of HCl

0.1            =      ———————-      x       10

40

0.1 x 40

Mass of HCl                        =      ———-            =      0.4g

10

 

Decision:

Since mass of NaOH is greater than HCl, therefore 0.1M NaOH is more concentrated.

 

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

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

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

Volume of solution      =      1dm        =      1000cm3

Molarity                     =      ?

 

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

0.00032                             1000

Molarity     =      ———————-      x       ——

78                                     1000

 

0.00032

Molarity     =      ———-            =      0.000004M         =      4×10-6M

78

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.

Case 1:

Mass of solute            =      0.85 g

Molar mass of NaCl      =      23+35.5     =      58.5 g/mol

 

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

0.85                  1000

Molarity     =      ——         x       ——         =      0.145 M

58.5                  1000

Case 2:

Molarity solute            =      5 g

Volume of Solution      =      100 cm3

Molar mass of glucose  =      12×6+1×12+16×6        =      180 g / mol

 

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

5                      1000

Molarity     =      ——         x       ——         =      0.278 M

80                     100

 

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

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

Volume of solution              =      100 cm3

Molarity (M)                        =      ?

Mass of solute                    1000

Molarity     =      ———————-      x       —————————-

Molar mass of solute           Volume of solution in cm3

 

1.5                    1000

Molarity     =      ——         x       ——         =      0.375 M

40                     100

 

THEORY

What is solute and solvent?

  • In a solution, substance that is present in lesser amount is called solute.
  • In a solution, substance that is present in larger amount is called solvent.

What is supersaturated solution?

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

What do you know about ozone?

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

Define the following terms:

Fog:

Fog is a solution of water vapours in air.

Rectified Spirit:

Fermentation of cane sugar produces 95% v/v of ethyl alcohol called rectified spirit.

Amalgam:

The metal mercury is the only metal that is liquid at room temperature. It dissolves a number of metals to give a solution called amalgam.

Solubility:

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

True Solution:

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

Suspension:

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

Colloid:

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

Alloy:

Most commercial metals are examples of solid solutions of various metals and are called alloys. For example brass.

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