Skip to main content

9th Atoms and Molecules - Mole Concept and Problems based on mole concept

Mole Concept
we can express the quantity of a substance is expressed in terms of mole.
Mole is a fundamental unit in the System International d' Unités, the SI system, and it is used to measure the amount of substance In Latin mole means a "massive heap" of material


Mole is defined as the amount of substance that contains as many specified elementary particles as the number of atoms in 12g of carbon-12 isotope.


One mole is also defined as the amount of substance which contains Avogadro number (6.023 x 1023) of particles.
Avogadro number: Number of atoms or molecules or ions present in one mole of a substance is called Avogadro number. Its value is 6.023 x 1023

Therefore, one mole of any substance contains Avogadro number of particles. The particles may be atoms, molecules, ions etc

E.g. 1 mole of oxygen atoms = 6.023 x 1023 atoms, molecules, ions etc  of oxygen
                                               = 32g of O2
Problems (based on mole concept)
1. When the mass of the substance is given:

a. Calculate the number of moles in

i) 81g of aluminium

Sol: 1. Atomic mass of Al= 27gm
           27g  of  aluminium = 1 mole of aluminium  
         81g  of aluminium   = 1/27 x 81=3 moles of aluminium  

OR, Use formula , Number of moles = given mass/atomic mass

Self: ii) 4.6g sodium   iii) 5.1g of Ammonia  iv) 90g of water v) 2g of NaOH

b. Calculate the mass of 0.5 mole of iron

Atomic mass of iron = 55.9 g
Mass of the 1 mole of iron   = 55.9 g
Mass of the 0.5 mole of iron   = 0.5 x 55.9 g = 27.95 g

Or, Using formula: mass = atomic mass x number of moles

FOLLOW UP: Find the mass of 2.5 mole of oxygen atoms

2. Calculation of number of particles when the mass of the substance is given:

Number of particles =   Avogadro number x given mass / gram molecular mass

Calculate the number of molecules in 11g of CO2
Solution: gram molecular mass of CO2 = 44g
 44g of CO= 6.023 x 1023  molecules
1 g of CO= ( 6.023 x 1023 ÷ 44 g ) molecules
11g of CO2 = ( 6.023 x 1023 ÷ 44 g ) x 11 =1.51 x 1023 molecules

FOLLOW UP: Calculate the number of molecules in 360g of glucose

3. Calculation of mass when number of particles of a substance is given:

Mass of a substance =  gram molecular mass x number of particles/ 6.023 x 1023
  
Eg. Calculate the mass of 18.069 x 1023  molecules of SO2
Sol: Gram molecular mass SO2 = 64g
6.023 x 1023 molecules of SO2  = 64 gm
1             molecules of SO2          = 64/(6.023 x 1023 )  gm
18.069 x 1023  molecules of SO2  = [ 64/(6.023 x 1023 )  x  18.069 x 1023  ]gm = 192 g

Calculate the mass of glucose in  2 x 1024 molecules
Gram molecular mass of glucose = 180g
Mass of glucose  [180 x 2 x 1024 ] / [6.023 x 1023  ] = 597.7g

FOLLOW UP: Calculate the mass of 12.046 x 1023 molecules in CaO.

4. Calculation of number of moles when you are given number of molecules:
Number of moles = Number of molecules/Avogadro Number
                       = [3.0115 x 1023 ] / [6.023 x 1023  ]      = 0.5 moles

FOLLOW UP:Calculate number of moles in 12.046x 1022 atoms of copper

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca