A guide for A level students KNOCKHARDY PUBLISHING

A guide for A level students KNOCKHARDY PUBLISHING
THE MOLE A guide for A level students 2015 SPECIFICATIONS KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING
THE MOLE INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... Navigation is achieved by using the left and right arrow keys on the keyboard

THE MOLE CONTENTS What is a mole and why do we use it?
Calculating the number of moles of a single substance Reacting mass calculations Solutions and moles Standard solutions Volumetric calculations Molar volume calculations

DON’T BE LEFT IN THE DARK!
THE MOLE Before you start it would be helpful to… know how to balance simple equations know how to re-arrange mathematical formulae DON’T BE LEFT IN THE DARK!

WHAT IS A MOLE ? THE MOLE it is just a number, a very big number
it is the standard unit of amount of a substance - it is just a number, a very big number it is a way of saying a number in words, just like... DOZEN for SCORE for 20 GROSS for 144

602200000000000000000000 (Approximately)... THAT’S BIG !!!
THE MOLE WHAT IS A MOLE ? it is the standard unit of amount of a substance - it is just a number, a very big number it is a way of saying a number in words, just like... DOZEN for SCORE for 20 GROSS for 144 HOW BIG IS IT ? (Approximately)... THAT’S BIG !!! It is a lot easier to write it as x 1023

WHAT IS A MOLE ? THE MOLE it is just a number, a very big number
it is the standard unit of amount of a substance - it is just a number, a very big number it is a way of saying a number in words, just like... DOZEN for SCORE for 20 GROSS for 144 HOW BIG IS IT ? (Approximately)... THAT’S BIG !!! It is a lot easier to write it as x 1023 It is also known as... AVOGADRO’S NUMBER It doesn’t matter what the number is as long as everybody sticks to the same value !

Yes, it would help if you can balance equations
THE MOLE WHY USE IT ? Atoms and molecules don’t weigh much so it is easier to count large numbers of them. In fact it is easier to weigh substances. Using moles tells you how many particles you get in a certain mass the mass of a certain number of particles DO I NEED TO KNOW ANYTHING ELSE ? Yes, it would help if you can balance equations AND Keep trying, you will get the idea ... EVENTUALLY!

THE MOLE – AN OVERVIEW WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METRE It is just a number, a very big number It is also a way of saying a number in words like DOZEN for 12 GROSS for 144

THE MOLE – AN OVERVIEW WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METRE It is just a number, a very big number It is also a way of saying a number in words like DOZEN for 12 GROSS for 144 HOW BIG IS IT ? (approx) - THAT’S BIG !!! It is a lot easier to write it as x 1023 And anyway it doesn’t matter what the number is as long as everybody sticks to the same value !

THE MOLE – AN OVERVIEW WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METRE It is just a number, a very big number It is also a way of saying a number in words like DOZEN for 12 GROSS for 144 HOW BIG IS IT ? (approx) - THAT’S BIG !!! It is a lot easier to write it as x 1023 And anyway it doesn’t matter what the number is as long as everybody sticks to the same value ! WHY USE IT ? Atoms and molecules don’t weigh much so it is easier to count large numbers of them. In fact it is easier to weigh substances. Using moles tells you :- how many particles you get in a certain mass the mass of a certain number of particles

MOLES = MASS MOLAR MASS THE MOLE molar mass g mol-1 or kg mol-1
CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE moles = mass / molar mass mass = moles x molar mass molar mass = mass / moles UNITS mass g or kg molar mass g mol-1 or kg mol-1 MOLES = MASS MOLAR MASS MASS MOLES x MOLAR COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED

MOLES OF A SINGLE SUBSTANCE
1. Calculate the number of moles of oxygen molecules in 4g MASS MOLES x MOLAR

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2 relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2 relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1 moles = mass = g = mol molar mass g mol -1 MASS MOLES x MOLAR

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2 relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1 moles = mass = g = mol molar mass g mol -1

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2 relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1 moles = mass = g = mol molar mass g mol -1 What is the mass of 0.25 mol of Na2CO3 ?

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2 relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1 moles = mass = g = mol molar mass g mol -1 What is the mass of 0.25 mol of Na2CO3 ? Relative Molecular Mass of Na2CO3 = (2x23) (3x16) = 106 Molar mass of Na2CO = 106g mol-1

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2 relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1 moles = mass = g = mol molar mass g mol -1 What is the mass of 0.25 mol of Na2CO3 ? Relative Molecular Mass of Na2CO3 = (2x23) (3x16) = 106 Molar mass of Na2CO = 106g mol-1 mass = moles x molar mass = x = g MASS MOLES x MOLAR

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2 relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1 moles = mass = g = mol molar mass g mol -1 What is the mass of 0.25 mol of Na2CO3 ? Relative Molecular Mass of Na2CO3 = (2x23) (3x16) = 106 Molar mass of Na2CO = 106g mol-1 mass = moles x molar mass = x = g

REACTING MASS CALCULATIONS
CaCO HCl ———> CaCl CO H2O 1. What is the relative formula mass of CaCO3? (3 x 16) = 100 2. What is the mass of 1 mole of CaCO g 3. How many moles of HCl react with 1 mole of CaCO3? 2 moles 4. What is the relative formula mass of HCl? = 5. What is the mass of 1 mole of HCl? g 6. What mass of HCl will react with 1 mole of CaCO3 ? 2 x 36.5g = 73g 7. What mass of CO2 is produced ? moles of CO2 = moles of CaCO3 moles of CO2 = moles mass of CO2 = x 44 = g

REACTING MASS CALCULATIONS
EQUATIONS give you the ratio in which chemicals react and are formed need to be balanced in order to do a calculation CaCO HCl ———> CaCl CO H2O 1. What is the relative molecular mass of CaCO3? (3 x 16) = 100 2. What is the mass of 1 mole of CaCO3? 100 g 3. What does 0.1M HCl mean? concentration is 0.1 mol dm-3 4. How many moles of HCl are in 20cm3 of 0.1M HCl? 0.1 x = moles 1000 5. How many moles of CaCO3 will react ? ½ x = moles 6. What is the mass of moles of CaCO3? mass = moles x molar mass = x 100 = 0.1 g 7. What mass of CO2 is produced ? moles of CO2 = moles of CaCO3 moles of CO2 = moles mass of CO2 = x 44 = g

CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION UNITS concentration mol dm-3 volume dm3 BUT IF... concentration mol dm-3 volume cm3 MOLES = CONCENTRATION x VOLUME MOLES CONC x VOLUME COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3) MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = mol dm-3 volume pipetted out into the conical flask = cm3 25cm3 250cm3 250cm3

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = mol dm-3 volume pipetted out into the conical flask = cm3 25cm3 250cm3 250cm3 The original solution has a concentration of mol dm-3

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = mol dm-3 volume pipetted out into the conical flask = cm3 25cm3 250cm3 250cm3 The original solution has a concentration of mol dm-3 This means that there are mols of solute in every 1 dm3 (1000 cm3) of solution

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = mol dm-3 volume pipetted out into the conical flask = cm3 25cm3 250cm3 250cm3 The original solution has a concentration of mol dm-3 This means that there are mols of solute in every 1 dm3 (1000 cm3) of solution Take out cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = mol dm-3 volume pipetted out into the conical flask = cm3 25cm3 250cm3 250cm3 The original solution has a concentration of mol dm-3 This means that there are mols of solute in every 1 dm3 (1000 cm3) of solution Take out cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm3 (1000cm3) = moles in 1cm3 = /1000 moles in 25cm3 = x / = x 10-3 mol

MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH

1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3 = moles

1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3 = moles 2 What volume of 0.1M H2SO4 contains moles ?

1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3 = moles 2 What volume of 0.1M H2SO4 contains moles ? volume = x moles (re-arrangement of above) (in cm3) conc = x = 20 cm3 0.1 mol dm-3

‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’
SOLUTIONS ‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’ Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution. WRONG Dissolve 1g of solute in 250cm3 of de-ionised water 250cm3 1g

‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’
SOLUTIONS ‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’ Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution. WRONG Dissolve 1g of solute in 250cm3 of de-ionised water RIGHT Dissolve 1g of solute in water and then add enough water to make 250cm3 of solution 250cm3 1g 250cm3 1g WATER

‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a clean The solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water.

STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a clean The solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ?

STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a clean The solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? mass of Na2CO3 in a 250cm3 solution = g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = g / 106g mol = mol

STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a clean The solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? mass of Na2CO3 in a 250cm3 solution = g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = g / 106g mol = mol Concentration is normally expressed as moles per dm3 of solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4

STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a clean The solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? mass of Na2CO3 in a 250cm3 solution = g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = g / 106g mol = mol Concentration is normally expressed as moles per dm3 of solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4 no. of moles in 1000cm3 (1dm3) = 4 x = mol ANS mol dm-3

How to work out how much to weigh out
STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?

STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = mol dm-3 How many moles will be in 1 dm3 ? = mol How many moles will be in 250cm3 ? = /4 = mol

STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = mol dm-3 How many moles will be in 1 dm3 ? = mol How many moles will be in 250cm3 ? = /4 = mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1

STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = mol dm-3 How many moles will be in 1 dm3 ? = mol How many moles will be in 250cm3 ? = /4 = mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1 What mass of Na2CO3 is in moles = x = g of Na2CO3 ? (mass = moles x molar mass)

STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = mol dm-3 How many moles will be in 1 dm3 ? = mol How many moles will be in 250cm3 ? = /4 = mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1 What mass of Na2CO3 is in moles = x = g of Na2CO3 ? (mass = moles x molar mass) ANS. The chemist will have to weigh out 2.650g, dissolve it in water and then make the solution up to 250cm3 in a graduated flask.

VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration mol dm-3.

Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration mol dm-3. 1. Write out a BALANCED equation NaOH HCl ——> NaCl H2O

Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration mol dm-3. 1. Write out a BALANCED equation NaOH HCl ——> NaCl H2O 2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl

Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration mol dm-3. 1. Write out a BALANCED equation NaOH HCl ——> NaCl H2O 2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl x 25/1000 (i) M is the concentration in mol dm-3 NaOH M x 20/1000 (ii)

Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration mol dm-3. 1. Write out a BALANCED equation NaOH HCl ——> NaCl H2O 2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl x 25/1000 (i) M is the concentration in mol dm-3 NaOH M x 20/1000 (ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/ = x 25/1000

Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration mol dm-3. 1. Write out a BALANCED equation NaOH HCl ——> NaCl H2O 2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl x 25/1000 (i) M is the concentration in mol dm-3 NaOH M x 20/1000 (ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/ = x 25/1000 5. Cancel the 1000’s M x = x 25 re-arrange the numbers to obtain M M = x 25 20

Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration mol dm-3. 1. Write out a BALANCED equation NaOH HCl ——> NaCl H2O 2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl x 25/1000 (i) M is the concentration in mol dm-3 NaOH M x 20/1000 (ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/ = x 25/1000 5. Cancel the 1000’s M x = x 25 re-arrange the numbers to obtain M M = x 25 20 6. Calculate the concentration of the NaOH = mol dm-3

2NaOH + H2SO4 ——> Na2SO4 + 2H2O
VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. 2NaOH H2SO4 ——> Na2SO H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 i.e moles of NaOH = x moles of H2SO4 or moles of H2SO4 = moles of NaOH 2 REMEMBER... IT IS NOT A MATHEMATICAL EQUATION 2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4 More examples follow

2NaOH + H2SO4 ——> Na2SO4 + 2H2O
VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. 2NaOH H2SO4 ——> Na2SO H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 i.e moles of NaOH = x moles of H2SO4 or moles of H2SO4 = moles of NaOH 2 2HCl Na2CO3 ——> 2NaCl CO H2O you need 2moles of HCl to react with every 1 mole of Na2CO3 i.e moles of HCl = x moles of Na2CO3 or moles of Na2CO3 = moles of HCl 2

MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+
VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. MnO4¯ H Fe2+ ——> Mn H2O Fe3+ you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯ i.e moles of Fe2+ = x moles of MnO4¯ or moles of MnO4¯ = moles of Fe

MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+
VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. MnO4¯ H Fe2+ ——> Mn H2O Fe3+ you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯ i.e moles of Fe2+ = x moles of MnO4¯ or moles of MnO4¯ = moles of Fe 2MnO4¯ H2O H+ ——> 2Mn O2 + 8H2O you need 5 moles of H2O2 to react with every 2 moles of MnO4¯ i.e moles of H2O2 = x moles of MnO4¯ 2 or moles of MnO4¯ = x moles of H2O

Calculate the volume of sodium hydroxide (concentration mol dm-3) required to neutralise 20cm3 of sulphuric acid of concentration mol dm-3. 2NaOH H2SO4 ——> Na2SO H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 therefore moles of NaOH = 2 x moles of H2SO4 moles of H2SO4 = x 20/ (i) moles of NaOH = x V/ (ii) where V is the volume of alkali in cm3

Calculate the volume of sodium hydroxide (concentration mol dm-3) required to neutralise 20cm3 of sulphuric acid of concentration mol dm-3. 2NaOH H2SO4 ——> Na2SO H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 therefore moles of NaOH = 2 x moles of H2SO4 moles of H2SO4 = x 20/ (i) moles of NaOH = x V/ (ii) where V is the volume of alkali in cm3 substitute numbers moles of NaOH = x moles of H2SO4 x V/ = x x 20/1000 cancel the 1000’s x V = x x 20 re-arrange Volume of NaOH (V) = x x = cm3 0.100

MOLAR VOLUME stp = standard temperature and pressure (273K and 105 Pa)
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 22.4dm3 at stp 1. Calculate the volume occupied by 0.25 mols of carbon dioxide at stp 1 mol of carbon dioxide will occupy a volume of dm3 at stp 0.25 mol of carbon dioxide will occupy a volume of x 0.25 dm3 at stp 0.25 mol of carbon dioxide will occupy a volume of 5.6 dm3 at stp stp = standard temperature and pressure (273K and 105 Pa) ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure

MOLAR VOLUME stp = standard temperature and pressure (273K and 105 Pa)
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 22.4dm3 at stp 1. Calculate the volume occupied by 0.25 mols of carbon dioxide at stp 1 mol of carbon dioxide will occupy a volume of dm3 at stp 0.25 mol of carbon dioxide will occupy a volume of x 0.25 dm3 at stp 0.25 mol of carbon dioxide will occupy a volume of 5.6 dm3 at stp 2. Calculate the volume occupied by 0.08g of methane (CH4) at stp Relative Molecular Mass of CH4 = (4x1) = 16 Molar Mass of CH4 = 16g mol-1 Moles = mass/molar mass 0.08g / 16g mol = mols 1 mol of methane will occupy a volume of dm3 at stp 0.005 mol of carbon dioxide will occupy a volume of x dm3 at stp 0.005 mol of carbon dioxide will occupy a volume of dm3 at stp stp = standard temperature and pressure (273K and 105 Pa) ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure

A guide for A level students KNOCKHARDY PUBLISHING

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