SSLC Chemistry: Chapter 02 Gas Laws and Mole Concept - Questions and Answers


Study Notes for Class 10 Chemistry (English Medium) വാതകനിയമങ്ങളും മോൾ സങ്കൽപ്പനവും | Text Books Solution Chemistry - Chapter 02 Gas Laws and Mole Concept

Class 10 Chemistry Questions and Answers - Chapter 02 Gas Laws and Mole Concept
* Study Notes
* Matter can be classified as solid, liquid, & gas. Compare to solid and liquid, gases have a lot of characteristics features. Some of the properties of gases are given below.   
• Each gas contains numerous minute molecules 
• when compare to the total volume of a gas, the real volume of a molecule is very less.
• The molecule of a gas is in a state of rapid random motion in all direction. 
• As a result of the random motion of the gas molecules, they colloid with each other and also colloid with the walls of the container, In which it is kept.
• As the collision of the molecules are perfectly elastic in nature, there is no loss of energy.
• There is no attraction between the gas molecules and the walls of the container. 
From the above statement, we can conclude that.
* Volume, Temperature, Pressure of a Gas:  
A) Volume of a gas.   
The volume of a gas is the volume of the container it occupies.  
If a gas that is kept in a cylinder having a volume of 1 litre, is completely transferred to another 5 litter cylinder then its new volume becomes 5 litres.  
1. Pull the piston of a syringe backwards. Press the piston after closing the nozzle of the syringe. What will happen to the volume of air inside the syringe?
When we press the piston after closing its nozzle, the volume of the gas inside the syringe decreases. 
B) Pressure of the gas.
Force exerted per unit area is called pressure.   
The pressure of the gas is the force experienced per unit area on the inner surface of the container. as a result of the collision of the gas molecules on the surface.    
Force on unit area= total force exerted on the surface/ surface area.   
C) Temperature of gas.   
When we heated a gas. The kinetic energy of a gas is increases and molecules are moving very fastly.    
The temperature of a gas is defined as the average kinetic energy of a molecule in a substance.

* Gas laws:  
A law that describes the properties of gases concerning temperature, volume, pressure, of gas re known as gas laws. The three important gas laws are,     
1) Boyle’s law  
2) Charle’s law  
3) Avogadro's law.  
A) Boyle’s law: (relation between volume and pressure of  gas)    
Boyle’s law state that, at a constant temperature, the volume of a definite mass of gas is inversely proportional to its pressure.  If P is the pressure and V is the volume of gas then, PxV is constant. 
V α 1 /P → mathematical expression.    
P xV= A constant → mathematical equation. 
P₁V₁=P₂V₂ → practical equation 
from the above diagram, picture A has 1L volume and 1atm pressure. T constant temperature, when we increase the pressure to 2 atm, the volume of cylinder become 0.5L. ie, (at a constant temperature, the volume of a gas is inversely proportional to the pressure of that gas).

* The size of the air bubbles rising from the bottom of an aquarium increases. Give reason. 
Here the temperature is constant. From bottom to top, the external pressure decreases. Hence the volume of the bubble increases. (Boyle’s law).
The situation related to Boyle’s law: 
1) The size of the air bubble rising from the bottom in an aquarium increases. 
2) The size of climate balloon increases as they move in the upward direction to the higher altitude in the atmosphere. 
3) When an inflated balloon is immersed in water, its size increases
B) Charel’s law (relation between volume and temperature of a gas) 
Charel’s law is defined as constant pressure, the volume of a definite ms of a gas is directly proportional to the temperature on the Kelvin scale. Let V is the volume and T is the temperature, then   
V α T  → mathematical expression.    
V/T =K  → mathematical equation. 
V₁ /  T₁=V₂ / T₂  → practical equation 

* Take a dry bottle (an injection bottle) having a rubber stopper. Fix an empty refill through the rubber stopper. Fill a drop of ink into the lower end of the refill tube, then close the bottle. Dip this arrangement in lukewarm water. What do you observe?
The ink rises up.
What is the reason for the rising of the ink upwards?
When the temperature increases, the volume of the gas inside the bottle increases. This will push the ink back.
What did you observe on cooling the bottle after taking it out? Why?
On cooling the bottle, the volume of the gas decreases. Then the ink goes down.

* If an inflated balloon is kept in sunlight, it will burst. What may be the reason for this?
When the temperature increases, the volume of the gas inside the balloon increases and finally it will burst. (Charle’s Law)

The situation related to Charles law,   
When an inflated balloon is kept in sunlight, it will burst.  
Vehicle tyres are not fully inflated during the summer season.  
When an inflated balloon kept in cold water, its size decreased.  

C) Avagadros law: (Relation between volume and number of moles) 
Avogadro's law states that at constant temperature and pressure the volume of a gas is directly proportional to the number of molecules. Let V is the volume and n is the number of moles, then 
V α n  → mathematical expression.  
V/n = k → mathematical equation.
V₁/n₁ = V₂/n₂  → practical equation 

* How is the number of minute particles calculated?
If the particles having the same size and mass, even though they are in crores, we can determine their accurate number based on mass.

* Relative atomic mass: 
The mass of an atom is compared to the mass of another atom and expressed as a number, which shows how many times it is heavier than the other atom. The atomic mass of elements is expressed by considering 1/12 as one unit. 

* Gram Atomic Mass & Gram Molecular Mass:
A) Gram atomic mass (GAM)
The atomic mass of the element hydrogen(₁H) is 1.  Hence 1g of hydrogen is known as 1GAM. 
Similarly, the atomic mass of carbon (C) is 12, and the atomic mass of Nitrogen(N) is 14, when we take 12g of carbon and 14g of nitrogen both are 1 GAM.
• The mass of an element in gram equal to its atomic mass is called 1 gram atomic mass (GAM) of the element. This may also be shorted as 1 gram atom 
We know the atomic mass of carbon is 12 when we take 12g of carbon which is 1GAM. 
One gram atomic mass(1GAM) of any element contains atoms. This number is known as the Avagadro number. Indicated as NA
Any substance that contains NA (Avagadro number) atoms we called 1 mole of atoms.
Eg: the atomic mass of Hydrogen(H) is 1.  
1g of H is known as 1GAM. It contains NA(6.022x10²³) atoms and we call it is one mole of hydrogen atoms. 
B) Molecular mass and gram molecular mass:
Method to find the molecular mass of a compound.
 Eg: H₂O 
We know the atomic mass of H=1 and O=16
H₂O contain 2 H and 1 Oxygen atom, so its molecular mass become 
2x1 + 1x16 = 2+16 =18 (molecular mass of H₂O=18g)
Similarly: the molecular mass of glucose(C₆H₁₂O₆) 
Glucose contain 6 carbon atoms 12 hydrogen atoms and 6 oxygen atoms. 
We know the atomic masses of C=12  O=16 H=1
Then the molecular mass of glucose become  
 6x12 + 12x1 + 6x16 = 72+12+96 =180  
* Gram Molecular Mass (GMM):
The amount of substance in grams equal to its molecular mass is called gram molecular mass. 
One gram of any substance contains an Avagadro number of molecules.
Any substance that contains NA(Avagadro number) molecules we can call it as on mole molecules.
Eg: molecular mass of O₂ =32g
How many GMM are there in 64g oxygen?
How many molecules are present in it?
a)   GMM=      𝐺𝐼𝑉𝐸𝑁 𝑀𝐴𝐴𝑆             =  64𝑔   = 2 GMM 
                   𝑀𝑂𝐿𝐸𝐶𝑈𝐿𝐴𝑅 𝑀𝐴𝑆𝑆            32𝑔   
We know 1 GMM contain NA molecules.
So, 2 GMM contain 2x6.022x10²³ molecules. 
* Relation between the volume of a gas and moles: 
Molar volume: one mole of any gas under the same condition of temperature and pressure will contain the same number of molecules, so their volume will also be the same. this is called molar volume.   
At STP one mole of any gas will occupy a volume of 22.4L. this called molar volume at STP. (273K temperature and 1 atm pressure are known as standard temperature and pressure (STP)). 
So, volume of L in STP= number of moles of gas at STP x 22.4 =2x22.4= 44.8 

Let’s Assess (TextBook Questions: Page 29, 30)

1. Examine the data given in the table (Temperature and number of molecules of the gas are kept constant).
a) Calculate P x V.
b) Which is gas law related to this?
Answer:
(a) = 8 L atm 
(b) = Boyle's Law

2. Analyse the situations given below and explain the gas law associated with it.
a) A balloon is being inflated.
b) When an inflated balloon is immersed in water, its size decreases.
Answer:
(a) = Avogadro's law, 
(b) = Boyle's law (Explanation of the example )

3. Certain data regarding various gases kept under the same temperature and pressure conditions are given below.
a) Complete the table.
b) Which gas law is applicable here?
(b) Avogadro's Law

4. a. Calculate the mass of 112 L CO₂ gas kept at STP (molecular mass =44).
b. How many molecules of CO₂ are present in it?
Answer:
(a) Number of moles of a gas at STP =  Volume of the gas in Litres at STP /22.4 L
= 112L / 22.4 L
= 5 moles
But, Number of moles = Mass given in grams / Gram Molecular Mass of the compound Mass in grams 
= Number of moles x Gram Molecular Mass of the compound
= 5 x 44 g
= 220 g
(b) Number of molecules of CO₂ = Number of moles x 6.022 x 10²³ 
= 5  x  6.022 x 10²³ 

5. Calculate the volume of 170g of ammonia at STP? (Molecular mass -17)
Answer:
Number of moles = Mass given in grams / Gram Molecular Mass of the compound
= 170 g / 17 g
= 10 moles
Number of moles of a gas at STP = Volume of the gas in litres at STP/ 22.4 L
Volume of the gas in litres at STP / 22.4 L = Number of moles at STP x 22.4 L
= 10 moles  x 22.4 L
= 10 x 22.4 L = 224 L

6. Find out the number of moles of molecules present in the samples given below.
(GMM - N₂ = 28g, H₂O = 18g)
a) 56g N₂         b) 90g H₂O
Answer:  (a) = 2      (b) = 5 

7. The Molecular mass of ammonia is 17.
a) How much is the GMM of ammonia?
b) Find out the number of moles of molecules present in 170g of ammonia.
c) Calculate the number of ammonia molecules present in the above sample of ammonia?
Answer: 

8. The molecular mass of oxygen is 32.
a) What is the GMM of O₂?
b) How many moles of molecules are there in 64 g of oxygen? How many
molecules are there in it?
c) Calculate the number of oxygen atoms present in 64g of oxygen?
Answer:

PRACTICE QUESTIONS WITH ANSWER
1. Find out the molecular mass of sulphuric acid (H₂SO₄),  Nitric acid (HNO₃),  Calcium carbonate (CaCO₃),  Water (H₂O), Ammonia (NH₃), Carbon dioxide (CO₂)
[Atomic mass: H -1, S-32, O- 16, N-14, Ca- 40, C-12, Cl – 35.5, Na - 23]
Answer:
H₂SO₄: 2x1+1x32+4x16 = 98            
HNO₃: 1x1 +1x14+3x16 = 63   
CaCO₃: 1x40+1x12+3x16 =100
H₂O: 2x1+1x16 = 18                
NH₃: 1x14 + 3x1 = 17             
CO₂: 1x12+2x16 = 44

2. How many molecules are present in one mole of water? What is known as this number?
Answer: 
i. 6.022x10²³
ii.  Avogadro number

3. Calculate the number of moles of molecules in 88g carbon dioxide (CO₂).
Answer:
Number of moles of (CO₂) molecules in 88 g of carbon dioxide = 88/44 = 2

4.  Find the mass of 2x6.022x10²³ water molecules.
Answer:
Mass of one mole of water(mass of 6.022x10²³ water molecules) = 18 g
Therefore the mass of  2x6.022x10²³ water molecules  = 2 x18 = 36 g

5. What will be the mass of  224 L of ammonia at STP?
Answer: 
Number of moles in 224 L ammonia gas at STP = 224/22.4  =10
Mass of one mole of ammonia  = 17 g
Therefore mass of 10 moles of ammonia = 10x17 =170 g

6. How many hydrogen molecules are present in 10 g of hydrogen?
Answer: 
Number of moles in 10 g of hydrogen = 10/2 = 5
Therefore the number of molecules in 10 g (5 moles) of hydrogen  
= 5x 6.022x10²³

7. What is the volume of 64 g of oxygen at STP?
Answer: Number of moles in 64 of oxygen = 64/32 = 2
Therefore volume 64 g of oxygen (2 mole oof oxygen) at STP =  2x22.4 = 44.8 L

8. Calculate the number of atoms in 16 of oxygen?
Answer: 
Number of moles of atoms in 16 g of oxygen = 16/GAM = 16/16 = 1
Number of atoms in one mole of atoms = 1x 6.022x10²³ = 6.022x10²³

9. What is the volume of 22 g of carbon dioxide gas at STP?
Answer:
 22 g of carbon dioxide = 22/44 = 1/2 mole.
Volume of  1/2 mole of  CO₂ gas = 1/2 x 22.4 = 11.2 L

10. The balanced equation of the formation of water is given. 
2H₂ + O₂ -----> 2H₂O
a) How many moles of water can be formed using one mole of hydrogen?
b) How many grams of oxygen is needed for reacting 2 g hydrogen? 
c) How many grams of oxygen are needed to obtain 18 g of water?
d) Which reactant is remained unreacted if 2 g of hydrogen is made to react with 32 g of oxygen? How many grams will be remained?
e) What should be the masses of hydrogen and oxygen to get 36 g of water? 
f) How many grams of hydrogen is needed for reacting with 64 g of oxygen? 
g) How many grams of water can be made from 20 g of hydrogen? 
h) How many moles of oxygen are required for reacting with one mole of hydrogen? 
Answer: 
a) One mole.
b) 2 g of hydrogen =1 mole.
One mole hydrogen needs 1/2 mole of oxygen.
1/2 mole oxygen = 1/2 x 32 = 16 g.
c) 18 g of water  = 18/18 = 1 mole.
To get 18 g ( one mole) of water 1/2 mole of oxygen is needed.
Mass of 1/2 mole oxygen = 1/2 x 32 = 16 g.
d) To react 2 g of hydrogen (one mole), it needs only 1/2 mole of oxygen (16 g of oxygen).
So 32 – 16 = 16 g of oxygen will remain unreacted.
e) 36 g of water is two moles. For getting two moles of water, it needs 2 moles of hydrogen and 1 mole of oxygen)
That is, 4 g of hydrogen and 32 g of water will be needed.
f) 64 g  of oxygen  = 2 moles.      
The equation shows that 4 moles of hydrogen are required for two moles of oxygen.
So mass of  hydrogen needed = 4x2 = 8 g.
g) 20 g of hydrogen  = 20/2 = 10 mole. From 10 mole of hydrogen, 10 moles of water can be made.
So the mass of water can be made from 20 g of hydrogen = 10 x 18 = 180 g    h) 1/2 mole. 
11. See the equation.  C + O₂ -----> CO₂
a. Calculate the mass of oxygen for the combustion of 6 g of carbon. 
b. How many litres of carbon dioxide  (at STP) is obtained when reacting one mole of carbon and oxygen?
c. How many grams of carbon and oxygen is required for getting 22 g of carbon dioxide?
d. How many carbon dioxide molecules are obtained by reacting 24 g of carbon and  64 g of oxygen?
e. Using  44.8 litre of oxygen at STP, how many grams of carbon dioxide can be prepared? 
Answer:
a. 16 g         
b. 22.4 L       
c. 22 g carbon dioxide = 22/44 = 1/2 mole.
As per the equation, forgetting 1/2  mole of carbon dioxide, 1/2  mole each of carbon and oxygen is needed.
Hence 6 g of carbon and 16 g of oxygen is required here. x 6.022x10²³       
d. Using 24 g (2 moles) of carbon and 64 g (2 moles) of oxygen, two moles of carbon dioxide can be obtained.
Two-mole carbon dioxide contains  2x 6.022x10²³ molecules.
e. 44.8 L  oxygen = 2 mole.  From two mole of oxygen, two moles of carbon dioxide can be prepared.
Two mole of carbon dioxide = 2 x 44 = 88 g

12. See the chemical equation.   N₂ + H₂ -----> NH₃      
a. Balance the equation.
b. How many moles of hydrogen is needed to react with one mole of nitrogen? 
c. To get one mole of ammonia how many moles of nitrogen are needed?
d. Calculate the masses of nitrogen and hydrogen for making 44.8 L ammonia at STP.
e. If 28 g of nitrogen is made to react with  10 g  of hydrogen, how many hydrogen molecules will remain unreacted when the reaction is completed? 
Answer:
a.  N₂ + 3H₂ -----> 2NH₃      
b.3 mole of hydrogen.     
c.1/2  mole nitrogen.  
d. 44.8 L ammonia = 44.8/22.4 = 2 mole.
Get 2 moles of ammonia, one mole of nitrogen and three moles of hydrogen are needed.
So 28 g of nitrogen and 6 g of hydrogen will be needed. 
e. Only 6 g (2 moles) of oxygen is needed for reacting with 28 g (one mole) of nitrogen.
So 10 – 6 = 4 g of hydrogen will remain unreacted when the reaction is completed.
The number of molecules in this hydrogen =  2x 6.022x10²³

13. C₆H₁₂O₆  is the chemical formula of glucose.
a. How many grams of glucose is one GMM glucose? 
b. How many molecules of glucose will be present in 1/2 litre of 1 molar solution. 
c. Can one prepare 1 molar solution of glucose using 45 g of glucose. Explain.
Answer:
a. 6x12+12x1+6x16 = 180 g
b. 1/2 x 6.022x10²³ = 3.011x10²³  
c. Yes. Put 45 g of glucose in a beaker and dissolve it insufficient water till its volume becomes 250 ml. Now he gets 250 ml of 1 molar solution of glucose. 

14. a. Calculate the molecular mass of NaCl.
b. How many grams of solute (NaCl) is required for preparing one litre of 1molar solution? 
Answer:
a. 1x23 + 1x35.5 = 58.5
b. 58.5 g

Extended Activities (TextBook Questions: Page 46)
1. How many grams of Carbon and Oxygen are required to get the same number of atoms as in one gram of Helium?
Answer:
Atomic Mass of Helium = 4
Number of atoms in 4 grams of Helium = 6.022 x 10²³
Number of atoms in 1 gram of Helium  = 6.022 x 10²³ / 4 
Atomic mass of carbon = 12 
Mass of 6.022 x 10²³ atoms of carbon = 12 g
Mass of 6.022 x 10²³ / 4 atoms of carbon = 12 g / 4 = 3 g
Atomic mass of oxygen = 16g 
Mass of 6.022 x 10²³ atoms of oxygen = 16 g
Mass of 6.022 x 10²³ / 4 atoms of oxygen = 16 g / 4 = 4 g

2.  Examine the samples given.
(Molecular mass  : He =4 , NH₃ = 17 , N₂ = 28 , H₂SO₄ = 98 , Water = 18)
a. 20 g He
b. 44.8 L of NH₃ at STP
c. 67.2 L of N₂ at STP
d. 1 mol Of H₂SO₄
e. 180 g of Water
(i) Arrange the samples in the increasing order of the number of molecules in each.
(ii) What will be the ascending order of the number of atoms in each?
(iii) What will be the mass of samples b, c and d
Answer:
Number of molecules in each sample 
The increasing order of number of molecules in a sample is 
1 mol Of H₂SO₄ < 44.8 L of NH₃ at STP < 67.2 L of N₂ at STP <  20 g He <180g of Water 
Number of atoms in each sample 
The ascending order of the number of atoms is
20g He < 67.2 L of N₂ at STP < 1 mol Of H₂SO₄ < 44.8 L of NH₃ at STP < 180g of Water
Mass of samples b, c and d

3. In 90 g of Water,
(a) How many molecules are present?
(b) What will be the total number of atoms?
(c)  What will be the total number of electrons in this sample?
Answer:


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