Class 10 Chemistry Chapter 4 Important Questions with Answers Kerala Syllabus

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SSLC Chemistry Chapter 4 Important Questions Kerala Syllabus

Gas Laws and Mole Concept Class 10 Important Questions

Question 1.
The gas in cylinder A of volume 2 L is completely transferred into cylinder B of volume 4 L without changing temperature.
a) What is the new volume of the gas?
b) Gas in which cylinder experiences more pressure?
Answer:
a) 4 L
b) Cylinder A

Question 2.
Give the SI unit of temperature
Answer:
Kelvin (K)

Question 3.
What is the relationship between the volume of a gas and its temperature at constant pressure?
Answer:
The volume (V) of a gas is directly proportional to the temperature (T)

Question 4.
Write the ideal gas equation and write what each letter represents.
Answer:
PV=n R T
P=Pressure
V=Volume
N=Number of moles
T=Temperature
R=Universal gas constant

Question 5.
Convert the given temperature to the Kelvin scale.
a) 0°C
b) 100°C
Answer:
a) 0°C = 0 + 273 = 273K
b) 100°C + 273 = 373K

Question 6.
One-gram atomic mass (1 GAM) of any element contains …………………….. atoms.
Answer:
6.022 × 10²³

Question 7.
Find the number of gram molecular mass (GMM) present in 56g of Nitrogen, of nitrogen is 28]
Answer:
Mass = \(\frac{\text { Given mass in gram }}{\text { Molecular mass of the element }}\) = \(\frac{56}{28}\) = 2GMM

Question 8.
Analyse the situations given below and write the one that is related to Boyle’s law.
a) If an inflated balloon is kept in sunlight, it will burst after some time
b) As the balloon is being inflated, its volume increases.
c) The size of air bubbles rising from the bottom of a water body gradually increases.
Answer:
c) The size of air bubbles rising from the bottom of a water body gradually increases.

Question 9.
Match the following

a) P ∝ 1/V	i) Avogadro’s Law
b) V ∝ T	ii) Boyle’s Law
c) V ∝ n	iii) Charles’s Law

Answer:
P ∝ 1/V: – Boyle’s Law
V ∝ T: – Charles’s Law
V ∝ n: – Avogadro’s Law

Question 10.
Select from the box the correct gas law related to the given situations.

Charles’s law, Boyle’s law, Avogadro’s law

(a) An inflated balloon kept in the sunlight burst after some time.
(b) The size of an air bubble rising from the bottom of an aquarium increases
Answer:
(a) Charles’s law
(b) Boyle’s Law

Question 11.
If a gas occupies 67.2 L at STP, calculate the number of moles it represents.
Answer:
Number of moles = \(\frac{\text { Volume at STP }}{22.4 \mathrm{~L}}\) = \(\frac{67.2}{22.4 \mathrm{~L}}\) = 3 moles

Question 12.
The volume of 1 mole of any gas at STP is 22.4 L. What will be the volume of 34g NH₃ at STP?
Answer:
No of moles = \(\frac{34}{17}\) = 2moles
Volume at STP =No of moles × 22.4 = 2 × 22.4 = 44.8L

Question 13.
Define absolute zero
Answer:
It was Lord Kelvin who identified that the lowest temperature that a gas can attain is – 273°C and named it Absolute zero. For practical purposes, -273°C is taken as the value of Absolute zero.

Question 14.
Convert the following temperature in Kelvin scale to Celsius scale
(i) 500K
(ii) 273K
(iii) 100K
(iv) 300K
Answer:
(i) 500K
500 – 273 = 227°C

(ii) 273K
273 – 273 = 0

(iii) 100K
100 – 273 = -173°C

(iv) 300K
300 – 273 = 27°C

Question 15.
Given Mass: 45 grams of water (H₂O) Molecular Mass of Water: 18 u Calculate:
a) The number of moles of water molecules.
b) The number of molecules in the given mass.
Answer:
a) Number of moles = \(\frac{45}{18}\) = 2.5 moles
b) Number of Molecules = 2.5 moles × 6.022 × 10²³molecules = 1.51 × 10²⁴molecules

Question 16.
Atomic mass of some elements are given [ Mg = 24, O = 16]
a) How many atoms are there in 120g Mg?
b) Find the mass of 4 × 6.022 × 10²³ oxygen atoms.
Answer:

Question 17.
a) If the volume of a gas at 127 °C and 1 atm pressure is 1600L, what will be the volume of the gas at 600K and 2 atm pressure?
b) The volume of a certain mass of gas at 1 atm is 1600 L. What will be the volume if the pressure is increased to 4 atm (temperature remains unchanged)?
Answer:
a) P₁V₁/T₁ = P₂V₂/T₂
P₁ = 1 atm
V₁ = 1600L
T₁ = 127°C = 127 + 273 = 400K
P₂ = 2atm
T₂ = 600K
V₂ = ?
\(\frac{1 \times 1600}{400}\) = \(\frac{2 \times V_2}{600}\)
V = \(\frac{960000}{800}\) = 1200L

c) P₁V₁ = P₂V₂
P₁ = 1 atm
V₁ = 1600L
P₂ = 4atm
V₂ = ?
1 × 1600 = 4 × V₂
V₂ = \(\frac{1600}{4}\) = 400L

Question 18.
The data of an experiment conducted on a fixed mass oigas at constant pressure are given.

Volume(V)L	Temperature(T)K
600	300
800	(a)
(b)	450

a) Find the values of (a) and (b)
b) Which gas law is illustrated here?
Answer:
(a) \(\frac{600}{300}\) = 2
\(\frac{\mathrm{V}}{\mathrm{~T}}\) = Constant
\(\frac{\mathrm{V}}{\mathrm{~T}}\) = 2
\(\frac{800}{2}\)
Therefore, T = 400K

(b) \(\frac{\mathrm{V}}{450}\) = 2
V = 2 × 450 = 900L

(c) Charles’s law

Question 19.
A gas kept in cylinder A, having a volume of 5 L at 4 atm pressure, is completely transferred to cylinder B of volume 10 L at constant temperature.
(a) What is the volume of the gas in cylinder B?
(b) What will be the pressure in cylinder B?
(c) Which Gas law is associated with this situation?
Answer:
(a) 10 L

(b) P₁V₁ = P₂V₂
P₁ = 4 atm
V₁ = 5L
V₂ = 10L
P₂ = ?
P₂ = P₁V₁/V₂
= \(\frac{5 \times 4}{10}\) = 2 atm

(c) Boyle’s law

Question 20.
The molecular mass of methane (CH₄) is 16.
a) What is the mass of 1 GMM of CH₄?
b) Calculate the number of moles in 160g of CH₄.
c) What is the mass of 5 × 6.022 10²³ CH₄ molecules?
Answer:
(a) 16g
(b) Number of moles = Mass given in grams / Gram molecular mass of the compound = 160 g / 16 g = 10 mole
(c) Number of moles × molecular mass= 5 × 16 = 80 g

Question 21.
The CO₂ gas kept at STP has a volume of 112 L. [Hint: Molecular mass – 44]
a) Find the number of moles of CO₂.
b) Calculate the mass of 112 L CO₂.
c) How many molecules of CO₂ are present in it?
Answer:
a) No of moles = \(\frac{\text { Volume at STP }}{22.4}\) = \(\frac{112}{22.4}\) = 5 moles
b) Mass = No of moles × Molar mass =5 moles × 44= 220g
c) No of molecules = No of moles × 6.022 × 10²³ = 5 × 6.022 × 10²³

Question 22.
Complete the following

Element/Compound	GMM	Given mass	No of moles	Volume at STP
O2	36	360	(a)	224
NH3	17	(b)	5	112
CO2	(c)	88	2	44.8
HCl	36.5	73	2	(d)

(a) 10
(b) 85
(c) 44
(d) 44.8

Question 23.
From the given statements, select and write down those related to gases:
a) The energy of the molecules is very high.
b) The force of attraction between the molecules is very high.
c) The distance between the molecules is very high.
d) The freedom of movement of the molecules is very low.
e) Since the collisions of the molecules are elastic in nature, there is no loss of energy.
f) As the molecules are so far apart, the volume of gaseous molecules is negligible in comparison with the total volume of the gas.
Answer:
a, c, e, f

Question 24.
Match the following.

(i) Relationship between volume and pressure	(a) Lord Kelvin
(ii) Relationship between volume and temperature	(b) Avogadro’s law
(iii) Relationship between the number of particles and volume	(c) Boyle’s law
(iv) Absolute zero	(d) Charles’s law

Answer:
(i) – c, (ii) – d, (iii) – b, (iv) – a

Question 25.
(a) Compare the following properties of a substance in its liquid and gaseous states:
• Energy
• The attractive force between molecules
• Freedom of movement of molecules
(b) Even though gas molecules are continuously colliding with each other, there is no loss of energy. Why?
Answer:
(a)

Properties	Liquid	Gas
Energy.	Low	Very High
The attractive force between molecules.	High	Very Low
Freedom of movement of molecules.	High	Very High

(b) The collisions are perfectly elastic

Question 26.
The volume of 2 mol of hydrogen gas at 1 atm pressure and 273 K is 44.8 L.
a) Name the scientist who established the relationship between the volume and pressure of a gas through experiment.
b) What will be the new pressure if the volume of hydrogen gas is changed to 22.4 L?
c) Suggest a method to increase the volume of this gas without changing the pressure and mass.
Answer:
a) Robert Boyle

b) P₁ V₁ = P₂V₂
P₁ = 1 atm
V₁ = 44.8L
V₂ = 22.4L
P₂ = P₁V₁/V₂
p₂ = \(\frac{1 \times 44.8}{22.4}\) = 2atm

c) Increase the temperature

Question 27.
The molecular mass of SO₂ is 64.
a) Find the mass of 1 GMM SO₂.
b) Find the number of molecules in 1 GMM SO₂.
c) Find out the number of moles of molecules in 320g of SO₂.
Answer:
Answer:
a) Mass of 1 GMM SO₂ = 64 g

b) No. of molecules in 1GMM SO₂ = 6.022 × 10²³ (N_A)

c) No or moles = \(\frac{\text { Given mass in gram }}{\text { Molecular mass }}\) = \(\frac{320}{64}\) = 5 moles

Question 28.
The chemical equation for the manufacture of ammonia is N₂(g) + 3H₂(g) → 2NH₃
a) Complete the following: 1 mol N₂ + H₂ → NH₃
b) Calculate the amount of H₂ required to react with 28 g of N₂ completely. [Hint: Molecular mass of N₂ = 28, H₂ = 2]
c) What will be the volume of NH₃ formed at STP if 22.4 L of N₂ is completely reacted?
Answer:
a) 1 mol N₂ + 3 mol H₂ → 2 mol NH₃

b) 28g N₂ = 1 mol N₂
1 mol N₂ requires 3 mol H₂
Mass of H₂ = 3 × 2g = 6g

c) 22.4 L N₂ = 1 mol N₂
When 1 mol N₂ undergo a reaction, 2 mol NH₃ is formed.
Volume of NH₃ formed at STP = 2 × 22.4 = 44.8 L

Question 29.
Complete the following

Answer:
a) Volume at STP = No of moles × 22.4 L =10 × 22.4 L = 224L

b) Number of molecules = No of moles × 6.022 × 10²³ = 10 × 6.022 × 10²³

c) Number of atoms = Atomicity × Number of moles (n) × Avogadro’s number (N_A)
= 4 × 10 × 6.022 × 10²³

d) Mass = Number of moles × molar mass =10 × 17 = 170g

Question 30.
The volume of the gas does not depend on the size of its molecules
a) Do you agree with this statement? Justify your answer
b) What are the factors on which the volume of the gas depends?
Answer:
(a) Yes. As the distance between the molecules is very large, the volume of the gas does not depend on the size of its molecules.
b) The volume of a gas depends on the number of molecules, pressure and temperature.

Question 31.
What is the SI unit of volume and temperature?
Answer:
Volume: m³
Temperature: K

Question 32.
Express the temperature 27°C in Kelvin (K)?
Answer:
K = °C + 273
K = 27 + 273
K = 300
Therefore, 27°C is equal to 300 K.

Question 33.
Explain how gas molecules exert pressure.
Answer:
Gas molecules in constant random motion collide with the container walls, exerting a force that we measure as pressure.

Question 34.
If two different gases at the same pressure and temperature have the same volume, then
a) Masses are equal
b) The number of molecules is equal
c) Characteristics are the same
d) None of these
Answer:
b) The number of molecules is equal.

Question 35.
The size of a weather balloon increases as it rises in the air. What may be the reason? Which gas law can be related to this?
Answer:
As height increases, the atmosphere pressure decreases and volume increases. Boyle’s law

Question 36.
When gas is filled in a cylinder at 1 atm pressure, its volume is 250 cm³. The volume decreased to 50 cm³ by pressing the piston. What will the pressure be inside?
Answer:
According to Boyle’s law
P₁V₁ = P₂V₂
1 × 250 = P₂ × 50
P₂ = \(\frac{250}{50}\) = 5 atm

Question 37.
Match those given in columns A, B, and C suitably.

A	B	C
Boyles law	V ∝ T	Temperature constant
Charles law	V ∝ n	Pressure constant
Avogadro’s law	V ∝ 1/P	Temperature and pressure are constant.

Answer:

A	B	C
Boyles law	V ∝ 1/P	Temperature constant
Charles law	V ∝ T	Pressure constant
Avogadro’s law	V ∝ n	Temperature and pressure are constant.

Question 38.
Find the relation in the first pair, then fill in the blank in the second pair.
44.8 litre gas at STP: 2 mol
224 litre gas at STP: ?
Answer:
44.8 litres of gas = 2 mol
Number or moles = \(\frac{\text { volume at STP }(\text { in litre })}{22.4}\)
= \(\frac{224}{22.4}\) = 10
224 litre gas at STP: 10

Question 39.
Calculate the gram molecular mass/gram formula mass of the following
a) HNO₃
b) CaCl₂
c) Na₂SO₄
d) NH₄NO₃
Answer:
a) HNO₃
(1 × 1 ) + (1 × 14) + (3 × 16)
1 GMM = 63 g

b) CaCl₂
(1 × 40) + (2 × 35.5)
1 GMM = 111g

c) Na₂ SO₄
(2 × 23) + (1 × 32) +(4 × 16)
1 GMM = 142 g

d) NH₄NO₃
(1 × 14) + (4 × 1) +(1 × 14) + (3 × 16)
1 GMM = 80g

Question 40.
Calculate the number of moles in 1 kg of water.
Answer:
lKg of water = 1000g of water
No of moles in 1 Kg of water = \(\frac{\text { Mass in gram }}{\text { GMM }}\) = \(\frac{1000}{18}\) = 55.55mol

Question 41.
Find out the number of molecules present in the 3 GMM water?
Answer:
The Number of molecules in 3 GMM water
1 GMM = 6.022 × 10²³ molecules
3 GMM = 3 × 6.022 × 10²³ molecules

Question 42.
a) What is meant by gram atomic mass?
b) Gram atomic mass of nitrogen is 14g. Then find out the following: the Number of moles of atoms in 70g of nitrogen.
Answer:
a) The mass in grams equal to the atomic mass of an element is called its gram atomic mass.

• Number of moles or atoms \(\frac{\text { Given Mass in gram }}{\text { Gram atomic mass }}\) = \(\frac{70}{14}\) = 5 mol

Question 43.
What would be the volume of 3.5 moles of methane (CH₄) gas at a temperature of 300K and a pressure of 1.5 atm? (Value of universal gas constant (R) = (0.0821L atm mol^-1K^-1)
Answer:
PV = nRT
T = 300K
n = 3.5mol
R = 0.0821L atm mol^-1K^-1
P = 1.5 atm
V = \(\frac{\mathrm{nRT}}{\mathrm{P}}\) = \(\frac{3.5 \times 0.0821 \times 300}{1.5}\) = 57.47L

Question 44.
If 0.50 moles of hydrogen gas (H₂) are contained in a 10.0 L container at a temperature of 25°C, what is the pressure of the gas in Pascal’s (Pa)? (Use R = 8.314Jmol^-1K^-1)
Answer:
P V = n R T
P = \(\frac{\mathrm{nRT}}{\mathrm{V}}\)
T = 25°C + 273 = 298K
n = 0.50mol
R = 8.314Jmol^-1K^-1
V = 10L = 0.01m³
P = \(\frac{0.50 \times 8.314 \times 298}{0.01}\)
P ≈ 123878.6Pa

Question 45.
Define ideal gas.
Answer:
The gases that obey the ideal gas equation at all temperatures and pressures are known as ideal gases

Question 46.
Consider the reaction for the formation of water: 2H₂ + O₂ → 2H₂O
If you have 8 grams of hydrogen gas, how many grams of oxygen gas are required for a complete reaction?
Answer:
2H₂ + O₂ → 2H₂O
4g H₂ + 32gO₂ → 36gH₂O
4g H₂ requires 32g of O₂
1g H₂ require = 32/4 g of O₂
8g H₂ require = \(\frac{32}{4}\) × 8 = 64g of O₂
64g of O₂ required.

Question 47.
Using the combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O
If 32 grams of oxygen react completely with methane, what mass of carbon dioxide (CO₂) is produced? (Molar mass of O₂ = 32g/mol, molar mass of CO₂ = 44g/mol).
Answer:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of oxygen that reacted = \(\frac{\text { Mass }}{\text { Molar mass }}\) = \(\frac{32 \mathrm{~g}}{32 \mathrm{~g} / \mathrm{mol}}\)
2mol O₂ produces 1 mol CO₂.
Therefore, 1 mol O₂ will produce 0.5 mol CO₂.
Mass of CO₂ = moles × molar mass = 0.5mol × 44g/mol = 22g
Therefore, 22 grams of carbon dioxide are produced.

Question 48.
Consider the combustion of butane (C₄H₁₀): 2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(g)
If 58 grams of butane are burned in excess oxygen, how many moles of carbon dioxide (CO₂) are produced? (Molar mass of C₄H₁₀ ≈ 58g/mol).
Answer:
The Number of moles of C₄H₁₀ = \(\frac{\text { Mass }}{\text { Molar mass }}\) = \(\frac{32 \mathrm{~g}}{32 \mathrm{~g} / \mathrm{mol}}\) = 1 mol
2 Moles of C₄H₁₀ produce 8 moles of CO₂
1 mole of C₄H₁₀ will produce = \(\frac{8}{2}\) = 4 mol CO₂