NCERT Solution Class 11th Chemistry Chapter – 5 States of Matter
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Chemistry |
| Chapter | Chapter – 5 |
| Chapter Name | States of Matter |
| Category | Class 11th Chemistry Question & Answer |
| Medium | English |
| Source | last doubt |
NCERT Solution Class 11th Chemistry Chapter – 5 States of Matter
?Chapter – 5?
✍States of Matter✍
?Question & Answer?
Q 1. What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
Answer: Initial pressure, P1 = 1 bar
Initial volume, V1 = 500 dm^{ 3 }
Final volume, V2 = 200 dm^{ 3 }
As the temperature remains the same, the final pressure (P2) can be calculated with the help of Boyle’s law.
Acc. Boyle’s law,
P1V1 = P2V2
P2 = \frac{P_{1}V_{1}}{V_{2}}
= \frac{1 \; \times \; 500}{200}
= 2.5 bar
∴ the minimum pressure required to compress is 2.5 bar.
Q 2. A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?
Answer: Initial pressure, P1 = 1.2 bar
Initial volume, V1 = 120 mL
Final volume, V2 = 180 mL
As the temperature remains the same, final pressure (P2) can be calculated with the help of Boyle’s law.
According to the Boyle’s law,
P1V1 = P2V2
P2 = \frac{P_{1}V_{1}}{V_{2}}
= \frac{1.2 \; \times \; 120}{180}
= 0.8 bar
Therefore, the min pressure required is 0.8 bar.
Q 3. Using the equation of state pV=nRT; show that at a given temperature density of a gas is proportional to gas pressure p.
Answer:
The equation of state is given by,
pV = nRT ……..(1)
Where, p = pressure
V = volume
N = number of moles
R = Gas constant
T = temp
\frac{n}{V} = \frac{p}{RT}
Replace n with \frac{m}{M}, therefore,
\frac{m}{MV} = \frac{p}{RT}……..(2)
Where, m = mass
M = molar mass
But, \frac{m}{V} = d
Where, d = density
Therefore, from equation (2), we get
\frac{d}{M} = \frac{p}{RT}
d = (\frac{M}{RT}) p
d \propto p
Therefore, at a given temp, the density of the gas (d) is proportional to its pressure (p).
Q 4. At 0°C, the density of a certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer: Density (d) of the substance at temp (T) can be given by,
d = \frac{Mp}{RT}
Now, density of oxide (d1) is as given,
d_{1} = \frac{M_{1}p_{1}}{RT}
Where, M1 = mass of the oxide
p1 = pressure of the oxide
Density of dinitrogen gas (d2) is as given,
d_{2} = \frac{M_{1}p_{2}}{RT}
Where, M2 = mass of the oxide
p2 = pressure of the oxide
Acc to the question,
d1 = d2
Therefore, M_{1}p_{1} = M_{2}p_{2}
Given:
p_{1} = 2 bar
p_{2} = 5 bar
Molecular mass of nitrogen, M_{2} = 28 g/mol
Now, M_{1}
= \frac{M_{ 2 }p_{2}}{p_{ 1 }}
= \frac{ 28 × 5 }{ 2 }
= 70 g/mol
Therefore, the molecular mass of the oxide is 70 g/mol.
Q 5. The pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Answer: For ideal gas A, the ideal gas equation is given by,
p_{X}V = n_{X}RT……(1)
Where p_{X} and n_{X} represents the pressure and number of moles of gas X.
For ideal gas Y, the ideal gas equation is given by,
p_{Y}V = n_{Y}RT……(2)
Where, p_{Y} and n_{Y} represent the pressure and number of moles of gas Y.
[V and T are constants for gases X and Y]
From equation (1),
p_{ X }V = \frac{m_{ X }}{M_{ X }} RT
\frac{p_{ X }M_{ X }}{m_{ X }} = \frac{ R T}{ V } ……(3)
From equation (2),
p_{ Y }V =\frac{m_{ Y }}{M_{ Y }} RT
\frac{p_{ Y }M_{ Y }}{m_{ Y }} = \frac{ R T}{V} …… (4)
Where, M_{ X } and M_{ Y } are the molecular masses of gases X and Y respectively.
Now, from equation (3) and (4),
\frac{p_{ X }M_{ X }}{m_{ X }} = \frac{p_{ Y }M_{ Y }}{m_{ Y }} ….. (5)
Given,
m_{ X } = 1 g
p_{ X } = 2 bar
m_{ Y } = 2 g
p_{ Y } = (3 – 2) = 1 bar (Since total pressure is 3 bar)
Substituting these values in equation (5),
\frac{2 \; \times \; M_{X} }{1} = \frac{1 \; \times \; M_{Y} }{2}
4 M_{ X } = M_{ Y }
Therefore, the relationship between the molecular masses of X and Y is,
4 M_{ X } = M_{ Y }
Q 6. The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts?
Answer:
The reaction of aluminum with caustic soda is as given below:
2Al + 2NaOH + 2H2O \rightarrow 2NaAlO2 + 3H2
At Standard Temperature Pressure (273.15 K and 1 atm), 54 g ( 2 × 27 g) of Al gives 3 ×22400 mL of H2.
Therefore, 0.15 g Al gives:
= \frac{3 \; \times \; 22400 \; \times \; 0.15}{54} mL of H2
= 186.67 mL of H2
At Standard Temperature Pressure,
p_{ 1 } = 1 atm
V_{ 1 } = 186.67 mL
T_{ 1 } = 273.15 K
Let the volume of dihydrogen be V_{ 2 } at p_{ 2 } = 0.987 atm (since 1 bar = 0.987 atm) and T_{ 2 } = 20^{\circ} C = (273.15 + 20) K = 293.15 K.
Now,
\frac{p_{ 1 }V_{ 1 }}{T_{ 1 }} = \frac{p_{ 2 }V_{ 2 }}{T_{ 2 }} V_{ 2 } = \frac{p_{ 1 }V_{ 1 }T_{ 2 }}{p_{ 2 }T_{ 1 }}
= \frac{1 \; \times \; 186.67 \; \times \; 293.15}{0.987 \; \times \; 273.15}
= 202.98 mL
= 203 mL
Hence, 203 mL of dihydrogen will be released.
Q 7. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °C?
Answer: It is known that,
p = \frac{m}{M} \frac{RT}{V}
For methane (CH4),
p_{CH_{ 4 }}
= \frac{ 3.2 }{ 16 } × \frac{8.314 \; \times \;300 }{9 \; \times \; 10^{-3 }} [Since 9 dm3 = 9 \; \times \; 10^{-3} m3 27^{\circ}C = 300 K]
= 5.543 × 10^{ 4 } Pa
For carbon dioxide (CO2),
p_{CO_{ 2 }}
= \frac{ 4.4 }{ 44 } × \frac{8.314 \; \times \;300 }{9 \; \times \; 10^{-3}
= 2.771 × 10^{4} Pa
Total pressure exerted by the mixture can be calculated as:
p = p_{CH_{ 4 }} + p_{CO_{ 2 }}
= (5.543 × 10^{ 4 } + 2.771 × 10^{4}) Pa
= 8.314 × 10^{ 4 } Pa
Q 8. What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?
Answer: Let the partial pressure of {H_{ 2 }} in the container be p_{H_{ 2 }}.
Now,
{p_{ 1 }} = 0.8 bar
{p_{ 2 }} = p_{H_{ 2 }} {V_{ 1 }} = 0.5 L
{V_{ 2 }} = 1 L
It is known that,
{p_{ 1 }} {V_{ 1 }} = {p_{ 2 }} {V_{ 2 }} {p_{ 2 }} = \frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }} p_{H_{ 2 }} = \frac{ 0.8 \; \times \; 0.5 }{ 1 }
= 0.4 bar
Now, let the partial pressure of O2 in the container be p_{O_{ 2 }}.
Now,
{p_{ 1 }} = 0.7 bar
{p_{ 2 }} = p_{O_{ 2 }} {V_{ 1 }} = 2.0 L
{V_{ 2 }} = 1 L
{p_{ 1 }} {V_{ 1 }} = {p_{ 2 }} {V_{ 2 }} {p_{ 2 }} = \frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }} p_{O_{ 2 }} = \frac{ 0.7 \; \times \; 20 }{ 1 }
= 1.4 bar
Total pressure of the gas mixture in the container can be obtained as:
p_{total} = p_{H_{ 2 }} + p_{O_{ 2 }}
= 0.4 + 1.4
= 1.8 bar
Q 9. The density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?
Answer: Given,
d1 = 5.46 g/dm3
p1 = 2 bar
T1 = 27^{\circ}C = (27 + 273) K = 300 K
p2 = 1 bar
T2 = 273 K
d2 = ?
The density ( d2 ) of the gas at STP can be calculated using the equation,
d = \frac{Mp}{RT} \frac{d_{1}}{d_{2}} = \frac{\frac{M \;p_{ 1 }}{R \; T_{ 1 }}}{\frac{M \; p_{ 2 }}{R \; T_{ 2 }}} \frac{d_{1}}{d_{2}} = \frac{p_{1} \; T_{2}}{p_{2} \; T_{1}}
d2 = \frac{p_{2} \; T_{1} \; d_{1}}{p_{1} \; T_{2}}
= \frac{1 \; \times 300 \; \times 5.46}{2 \; \times 273}
= 3 g dm-3
Hence, the density of the gas at STP will be 3 g dm-3
Q 10. 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer: Given,
p = 0.1 bar
V = 34.05 mL = 34.05 × 10-3 dm^{ 3 }
R = 0.083 bar dm^{ 3 } at K-1 mol-1
T = 546^{\circ} C = (546 + 273) K = 819 K
The no of moles (n) can be calculated using the ideal gas equation as:
pV = nRT
n = \frac{ pV }{ RT }
= \frac{ 0.1 \times 34.05 × 10^{-3} }{ 0.083 \times 819 }
= 5.01 × 10-4 mol
Therefore, molar mass of phosphorus = \frac{ 0.0625 }{ 5.01 \times 10^{-5} }
= 125 g mol -1
Q 11. A student forgot to add the reaction mixture to the container at 27^{\circ} C but instead, he placed the container on the flame. After a lapse of time, he came to know about his mistake, and using a pyrometer he found the temp of the container 477^{\circ} C. What fraction of air would have been expelled out?
Answer: Let the volume of the container be V.
The volume of the air inside the container at 27^{\circ} C is V.
Now,
V1 = V
T1 = 27^{\circ} C = 300 K V2 = ?
T2 = 477^{\circ} C = 750 K
Acc to Charles’s law,
\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} V_{ 1 } = \frac{V_{ 1 }T_{ 2 }}{T_{ 1 }}
= \frac{750V }{300 }
= 2.5 V
Therefore, volume of air expelled out
= 2.5 V – V = 1.5 V
Hence, fraction of air expelled out
= \frac{1.5V }{ 2.5V }
= \frac{3 }{ 5 }
Q 12. Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm3 K–1 mol–1).
Answer: Given,
N= 4.0 mol
V = 5 dm^{ 3 }
p = 3.32 bar
R = 0.083 bar dm^{ 3 } at K-1 mol-1
The temp (T) can be calculated using the ideal gas equation as:
pV = nRT
T = \frac{ pV }{ nR }
= \frac{3.32 \; \times \; 5 }{ 4 \;\times \; 0.083 }
= 50 K
Therefore, the required temp is 50 K.
Q 13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Answer: Molar mass of dinitrogen (N2) = 28 g mol-1
Thus, 1.4 g of N2
= \frac{ 1.4 }{ 28 }
= 0.05 mol
= 0.05 × 6.02 × 1023 no. of molecules
= 3.01 × 1023 no. of molecules
Now, 1 molecule of N2 has 14 electrons.
Therefore, 3.01 × 1023 molecules of N2 contains,
= 14 × 3.01 × 1023
= 4.214 × 1023 electrons
Q 14. How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second?
Answer: Avogadro no. = 6.02 × 1023
Therefore, time taken
= \frac{6.02 \; \times \; 10^{23}}{10^{10}} s
= 6.02 × 1013 s
= \frac{6.02 \; \times \; 10^{23}}{60 \; \times \; 60 \; \times \; 24 \; \times \; 365 } years
= 1.909 × 106 years
Therefore, the time taken would be 1.909 × 106 years.
Q 15. Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1
Answer: Given:
Mass of O2 = 8 g
No. of moles
= \frac{ 8 }{ 32 }
= 0.25 mole
Mass of H2 = 4 g
No. of moles
= \frac{ 4 }{ 2 }
= 2 mole
Hence, total no. of moles in the mixture
= 0.25 + 2
= 2.25 mole
Given:
V = 1 dm^{ 3 }
n = 2.25 mol
R = 0.083 bar dm^{ 3 } at K-1 mol-1
T = 27^{\circ} C = 300 K
Total pressure :
pV = nRT
p = \frac{ nRT }{ V }
= \frac{225 \; \times \; 0.083 \; \times \; 300 }{ 1 }
= 56.025 bar
Therefore, the total pressure of the mixture is 56.025 bar.
Q 16. Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 and R = 0.083 bar dm3 K–1 mol–1)
Answer: Given:
r = 10 m
Therefore, volume of the balloon
= \frac{4}{3} πr3
= \frac{ 4 }{ 3 }\; \times \; \frac{ 22 }{ 7 } \; \times \; 10^{3}
= 4190.5 m3 (approx.)
Therefore, the volume of the displaced air
= 4190.5 × 1.2 kg
= 5028.6 kg
Mass of helium,
= \frac{ MpV }{ RT }
Where, M = 4 × 10-3 kg mol-1
p = 1.66 bar
V = volume of the balloon
= 4190.5 m3
R = 0.083 0.083 bar dm^{ 3 } at K-1 mol-1
T = 27 °C = 300 K
Then,
m = \frac{ 4 \; \times \; 10^{-3} \; \times \; 1.66 \; \times \; 4190.5 \; \times \; 10^{3}}{0.083 \; \times \; 300}
= 1117.5 kg (approx.)
Now, total mass with helium,
= (100 + 1117.5) kg
= 1217.5 kg
Therefore, pay load,
= (5028.6 – 1217.5)
= 3811.1 kg
Therefore, the pay load of the balloon is 3811.1 kg.
Q 17. Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar dm3 K–1 mol–1.
Answer: pVM = mRT
V = \frac{mRT}{Mp}
Given:
m = 8.8 g
R = 0.083 bar dm^{ 3 } at K-1 mol-1.
T = 31.1 °C = 304.1 K
M = 44 g
p = 1 bar
Thus, Volume (V),
= \frac{8.8 \; \times \; 0.083 \; \times \; 304.1}{ 44 \; \times \; 1}
= 5.04806 L
= 5.05 L
Therefore, the volume occupied is 5.05 L.
Q 18. 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?.
Answer: Volume,
V = \frac{mRT}{Mp}
= \frac{0.184 \; \times \; R \; \times \; 290}{ 2 \; \times \; p}
Let M be the molar mass of the unknown gas.
Volume occupied by the unknown gas is,
= \frac{mRT}{Mp}
= \frac{2.9 \; \times \; R \; \times \; 368}{ M \; \times \; p}
According to the ques,
\frac{0.184 \; \times \; R \; \times \; 290}{ 2 \; \times \; p} = \frac{2.9 \; \times \; R \; \times \; 368}{ M \; \times \; p} \frac{0.184 \; \times \; 290}{ 2 } = \frac{2.9 \; \times \; 368}{ M }
M = \frac{2.9 \;\times \; 368 \; \times \; 2}{0.184 \; \times \; 290}
= 40 g mol-1
Therefore, the molar mass of the gas is 40 g mol-1
Q 19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer: Let the weight of dihydrogen be 20 g.
Let the weight of dioxygen be 80 g.
No. of moles of dihydrogen (nH2),
= \frac{20}{2}
= 10 moles
No. of moles of dioxygen (nO2),
= \frac{80}{32}
= 2.5 moles
Given:
ptotal = 1 bar
Therefore, partial pressure of dihydrogen (pH2),
= \frac{ n_{H_{2}} }{ n_{H_{2}} \; + \; n_{O_{2}} } × ptotal
= \frac{ 10 }{ 10 \; + \; 2.5 } \; \times \; 1
= 0.8 bar
Therefore, the partial pressure of dihydrogen is 0.8 bar.
Q 20. What will be the SI unit for the quantity \frac{ pV^{ 2 }T^{ 2 } }{ n }?
Answer: SI unit of pressure, p = Nm^{ -2 }
SI unit of volume, V = m^{ 3 }
SI unit of temp, T = K
SI unit of number of moles, n = mol
Hence, SI unit of \frac{ pV^{ 2 }T^{ 2 } }{ n } is,
= \frac{(Nm^{ -2 }) \; (m^{3})^{2} \; (K)^{2}}{mol}
= Nm^{ 4 }K^{ 2 }mol^{ -1 }
Q 21. In terms of Charles’ law explain why -273^{\circ} C is the lowest possible temp.
Answer: Charles’ Law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.
Charles found that for all gases, at any given pressure, the graph of volume vs temperature (in Celsius) is a straight line and on extending to zero volume, each line intercepts the temperature axis at – 273.15°C.
We can see that the volume of the gas at – 273.15°C will be zero. This means that gas will not exist. In fact, all the gases get liquified before this temperature is reached.
Q 22. Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?
Answer: If the critical temperature of a gas is higher then it is easier to liquefy. That is the intermolecular forces of attraction among the molecules of gas are directly proportional to its critical temp.
Therefore, in CO2 intermolecular forces of attraction are stronger.
Q 23. Explain the physical significance of Van der Waals parameters?
Answer: After accounting for pressure and volume corrections, the van der Waals equation is
(p+\frac{an^{2}}{V^{2}}) (V-nb) = nRT
The van der Waals constants or parameters are a and b.
The relevance of a and b is crucial here:
The magnitude of intermolecular attractive forces within the gas is measured by the value of ‘a,’ which is independent of temperature and pressure.
The volume occupied by the molecule is represented by ‘b,’ while the total volume occupied by the molecules is represented by ‘nb.’