Chapter 5 States of matter Questions and Answers: NCERT Solutions for Class 11th Chemistry (Partnership Chemistry )

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Class 11 Chemistry (Partnership Chemistry ) Chapter 5: States of matter - Questions and Answers of NCERT Book Solutions.

1. What will be the minimum pressure required to compress 500 dm3dm3of air at 1 bar to 200 dm3dm3 at 30oC30oC?

Ans: Given,
Initial pressure,p1p1 = 1 bar
Initial volume, V1V1 = 500 dm3dm3
Final volume,V2V2 = 200 dm3dm3
Since the temperature remains constant, the final pressure (p2p2) can be calculated using Boyle’s law.
According to Boyle’s law,
p1V1=p2V2
p1V1=p2V2
⇒p2=p1V1V2
⇒p2=p1V1V2
=1×500200bar
=1×500200bar
= 2.5 bar
= 2.5 bar
Therefore, the minimum pressure required is 2.5 bar.

2. A vessel of 120 mL capacity contains a certain amount of gas at 35oC35oC and 1.2 bar pressure. The gas is transferred to an answer vessel of volume 180 mL at 35oC35oC. What would be its pressure?

Ans: Given,
Initial pressure,p1p1 = 1.2 bar
Initial volume,V1V1 = 120 mL
Final volume,V2V2 = 180 mL
Since the temperature remains constant, the final pressure (p2p2) can be calculated using Boyle’s law.
According to Boyle’s law,
p1V1=p2V2
p1V1=p2V2
⇒p2=p1V1V2
⇒p2=p1V1V2
=1.2×120180bar
=1.2×120180bar
= 0.8 bar
= 0.8 bar

3.Using the equation of state pV = nRT shows that at a given temperature density of a gas is proportional to gas pressure ep.

Ans: The equation of state is given by,
pV=nRT........(i)
pV=nRT........(i)
Where,
p = Pressure of gas
V = Volume of gas
n = Number of moles of gas
R = Gas constant
T = Temperature of gas
From equation (i) we have,
nV=pRT
nV=pRT
Replacing n withmMmM , we have
mMV=pRT......(ii)
mMV=pRT......(ii)
Where,
m = Mass of gas
M = Molar mass of gas
But, mV=dmV=d (d = density of gas)
Thus, from equation (ii), we have
dM=pRT
dM=pRT
⇒d=(MRT)p
⇒d=(MRT)p
Molar mass (M) of gas is always constant and therefore, at constant temperature (T),MRTMRT = constant,
d = (constant) p
⇒d∝p⇒d∝p
Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p).

4. At 0oC0oC , the density of 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?

Ans: Density (d) of substance at temperature (T) can be given by the expression,
d=MpRT
d=MpRT
Now, density of oxide (d1d1) is given by,
d1=M1p1RT
d1=M1p1RT
Where, M1M1 and p1p1 are the mass and pressure of the oxide respectively.
Density of dinitrogen gas (d2d2) is given by,
d2=M2p2RT
d2=M2p2RT
Where, M2M2 and p2p2 are the mass and pressure of the oxide respectively.
According to the given question,
d1=d2
d1=d2
Therefore,
M1p1=M2p2
M1p1=M2p2
Given,
p1p1 = 2 bar
p2p2 = 5 bar
Molecular mass of nitrogen, M2M2 = 28 g/mol
Now, M1=M2p2p1M1=M2p2p1
=28×52
=28×52
=70g/mol
=70g/mol
Hence, the molecular mass of the oxide is 70 g/mol.

5. Pressure of 1 g of an ideal gas A 27oC27oC 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.

Ans: For ideal gas A, the ideal gas equation is given by,
pAVA=nRT....(i)
pAVA=nRT....(i)
Where, pApA and VAVA represent the pressure and number of moles of gas A.
For ideal gas B, the ideal gas equation is given by,
pBVB=nRT....(ii)
pBVB=nRT....(ii)
Where, pBpB andVBVB represent the pressure and number of moles of gas B.
VandTareconstantsforgasesAandBVandTareconstantsforgasesAandB

From equation (i), we have
pAV=mAMART⇒pAMAmA=RTV......(iii)
pAV=mAMART⇒pAMAmA=RTV......(iii)
From equation (ii), we have
pBV=mBMBRT⇒pBMBmB=RTV......(iv)
pBV=mBMBRT⇒pBMBmB=RTV......(iv)
Where, MAMA and MBMB are the molecular masses of gases A and B respectively.
Now, from equations (iii) and (iv), we have
pAMAmA=pBMBmB.......(v)
pAMAmA=pBMBmB.......(v)
Given,
mAmA =1g
pApA =2bar
mBmB = 2g
pBpB = (3-2) = 1bar
(Since total pressure is 3 bar)
Substituting these values in equation (v), we have
2×MA1=1×MB2
2×MA1=1×MB2
⇒4MA=MB
⇒4MA=MB
Thus, a relationship between the molecular masses of A and B is given by 4MA=MB4MA=MB.

5. The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20oC20oC and one bar will be released when 0.15g of aluminum reacts?

Ans: The reaction of aluminum with caustic soda can be represented as:
2Al+2NaOH+2H2O→2NaAlO2+3H2
2Al+2NaOH+2H2O→2NaAlO2+3H2
At STP (273.15 K and 1 atm), 54 g (2 ×× 27 g) of Al gives 3×× 22400 mL of H2H2.
Therefore, 0.15 g Al gives3×22400×0.1554mL3×22400×0.1554mL ofH2H2 i.e., 186.67 mL of H2H2.
At STP,
p1p1 = 1atm
V1V1 = 186.67 mL
T1T1 = 273.15 K
Let the volume of dihydrogen be V2V2 at p2p2 = 0.987 atm (since 1 bar = 0.987 atm) and T2T2 = 20oC20oC = (273.15 + 20) K = 293.15 K.
p1V1T1=p2V2T2
p1V1T1=p2V2T2
V2=p1V1T2p2T1
V2=p1V1T2p2T1
=1×186.67×293.150.987×273.15
=1×186.67×293.150.987×273.15
=202.98mL
=202.98mL
=203mL=203mL
Therefore, 203 mL of dihydrogen will be released.

6. 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 dm3dm3 flask at 27oC27oC ?

Ans: It is known that,
p=mMRTVp=mMRTV
For methane (CH4CH4),
pCH4=3.216×8.314×3009×10−3
pCH4=3.216×8.314×3009×10−3
since 9dm3=9×10−3m3
9dm3=9×10−3m3
27oC=300K
27oC=300K
=5.543×104Pa
=5.543×104Pa
For carbon dioxide (CO2CO2),
pCO2=4.444×8.314×3009×10−3
pCO2=4.444×8.314×3009×10−3
=2.771×104Pa
=2.771×104Pa
The pressure exerted by the mixture can be obtained as:
p=pCH4+pCO2
p=pCH4+pCO2
=(5.543×104+2.771×104)Pa
=(5.543×104+2.771×104)Pa
=8.314×104Pa
=8.314×104Pa
Hence, the total pressure exerted by the mixture is =8.314×104Pa=8.314×104Pa.

7. What will be the pressure of the gaseous mixture when 0.5 L of H2H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27oC27oC ?

Ans: Let the partial pressure of H2H2in the vessel be pH2pH2.
Now,
p1p1 = 0.8bar
V1V1 = 0.5L
V2V2 = 1L
p2=pH2p2=pH2= ?
It is known that,
p1V1=p2V2
p1V1=p2V2
⇒p2=p1V1V2
⇒p2=p1V1V2
⇒pH2=0.8×0.51
⇒pH2=0.8×0.51
=0.4bar
=0.4bar
Now, let the partial pressure of O2O2in the vessel bepO2pO2.
p1p1 = 0.7bar
V1V1 = 2.0L
V2V2 = 1L
p2=pO2p2=pO2 = ?
Since,
p1V1=p2V2
p1V1=p2V2
⇒p2=p1V1V2
⇒p2=p1V1V2
⇒pO2=0.7×201
⇒pO2=0.7×201
=1.4bar
=1.4bar
Total pressure of the gas mixture in the vessel can be obtained as:
ptotal=pH2+pO2
ptotal=pH2+pO2
=0.4+1.4
=0.4+1.4
=1.8bar
=1.8bar

8. Density of a gas is found to be 5.46 g/dm3g/dm3 at 27oC27oC at 2 bar pressure. What will be its density at STP?

Ans: Given,
d1d1 = 5.46g/dm3g/dm3
p1p1 =2 bar
T1=27oCT1=27oC = (27+273)K=300K
p2=p2= 1bar
T2T2 =273K
d2d2 =?
The density (d2d2) of the gas at STP can be calculated using the equation,
d=MpRT
d=MpRT
∴d1d2Mp1RT1Mp2RT2
∴d1d2Mp1RT1Mp2RT2
⇒d1d2=p1T2p2T1
⇒d1d2=p1T2p2T1
⇒d2=p2T1d1p1T2
⇒d2=p2T1d1p1T2
=1×300×5.462×273
=1×300×5.462×273
=3gdm−3
=3gdm−3
Hence, the density of the gas at STP will be 3 gdm−3dm−3.

9. 34.05 mL of phosphorus vapour weighs 0.0625 g at 546oC546oC and 0.1 bar pressure. What is the molar mass of phosphorus?

Ans: Given,
p = 0.1 bar
V = 34.05 mL = 34.05×10−3L=34.05×10−3dm−334.05×10−3L=34.05×10−3dm−3
R =0.083 bar dm3K−1mol−1dm3K−1mol−1
T = 546oC546oC = (546 + 273) K = 819 K
The number of moles (n) can be calculated using the ideal gas equation as:
pV=nRT
pV=nRT
⇒n=pVRT
⇒n=pVRT
=0.1×34.05×10−30.083×819
=0.1×34.05×10−30.083×819
=5.01×10−5mol
=5.01×10−5mol

10. A student forgot to add the reaction mixture to the round bottomed flask at 27oC27oC but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477oC477oC . What fraction of air would have been expelled out?

Ans: Let the volume of the round bottomed flask be V.
Then, the volume of air inside the flask at 27oC27oCis V.
Now,
V1V1 = V
T1T1 = 27oC27oC = 300 K
V2V2 = ?
T2T2 = 477oC477oC = 750 K
According to Charles’s law,
V1T1=V2T2
V1T1=V2T2
⇒V2=V1T2T1
⇒V2=V1T2T1
=750V300
=750V300
=2.5V
=2.5V

11. Calculate the temperature of 4.0 mol of gas occupying 5 dm3dm3 at 3.32 bar. (R = 0.083 bardm3K-1mol-1dm3K-1mol-1).

Ans: Given,
n = 4.0 mol
V = 5 dm3dm3
p = 3.32 bar
R = 0.083 bar dm3K−1mol−1dm3K−1mol−1
The temperature (T) can be calculated using the ideal gas equation as:
pV=nRT
pV=nRT
⇒T=pVnR
⇒T=pVnR
=3.32×54×0.083
=3.32×54×0.083
=50K
=50K
Hence, the required temperature is 50 K.

12. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Ans: Molar mass of dinitrogen (N2N2) = 28 gmol−1gmol−1
Thus, 1.4 g of N2N2 1.4281.428 = 0.05 mol
= 0.05×6.02×1023×6.02×1023 number of molecules
= 3.01×1023×1023 number of molecules
Now, 1 molecule of N2N2contains 14 electrons.
Therefore, 3.01×1023×1023 molecules of N2N2contains = 14×3.01×102314×3.01×1023 = 4.214×10234.214×1023 electrons.

13. How much time would it take to distribute one Avogadro number of wheat grains, if 10101010 grains are distributed each second?

Ans: Avogadro number = 6.02×10236.02×1023
Thus, time required
=6.02×10231010s
=6.02×10231010s
=6.02×1023s
=6.02×1023s
=6.02×102360×60×24×365years
=6.02×102360×60×24×365years
=1.909×106years
=1.909×106years
Hence, the time taken would be=1.909×106years=1.909×106years.

14. Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3dm3 at 27oC27oC . (R = 0.083 bar dm3K-1mol-1dm3K-1mol-1 ).

Ans: Given, Mass of dioxygen (O2O2) = 8 g Thus, number of moles of O2O2 =832832 = 0.25 mole Mass of dihydrogen (H2H2) = 4 g H2=42H2=42 = 2mole Therefore, total number of moles in the mixture = 0.25 + 2 2.25 mole Given, V = 1 dm3dm3 n = 2.25 mol R = 0.083 bar dm3K−1mol−1dm3K−1mol−1 T = 27oC27oC = 300 K Total pressure (p) can be calculated as: pV=nRT pV=nRT ⇒p=nRTV ⇒p=nRTV =225×0.083×3001 =225×0.083×3001 =56.025bar =56.025bar Hence, the total pressure of the mixture is 56.025 bar.

15. Pay load 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 27oC27oC . (Density of air = 1.2kgm-3kgm-3 . And R = 0.083 bar dm3K-1mol-1dm3K-1mol-1 ).

Ans: Given,
Radius of the balloon, r = 10 m
Therefore, Volume of the balloon =43πr343πr3
=43×227×1023
=43×227×1023
=4190.5m3(approx)
=4190.5m3(approx)
Thus, the volume of the displaced air is 4190.5 m3m3.
Given,
Density of air = 1.2 kgm−3kgm−3
Then, mass of displaced air = 4190.5m3m3 ×× 1.2kgm−3kgm−3
= 5028.6 kg
Now, mass of helium (m) inside the balloon is given by,
m=MpVRT
m=MpVRT
Here,
M = 4×10−3kgmol−1×10−3kgmol−1
p = 1.66 bar
V = Volume of the balloon
= 4190.5 m3m3
R = 0.083 bar dm3K−1mol−1dm3K−1mol−1
T = 27oC27oC = 300 K
Then,
m=4×10−3×1.66×4190.5×1030.083×300
m=4×10−3×1.66×4190.5×1030.083×300
=1117.5kg(approx)
=1117.5kg(approx)
Now, total mass of the balloon filled with helium = (100 + 1117.5) kg
= 1217.5 kg
Hence, payload = (5028.6 – 1217.5) kg
=3811.1 kg
Hence, the payload of the balloon is 3811.1 kg.

16. Calculate the volume occupied by 8.8 g of CO2CO2 at 31.1oC31.1oC and 1 bar pressure. R = 0.083 bar LK-1mol-1K-1mol-1.

Ans: It is known that,
pV=mNRT
pV=mNRT
⇒V=mRTMp
⇒V=mRTMp
Here,
m = 8.8 g
R = 0.083 bar LK−1mol−1K−1mol−1
T = 31.1oC31.1oC = 304.1 K
M = 44 g
p = 1 bar
Thus, Volume (V) =8.8×0.083×304.144×1=8.8×0.083×304.144×1
= 5.04806 L
= 5.04806 L
= 5.05 L
= 5.05 L
Hence, the volume occupied is 5.05 L.

17. 2.9 g of gas at 95oC95oC occupied the same volume as 0.184 g of dihydrogen at 17oC17oC , at the same pressure. What is the molar mass of the gas?

Ans: Volume (V) occupied by dihydrogen is given by,
V=mMRTp
V=mMRTp
=0.1842×R×290p
=0.1842×R×290p
Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:
V=mMRTp
V=mMRTp
=2.9M×R×368p
=2.9M×R×368p
According to the equation,
0.1842×R×290p=2.9M×R×368p
0.1842×R×290p=2.9M×R×368p
⇒0.184×2902=2.9×368M
⇒0.184×2902=2.9×368M
M=2.9×368×20.184×290
M=2.9×368×20.184×290
=40gmol−1
=40gmol−1
Hence, the molar mass of the gas is40gmol−140gmol−1.

18. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Ans: Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.
Then, the number of moles of dihydrogen,nH2=202nH2=202 =10 moles and the number of moles of dioxygen,nO2=8032nO2=8032 =2.5 moles.
Given,
Total pressure of the mixture, PtotalPtotal = 1bar
Then, partial pressure of dihydrogen, pH2=nH2nH2+nO2×Ptotal
pH2=nH2nH2+nO2×Ptotal
=1010+2.5×1
=1010+2.5×1
=0.8bar
=0.8bar

19. What would be the SI units for the quantity pV2T2/npV2T2/n?

Ans: The SI units for pressure, p isNm−2Nm−2 . The SI unit for volume, V ism3m3. The SI unit for temperature, T is K. The SI unit for the number of moles, n is mol. Therefore, the SI unit for quantity pV2T2npV2T2n is given by, =(Nm−2)(m3)2(K)2mol =(Nm−2)(m3)2(K)2mol =Nm4K2mol−1 =Nm4K2mol−1

20. In terms of Charles’ law explain why -273oC-273oC is the lowest possible temperature.

Ans: Charles’s law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
(Image Will Be Updated Soon)
It was found that for all gases (at any given pressure), the plots of volume vs. temperature (inoCoC ) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at−273oC−273oC . In other words, the volume of any gas at 273oC273oC is zero. This is because all gases get liquefied before reaching a temperature of273oC273oC . Hence, it can be concluded that −273oC−273oCis the lowest possible temperature.

21. Critical temperature for carbon dioxide and methane are 31.1oC31.1oC and –81.9oC81.9oC respectively. Which of these has stronger intermolecular forces and why?

Ans: Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case ofCO2CO2 .

22. Explain the physical significance of Van der Waals parameters.

Ans: The Van der waals equation is an equation of state for a fluid composed of particles that have a non-zero volume and a pair wise attractive inter-particle force ( Van der waals force) The equation is :
(p+n2aV2)(V−nb)=nRT
(p+n2aV2)(V−nb)=nRT
Physical significance of ‘a’:
‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas.
Physical significance of ‘b’:
‘b’ is a measure of the volume of a gas molecule.
V is the total volume of the container containing the fluid.
Let us Quickly Give a Glance at the Properties of the Four States of Matter
Solid: In solids, the atoms, molecules, and ions are closely packed together and so the particles can vibrate at their own position but are not
free to move. The volume and the shape of a solid can change when an external force is applied to them.
Liquid: The atoms, ions, and molecules are incompressible in liquid. They do not depend on pressure. They have a fixed volume when the pressure and temperature are constant. When solid is exposed to very high temperatures they transform into the liquid state, subject to the properties of pressure.
Gas: The intermolecular force of attraction between the interacting particles has enough kinetic energy that changes the molecular forces to zero. The intermolecular space between the molecules is very large and so a liquid can be transformed into a gaseous state.
Plasma: Plasma is a state of matter in which the ions are highly electrically conducive. The magnetic fields and currents produced by them dominate the behaviour of the matter.
So we see that the behaviour of matter in different states of matter is governed by factors like temperature, pressure, mass, and volume. The chemical properties of a substance do not change with the change of state but chemical reactions depend on the physical state of the matter.

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Last Updated on: Mar 28, 2024