NCERT Solutions for Class 11 Chemistry States of Matter & Solid State [Old Book]

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Chemistry
Chapter : 5
Chapter Name : States of Matter

Q5.1 What will be the minimum pressure required to compress 500 dm of air at 1 bar to 200
3

dm at 30°C?
3

Answer. Given,
Initial pressure, P = 1 bar1

Initial volume, V = 500 dm 3
1

Final volume, V = 200 dm 3
2

Since the temperature remains constant, the nal pressure (P ) can be calculated using Boyle's
2

law.
According to Boyle's law,
p1 V1 = p2 V2

p1 V1
⇒ p2 =
V2

1 × 500
= bar
200

= 2.5bar

Therefore, the minimum pressure required is 2.5 bar.

Q5.2 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. Given,
Initial pressure, P = 1.2 bar
1

Initial volume, V = 120 mL
1

Final volume, V = 180 mL Since the temperature remains constant, the nal pressure (p ) can
2 2

be calculated using Boyle's law
According to Boyle's law,
p1 V1 = p2 V2

p1 V1
p2 =
V2

1.2 × 120
= bar
180

= 0.80

Therefore, the pressure would be 0.8 bar.

Q5.3 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,
ρV = nRT … … …

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,
n p
=
V RT

m
Replacing n with , we have
M

m p
= … … … ( ii )
MV RT

Where,

m → Mass of gas

M → Molar mass of gas
m
= d
V

Thus, from equation (ii), we have

d p
=
M RT

M
⇒ d = ( )P
RT

Molar mass (M ) of a gas is always constant and therefore, at constant temperature

M
(T ), = constant.
RT

d = ( constant )p

⇒ d ∝ p

Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p)

Q5.4 At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar.
What is the molecular mass of the oxide?

Answer.
Density (d) of the substance at temperature (T) can be given by the expression,

Mp
d =
RT

Now, density of oxide (d1 ) is given by,

M1 p1
d1 =
RT

Where, M2 and p1 are the mass and pressure of the oxide respectively.

Density of dinitrogen gas (d2 ) ls given by,

M2 p2
d2 =
RT

Where, M2 and p2 are the mass and pressure of the oxide respectively.

According to the given question,

d1 = d2

∴ M1 p1 = M2 p2

Given,

p1 = 2bar

p2 = 5 bar

Molecular mass of nitrogen, M2 = 28g/mol

M2 p2
Now, M1 =
p1

28 × 5
=
2

= 70g/mol

Hence, the molecular mass of the oxlde is 70g/mol

Q5.5 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 ask at same temperature the pressure becomes 3 bar. Find a
relationship between their molecular masses.

For ideal gas A, the ideal gas equation is given by,
Answer.
pA V = nA RT … … (i)

Where, p and n represent the pressure and number of moles of gas A.
A B

For ideal gas B, the ideal gas equation is given by,
pB V = nB RT … … .

Where, p and n, represent the pressure and number of moles of gas B.
[ V and T are constants for gases A and B ]
From equation (i), we have
mA pA MA RT
pA V = RT ⇒ = ………
MA mA V

From equation (ii), we have
mB pB MB RT
pB V = RT ⇒ = ………
MB mB V

Where, M and M are the molecular masses of gases A and B respectively.
A B

Now, from equations (iii) and (iv) we have
pA MA pg MB
= … … … (v)
mA mB

Given,

mA = 1g

pA = 2 bar

mB = 2g

pB = (3 − 2) = 1bar

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

2×MA 1×MB
=
1 2

⇒ 4MA = MB

Thus, a relationship between the molecular masses of A and B is given by

4MA = MB

Q5.6 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?

The reaction of aluminium with caustic soda can be represented as:
Answer.
2Al + 2NaOH + 2H2 O ⟶ 2NaAlO2 + 3H2

2 × 27g

STP(273.15K and 1atm), 54g(2 × 27g) of Al gives 3 × 22400mL of H2..
3×22400×0.15
∴ 0.15 g Al gives mL of H2
54

At STPr

p1 = 1atm

V1 = 186.67mL

T1 = 273.15K

Let the volume of dihydrogen be V2 at ρ2 = 0.987 atm (since 1 bar = 0.987 atm ) and T2
∘
= 20 C = (273.15 + 20)K = 293.15K

Now,

p1 V1 p2 V2
=
T1 T2

p1 V1 T2
⇒ V2 =
p2 T1

1 × 186.67 × 293.15
=
0.987 × 273.15

= 203mL

= 203mL

Therefore, 203 mL of dihydrogen will be released.

Q5.7 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 ask at 27 °C ?

It is known that,

m RT
p =
M V

Answer. For methane (CH4 )
t

3.2 8.314×300 3 −3 3
pCH = × −3
[ since 9dm = 9 × 10 m ]
4 16 9×10

4
= 5.543 × 10 Pa

For carbon dioxide (CO2 )

4.4 8.314 × 300
pCC = ×
2 −3
44 9 × 10
4
= 2.771 × 10 Pa

Total pressure exerted by the mixture can be obtained as :
p = pCH + pCO
4 2

4 4
= (5.543 × 10 + 2.771 × 10 ) Pa

4
= 8.314 × 10 Pa

Hence, the total pressure exerted by the mixture is 8.314 x 10 Pa 4

Q5.8 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 vessel be PH2 .
Now,
p1 = 0.8bar p2 = pH1 =?

V1 = 0.5L V2 = 1L

It is known that,

p1 V1 = p2 V2

ρ 1 V1
⇒ p2 =
V2

0.8 × 0.5
⇒ pM2 =
1

= 0.4bar

Now, let the partial pressure of O In the vessel be P O .2 2

Now,
p1 = 0.7 bar p2 = p02 =?

V1 = 2.0L V2 = 1L

p1 V1 = p2 V2

p1 V1
⇒ p2 =
V2

0.7×20
⇒ p02 =
1

= 0.4bar

Total pressure of the gas mixture in the vessel can be obtained as:
p total = pH + po
2

= 0.4 + 1.4

= 1.8bar

Hence, the total pressure of the gaseous mixture in the vessel is 1.8 bar.

Q5.9 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,
3
d1 = 5.46g/dm

p1 = 2bar

∘
T1 = 27 C = (27 + 273)K = 300K

p2 = 1bar

T2 = 273K

d2 =?

The density d of the gas at STP can be calculated using the equation,
2
Mp
d =
RT

Mp
1

d1 RT
1
∴ =
d2 M P2

RT2
d1 p1 T2
⇒ =
d2 p2 T1

p2 T1 d1
⇒ d2 =
p1 T2

1 × 300 × 5.46
=
2 × 273
−3
= 3gdm

Hence, the density of the gas at STP will be 3 g dm −3
.

Q5.10 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

−3 −3 3
V = 34.05mL = 34.05 × 10 L = 34.05 × 10 dm

3 −1 −1
R = 0.083bardm K mol
∘
T = 546 C = (546 + 273)K = 819K

The number of moles (n) can be calculated using the ideal gas equation as:
pV = nRT

pV
⇒ n =
RT
−3
0.1 × 3.05 × 10
=
0.083 × 819
−5
=5.01 × 10 mol

0.0625
−1
Therefore, molar mass of phosphorus = = 1247.5gmol
−5
5.01 × 10

Hence, the molar mass of phosphorus is 1247.5 g mol −1

Q5.11 A student forgot to add the reaction mixture to the round bottomed ask at 27 °C but
instead he/she placed the ask on the ame. After a lapse of time, he realized his mistake, and
using a pyrometer he found the temperature of the ask was 477 °C. What fraction of air would
have been expelled out?

Answer. Let the volume of the round bottomed ask be V.
Then, the volume of air inside the ask at 27 C is V.
Now,
V1 = V
∘
T1 = 27 C = 300K

V2 =?
∘
T2 = 477 C = 750K

According to Charles's law,

V1 V2
=
T1 T2

V 1 T1
⇒ V2 =
T1

750V
=
300

= 2.5V

Therefore, volume of air expelled out = 2.5 v - v = 4.5 v.
Hence, fraction of air expelled out = .
1.5V 3
=
2.5V 5

Q5.12 Calculate the temperature of 4.0 mol of a gas occupying 5 dm at 3.32 bar. (R = 0.083 bar 3

)).
3 −1 −1
dm K mol

Answer. Given,
n = 4.0mol

3
V = 5dm

p = 3.32bar

3 −1 −1
R = 0.083bardm K mol

The temperature (T) can be calculated using the ideal gas equation as:
pV = nRT

pV
⇒ T =
nR

3.32 × 5
=
4 × 0.083

= 50K

Hence, the required temperature is 50 K.

Q5.13 Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Answer. Molar mass of dinitrogen (N ) = 28gmol
−1
2

1.4
N2 = = 0.05mol
28

23
= 0.05 × 6.02 × 10 number of molecules

23
= 3.01 × 10 number of molecules

Now, 1 molecule of N contains 14 electrons.
2
23
Therefore, 3.01 × 10 molecules of N2 contains = 14 × 3.01 × 1023

23
= 4.214 × 10 electrons

Q5.14 How much time would it take to distribute one Avogadro number of wheat grains, if 1010
grains are distributed each second ?

23
Avogadro number = 6.02 × 10
Answer.
Thus, time required
23
6.02 × 10
s
16
10
25
=6.02 × 10 s
23
6.02 × 10
= years
23
60 × 60 × 10
6
= 1.909 × 10 years

Hence, the time taken would be 1.909 × 10 years.
6

Q5.15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen con ned
in a vessel of 1 dm at 27°C. R = 0.083 bar dm K mol .
3 3 –1 –1

Answer. Given,
Mass of dioxygen (O2 ) = 8g

8
Thus, number of moles of O2 = = 0.25 mole
32

Mass of dihydrogen (H2 ) = 4g

Then, mass of displaced air = 4190.5 × 1.2kg

= 5028.6kg

Now, mass of helium (m) inside the balloon is given by,

M pV
m =
RT

Here

−3 −1
M = 4 × 10 kgmol

p = 1.66bar

V = Volume of the balloon

3
= 4190.5m
3 −1 −1
R = 0.083bardm K mol
∘
T = 27 C = 300K
3 3
4 × 10 × 1.66 × 4190.5 × 10
Then, m =
0.083 × 300

= 1117.5kg( approx )

Now, total mass of the balloon filled with helium = (100 + 1117.5)kg

= 1217.5kg

Hence, pay load = (5028.6 − 1217.5)kg

= 3811.1kg .
Hence, the pay load of the balloon is 3811.1kg .

Q5.16 Pay load is de ned as the difference between the mass of displaced air and the mass of the
balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is lled with helium
at 1.66 bar at 27°C. (Density of air = 1.2 kg m and R = 0.083 bar dm K mol )
−3 3 −1 −1

Given,

Answer. Radius of the balloon, r = 10m

4 3
∴ Volume of the balloon = πr
3
4 22 3
= × × 10
3 7

3
= 4190.5m ( approx )

3
Thus, the volume of the displaced air is 4190.5m .

Given,

−3
Density of air = 1.2kgm
∘
T = 31.1 C = 304.1K

M = 44g

P = 1bar

8.8 × 0.083 × 304.1
Thus, volume (V ) =
44 × 1

= 5.04806L

= 5.05L

Hence, the volume occupied is 5.05L

Q5.17 Calculate the volume occupied by 8.8 g of CO at 31.1°C and 1 bar pressure. R = 0.083 bar
2

L K mol .
−1 −1

It is known that,
m

Answer. pV = M
RT

mRT
⇒ V =
Mp

Here,

m = 8.8g

−1 −1
R = 0.083barLK mol
4
H2 = =2mole
Thus, number of moles of 2

Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25mole

Given,

3
V = 1dm

n = 2.25mol

3 −1 −1
R = 0.083bardm K mol
∘
T = 27 C = 300K

Total pressure(p) can be calculated as:
P V =nRT

nRT
⇒ p =
V

225 × 0.083 × 300
=
1

= 56.025bar

Hence, the total pressure of the mixture is 56.025 bar.

Q5.18 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?

Volume (V) occupied by dihydrogen is given by,

m RT
Answer. V = M p

0.184 R×290
= ×
2 p

Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be
calculated as:
m RT
V =
M p

2.9 R × 368
= ×
M p

According to the question,
0.184 R×290 2.9 R×368
× = ×
2 p M p

0.184×290 2.9×368
⇒ =
2 M

2.9×368×2
⇒ M =
0.184×290

−1
= 40gmol

Hence, the molar mass of the gas is 40 g mol −1

Q5.19 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 number of moles of dihydrogen be 20 g and the weight of dioxygen be 80 g.
Then, the number of moles of dihydrogen, n = = 10moles and the number of moles of
Hz
20

2

dioxygen, n 02 =
80

32
= 2.5moles .
Given,

Total pressure of the mixture, p tead = 1 bar

Then, partial pressure of dihydrogen,
nH2
pH2 = × Pwul
nH + nO
2 2

10
= × 1
10 + 2.5

= 0.8bar

Hence, the partial pressure of dihydrogen is 0.8 bar.

Q5.20 What would be the SI unit for the quantity pV 2
T
2
/n?

Answer. The SI unit for pressure, p is Nm .
−2

The SI unit for volume, V is m 3

The SI unit for temperature, T is K.

The SI unit for the number of moles, n is mol.
2 2

Therefore, the SI unit of quantity is given by,
pV T

n

−2 2
3 2
(Nm ) (m ) (K)

=
mol
4 2 −1
= Nm K mol

Page : 159 , Block Name : Exercise

Q5.21 In terms of Charles’ law explain why –273 °C is the lowest possible temperature

Answer. Charles' law states that at constant pressure, the volume of xed mass of gas is directly
proportional to its absolute temperature.

It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in C∘

) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis
at -273 C. In other words, the volume of any gas at - 273 Cis zero, This is because all gases get
∘ ∘

lique ed before reaching temperature of - 273 C. Hence, it can be concluded that - 273 C is the
∘ ∘

lowest possible temperature.

Page : 159 , Block Name : Exercise

Q5.22 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. 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 of CO . 2

Page : 159 , Block Name : Exercise

Q5.23 Explain the physical signi cance of van der Waals parameters.

Answer. Physical signi cance of 'a':
‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas. Physical
signi cance Of 'b':

‘b’ is a measure of the volume of gas molecule.

Page : 159 , Block Name : Exercise