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States of Matter: Gases and Liquids

NEET Chemistry · 61 questions · Page 3 of 7 · Click an option or "Show Solution" to reveal answer

Q21
Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be
A 50.00 u
B 12.25 u
C 6.50 u
D 25.00 u
Correct Answer
Option B
Solution

We know that

rArB=v/tAv/tB=MBMA{{{r_A}} \over {{r_B}}} = {{v/{t_A}} \over {v/{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}}
⇒tAtB=MBMA⇒1020=MB49\Rightarrow {{{t_A}} \over {{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}} \Rightarrow {{10} \over {20}} = \sqrt {{{{M_B}} \over {49}}}
⇒(1020)2=MB49⇒100400= MB49\Rightarrow {\left( {{{10} \over {20}}} \right)^2} = {{{M_B}} \over {49}} \Rightarrow {{100} \over {400}} = \,{{{M_B}} \over {49}}
⇒MB=49×100400=12.25\Rightarrow {M_B} = {{49 \times 100} \over {400}} = 12.25

u

Q22
By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled ?
A 2.0
B 2.8
C 4.0
D 1.4
Correct Answer
Option D
Solution

Average velocity (v AV ) =

8RTπM\sqrt {{{8RT} \over {\pi M}}}

⇒\Rightarrow v AV ∝\propto

T\sqrt T

∴\therefore

(vAV)2(vAV)1=2TT{{{{\left( {{v_{AV}}} \right)}_2}} \over {{{\left( {{v_{AV}}} \right)}_1}}} = \sqrt {{{2T} \over T}}

= 1.4

Q23
A gaseous mixture was prepared by taking equal mole of CO and N 2 . If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N 2 ) in the mixture is
A 0.5 atm
B 0.8 atm
C 0.9 atm
D 1 atm
Correct Answer
Option A
Solution

Number of moles ⇒\Rightarrow n CO = n N 2 Volume of container is same.

So, V CO = V N 2 .

Also, temperature is same for both the gases thus, T CO = T N 2 According to ideal gas equation, PV = nRT Now, V, n, R and T for both gases are same.

So, P CO = P N 2 Now, total pressure is 1 atm and according to Dalton’s law of partial pressure, P CO + P N 2 = 1 atm ⇒\Rightarrow 2P N 2 = 1 atm {∵\because P CO = P N 2 } ⇒\Rightarrow P N 2 = 0.5 atm

Q24
The pressure exerted by 6.0 g of methane gas in a 0.03 m 3 vessel at 129 o C is (Atomic masses : C = 12.01, H = 1.01 and R = 8.314 J K −-1 mol −-1 )
A 215216 Pa
B 13409 Pa
C 41648 Pa
D 31684 Pa
Correct Answer
Option C
Solution

PV = nRT ⇒\Rightarrow

P=nRTVP = {{nRT} \over V}

=

wmRTV{w \over m}{{RT} \over V}

=

616.05×8.314×4020.03{6 \over {16.05}} \times {{8.314 \times 402} \over {0.03}}

= 41648 Pa

Q25
The energy absorbed by each molecule (A 2 ) of a substance is 4.4 × \times 10 −-19 J and bond energy per molecule is 4.0 × \times 10 −-19 J. The kinetic energy of the molecule per atom will be
A 2.2 × \times 10 −-19 J
B 2.0 × \times 10 −-19 J
C 4.0 × \times 10 −-20 J
D 2.0 × \times 10 −-20 J
Correct Answer
Option D
Solution

Energy absorbed by each molecule = Bond energy per molecule + Kinetic energy per molecule ⇒\Rightarrow 4.4 × 10 –19 J = 4.0 × 10 –19 J + Kinetic energy per molecule ⇒\Rightarrow 0.4 × 10 –19 = Kinetic energy per molecule Kinetic energy per atom = Kinetic energy per molecule 2 = 0.4 × 10 –19 2 = 0.2 × 10 –19 J = 2 × 10 –20 J

Q26
Volume occupied by one molecule of water (density = 1 g cm −-3 ) is
A 3.0 × \times 10 −-23 cm 3
B 5.5 × \times 10 −-23 cm 3
C 9.0 × \times 10 −-23 cm 3
D 6.023 × \times 10 −-23 cm 3
Correct Answer
Option A
Solution

1 mole of water contains 6.023 × 10 23 molecules of water 6.023 × 10 23 molecules of water weigh = 18 g So, 1 molecule of water weighs =

186.023×1023{{18} \over {6.023 \times {{10}^{23}}}}

Now, volume of 1 molecule of water = Mass of 1 molecule of water Density of water . =

186.023×1023×11 g cm−3{{18} \over {6.023 \times {{10}^{23}}}} \times {1 \over {1\,g\,c{m^{ - 3}}}}

= 3 ×\times 10 -23 cm 3

Q27
If a gas expands at constant temperature, it indicates that
A kinetic energy of molecules remains the same
B number of the molecules of gas increases
C kinetic energy of molecules decreases
D pressure of the gas increases.
Correct Answer
Option A
Solution

Kinetic energy of a gas is expressed as K.E =

32nRT{3 \over 2}nRT

Thus, on expansion of fixed amount of gas at constant temperature the kinetic energy remains constant.

Q28
In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture underthe aforesaid condition in the end ?
A 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
B 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
C 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
D 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
Correct Answer
Option B
Solution

Balanced chemical equation for Haber's process is as follows : 3H 2 + N 2 →\to 2NH 3 It is given that only 50% of the expected product is formed hence only 10 litre of NH 3 is formed.

Therefore, composition of gaseous mixture at the end is as follows : N 2 used = 5 litres N 2 left = 30 L – 5 L = 25 L H 2 used = 15 litres, H 2 legt = 30 L – 15 L = 15 L NH 3 = 10 L

Q29
Van der Waal's real gas, acts as an ideal gas, at which conditions?
A High temperature, low pressure
B Low temperature, high pressure
C High temperature, high pressure
D Low temperature, low pressure
Correct Answer
Option A
Solution

At higher temperature and low pressure real gas acts as an ideal gas.

Q30
The beans are cooked earlier in pressure cooker because
A boiling point increases with increasing pressure
B boiling point decreases with increasing pressure
C extra pressure of pressure cooker softens the beans
D internal energy is not lost while coocking in pressure cooker.
Correct Answer
Option B
Solution

When water pressure increases in the pressure cooker, water boils at lower temperature and the beans in pressure cooker are cooked earlier.

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