States of Matter : Gases & Liquids Test

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States of Matter : Gases & Liquids Test - 2

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Question 1
4 / -1

Direction (Q. Nos. 1-12) This section contains 12 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q.
The vapour pressure of liquid water is 23.8 torr at 25°C. By what factor does the molar volume of water increases as it vaporises to form an ideal gas under these conditions? The density of liquid water is 0.997 g cm^-3 at 298 K ?

A
4.32 x 10⁴

B
2.31 x 10^-5

C
4.32 x 10⁶

D
2.31 x 10^-7
Question 2
4 / -1

A gas at a pressure of 5.0 bar is heated from 0°C to 546°C and is simultaneously compressed to one-third of its original volume. Hence, final pressure is

A
5/9 bar

B
15.0bar

C
30.0 bar

D
45.0 bar

Solution

P₁V₁/T₁ = P₂V₂/T₂
Let the original volume be 3 volume units . The final volume will be 1/3 of 3 which is 1 volume unit
Temperature must be in K - 0 DEG C = 273K and 546 DEG C = 819 K
We want to solve for P1
P₁*1/819 = 5*3/273
P₁ = 819*5*3/273
P₁ = 45
Question 3
4 / -1

A 8.3 L cylinder contains 128 . 0 g O₂ gas at 27°C. What mass of O₂ gas must be released to reduce the pressure to 6.0 bar ?

A
64.0 g

B
32.0 g

C
16.0 g

D
8.0 g

Solution

By applying ideal gas eqn. PV = nRT
Further solving it we get an eq.
w = PVM/RT
Putting values, we get 64g.
Hence, the correct answer is Option A.
Question 4
4 / -1

N₂ gas under the given state I is to be changed to state II. Thus, N₂ escaped from I is

A
16.67%

B
25.00%

C
75.00%

D
83.33%

Solution

Applying pV = nRT in state 1, we have n₁ = 0.110
Applying pV = nRT in state 1, we have n₂ = 0.0184
Mole fraction(this fraction is present) = 0.0184/0.110 = 0.16727
So, fraction escaped = (1-0.16727)*100 = 83.33%
Question 5
4 / -1

1 g H₂ gas and x g O₂ gas exert a total pressure of 5 atm. At a given temperature partial pressure of O₂ gas is 4 atm. This O₂ in the mixture is

A
32g

B
64 g

C
96 g

D
125 g
Question 6
4 / -1

The average molar mass of the vapour above solid NH₄CI is nearly 26.75 g mol^-1.
Thus, vapour consists of (mole fraction)

A
a

B
b

C
c

D
d

Solution

The correct answer is Option C.
The average molar mass of the vapour above solid NH₄Cl = 26.5 mol^-1
Evaporation product of NH4Cl --->
NH₄Cl (s) ----> NH₃(g) + HCl(g)
Let Mole fraction of NH₃ = X_L
Mole fraction of HCl = X₂
∴ Average molar mass of NH₃ = 14 + 31 = 17
=> 26.5 = m₁x₁ + m2x2
=> 26.5 = 17x1 + 36.5x₂ ---(i)
And we know;
x₁ + x₂ = 1
From (i) and (ii)
x1 = 0.5 and x₂ = 0.5
Question 7
4 / -1

Pressure of 1 g of an ideal gas A at 300 K is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature, pressure becomes 3 bar. Thus,

A
m_A = m_B

B
m_A = 3m_B

C
m_A = 4m_B

D
m_B = 4 m_A

Solution

Given pressure(P) of gas A= 2bar Total pressure = 3bar then pressure(P) of gas B = 3-2 =1bar
no. of mole (n) = given mass(m) / molar mass(M)
Gas A
PV = nRT
2×V =( mₐ/Mₐ) RT ------------- (1)
Gas B
PV = nRT
1V = (mb/Mb)RT ---------------- (2)
On dividing eqn 1 by eqn 2
2V/1V = (1/Mₐ) ÷(2/Mb)
2/1 = (Mb/Mₐ) × 1/2
(M_B/Mₐ) = (2/1)×(2/1)
M_B/Mₐ = 4/1
M_B = 4Mₐ
Question 8
4 / -1

Total pressure of a mixture of H₂ and O₂ is 1.00 bar. The mixture is allowed to react to form water, which is completely removed to leave only pure l-l₂ at a pressure of 0.35 bar. Assuming ideal gas behaviour and that all pressure measurements were made under the same temperature and volume conditions, the composition of the original mixture is

A
a

B
b

C
c

D
d

Solution

Correct answer is option B
Avagadro′s Law says:
Equal volume of two gases at same pressure and temperature conditions have equal number of molecules.
∴1 O₂ molecule consumes 2 H₂ molecules.
∴Since after the reaction only H₂ is left, we know all of O₂ is consumed.
∴This implies what all consumed 1/3 of it was O₂.
Hence, O₂ in the mix was at a partial pressure of (1 - 0.35)* 1/3 = 0.22
And rest was H₂ = (1 - 0.22) = 0.78
Hence oxygen- hydrogen composition in the original mix was:0.22 bar O2 and 0.78 bar H₂.
Question 9
4 / -1

The composition of air inhaled by a human is 21 % by volume O₂ and 0.03% CO₂ and that of exhaled air is 16% O₂ and 4.03% CO₂. Assuming a typical volume of 6200 L day ^-1, moles of O₂ used and CO₂ generated by the body at 37°C and 1.00 bar are (assume ideal behaviour)

A
a

B
b

C
c

D
d
Question 10
4 / -1

The following arrangement of flasks is set up. Assuming no temperature change, determine the final pressure inside the system after all stopcocks are opened. The connecting tube has zero volume.

A
1.62 atm

B
4.53 atm

C
1.51 atm

D
3.24 atm

Solution

Applying p1V1 + p2V2 + p3V3 = pnet × Vnet
(0.792)×4 + (1.23)×3 + (2.51)×5 = pnet × 12
On solving, we get p = 1.617 = 1.62 atm
Question 11
4 / -1

1.00 mole sample of O₂ and 3.00 moles sample of H₂ are mixed isothermally in a 125.3 L container at 125°C. Gaseous mixture undergoes reaction to form water vapours. Assume ideal gas behaviour total pressure before and after H₂O (g) is formed is are respectivel

A
1.0545 bar, 0.7909 bar

B
1.056 bar, 1.056 ba

C
0.792 bar, 1.056 bar

D
0.264 bar, 0.528 bar

Solution

Correct option is not provided.
2H2 + O2 ------> 2H2O
3    1     0
(-) 2    1     2 mole H₂O formed
___________________________
1 0     Total mole = 1+2 = 3
PV = nRT / 100
n = PV / RT
P = nRT/ V
= [4 * 0.083 * (125 + 273)] / 125.3
Pressure = 1.0545 bar after water formed is 2 mole.
P = [3 * 0.083 * 398] /125.3
= 0.7909 bar
= 0.791 bar (appx)
Hence, the answer is 1.0545 bar, 0.7909 bar.
Question 12
4 / -1

When NO₂ is cooled to room temperature, some of it reacts to form a dimer N₂O₄ through the reaction,
In a reaction, 15.2 g of NO₂ is placed in a 10.0 L flask at high temperature and the flask is cooled to 25°C. The total pressure is measured to be 0.500 atm. Thus, partial pressure of NO₂ is found to be

A
0.19 atm

B
0.31 atm

C
0.10 atm

D
0.40 atm

Solution

n_NO2 + 2n_N2O4= 0.330 mol
Subtracting n_NO2+ n_N2O4 = 0.204 mol
From n_NO2+ 2n_N2O4 = 0.330 mol
n_NO2 = 0.126 mol
n_N2O4 = 0.078 mol
From these results, NO₂ has a mole fraction of 0.62and a partial pressure of 0.62(0.50 atm) = 0.31 atm;N2O4has a mole fraction of 0.38 and a partial pressure of 0.38(0.50 atm) = 0.19 atm
Question 13
4 / -1

A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.
During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)

A
9

B
11

C
8

D
12

Solution

We know that
pV = nRT
A/Q , p and T are constant;
So n₁/n₂ = V₁/V₂
(¼)/(0.25/4) = (10)3/r3
r3 = 750
r ∼ 9
Question 14
4 / -1

What will be the temperature difference needed in a hot air balloon to lift 1.00 kg of mass ? Assume that the volume of the balloon is 100 m³, the temperature of the ambient air is 298 K, the pressure is 1.00 bar, and the air is an ideal gas with average molar mass of 29 g mol^-1.

A
5

B
4

C
8

D
2

Solution

The correct answer is 2
The solution is as follows:
Δm = m_cold − m_hot
=[n_cold−n_hot]M
= PVMR[1T1− 1T2]
Solving we get:
T2−T1 = ΔT = 2
Question 15
4 / -1

Iron forms a series of compounds of the type Fe_x (CO)_y. In air, they are oxidised to Fe₂O₃ and CO₂ gas. After heating a 0.142 g sample in air, you isolate the CO₂ in a 1.50 L flask at 25°C. The pressure of the gas is 44.9 mm Fig. What will be the value of y/x in the simplest empirical formula of Fe_x(CO)_y? [Atomic mass of Fe = 56]

A
15

B
12

C
4

D
5

Solution

The empirical formula is = Fe(CO)₂
Carbon dioxide obtained:
Pressure = 44.9 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 44.9 / 760 = 0.0591 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25 + 273.15) K = 298.15 K
V = 1.50 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value
= 0.0821 L.atm/K.mol
Applying the equation as:
0.0591 atm × 1.50 L
= n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 0.0036 moles
1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,
Moles of C = 0.0036 moles
Molar mass of C atom
= 12.0107 g/mol
Mass of C in molecule
= 0.0036 x 12.0107 = 0.0432 g
Given that the compound only contains iron and carbon. So,
Mass of Fe in the sample
= Total mass - Mass of C
Mass of the sample = 0.142 g
Mass of Fe in sample
= 0.142 g - 0.0432 g = 0.0988 g
Molar mass of Fe = 55.845 g/mol
Moles of Fe
= 0.0988 / 55.845 = 0.0018 moles
Taking the simplest ratio for Fe and C as:
0.0018 : 0.0036
= 1 : 2
Question 16
4 / -1

To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mole of Fie and 0.1 mole of an unknown compound (vapour pressure 0.68 atm at 0°C) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases 0°C is close to...

[IITJEE 2011]

A
7

B
13

C
14

D
5

Solution

Let nx mole of unknown compound be present in vapour form so its partial pressure is
0.68 = (nx/nx+0.1)×1 atm
nx+0.1/nx = 1/0.68 = 1.47
0.1/nx = 0.47
nx = 0.212
Applying ideal gas equation,
1×V = (0.212/0.1)×0.0821×273
V = 7