Gaseous States Test

Result

Gaseous States Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

The kinetic energy for 14 grams of nitrogen gas at 127⁰C is nearly (mol. mass of nitrogen = 28 and gas constant = 8.31 JK⁻¹mol⁻¹)

A
1230 J

B
4015 J

C
2493 J

D
990 J

Solution

KE = 3/2 nRT n = 14/28 = 0.5

= 3/2 × 0.5 × 8.314 × (127 + 273)

= 2494.2J
Question 2
1 / -0

How many moles of He gas occupy 22.4 litres at 30^oC and one atmospheric pressure?

A
0.90

B
1.11

C
0.11

D
1.0

Solution

We know for ideal condition

pV = nRT

n = pV/RT

o^oC = 273K

30^oC = 30 + 273 = 303K

n = 0.90 mole
Question 3
1 / -0

At N.T.P. the volume of a gas is found to be 273 ml. What will be the volume of this gas at 600 mm Hg and 273^∘C

A
391.8 ml

B
380 ml

C
691.6 ml

D
750 ml

Solution
Question 4
1 / -0

A 1:1 mixture (by weight) of hydrogen and helium is enclosed in a one-litre flask at 0^∘C. Assuming ideal behaviour, the partial pressure of helium is found to be 0.42 atm, then the concentration of hydrogen would be:

A
0.0375 M

B
0.028 M

C
0.0562 M

D
0.0187 M

Solution

Let the weights of H₂ and He in the mixture be x each.

Then the number of moles of H₂= x/2

and the number of moles of He = x/4
Question 5
1 / -0

3.2 g of oxygen (At. wt. = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12L flask at 0^∘C. The total pressure of the gas mixture will be

A
1 atm

B
4 atm

C
3 atm

D
2 atm

Solution
Question 6
1 / -0

A gaseous mixture contains 56 g of N₂, 44 g of CO₂ and 16 g of CH₄. total pressure of the mixture is 720 mm of Hg. The partial pressure of methane is

A
75 atm

B
100 atm

C
180 atm

D
215 atm

Solution
Question 7
1 / -0

The volume of 0.0168 mol of O₂ obtained by decomposition of KClO₃ and collected by displacement of water is 428mL at a pressure of 754 mm Hg at 25^∘C. The pressure of water vapour at 25^∘C is

A
18 mm Hg

B
20 mm Hg

C
22 mm Hg

D
24 mm Hg

Solution
Question 8
1 / -0

The density of a gas at 27^∘C and 1 atm is ′d′. Pressure remaining constant, the temperature at which its density becomes 0.75 d is:

A
20^∘C

B
30^∘C

C
400 K

D
300 K

Solution
Question 9
1 / -0

0.5 mole of each of H, SO₂ and CH₄, are kept in a container. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be:

A
P(CH₄) > P(SO₂) > P(H₂)

B
P(H₂) > P(CH₄) > P(SO₂)

C
P(SO₂) > P(CH₄) > P(H₂)

D
P(H₂) > P(SO₂) > P(CH₄)

Solution

SO₂ Molar mass= 32+32 = 64g/mol

CH₄ Molar mass= 12+4 = 16g/mol

H₂ Molar mass= 2×1 = 2g/mol

Rate of diffusion is inversely related to molecular weight. Lighter the gas, more is the diffusion. The rate of diffusion is given by,

^rH₂ > ^rCH₄ > ^rSO₂

Hence the order of partial pressure will be

^PSO₂ > ^PCH₄ > ^PH₂
Question 10
1 / -0

The number of moles of H₂ in 0.224L of hydrogen gas at STP (273K,1atm) assuming ideal gas behaviour is:

A
1

B
0.1

C
0.01

D
0.001

Solution

Number of moles