Gaseous State Test

Result

Gaseous State Test - 17

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Weekly Quiz Competition

Question 1
4 / -1

Two grams of hydrogen diffuse from a container in 10 minutes. How many grams of oxygen would diffuse through the same container in the same time under similar conditions?

A
0.5 g

B
4 g

C
6 g

D
8 g

Solution
Question 2
4 / -1

Volume of the gases produced at STP from 48.0g of carbon:

A
179.2 L

B
89.6 L

C
44.8 L

D
22.4 L

Solution

C + H₂O → CO_(g) + H_2(g)

12gm → 1mol + 1mol

12 gm C produces 2mole of gases (1mole CO & 1 mole of H₂ )

∴ 48 gm C may produce 48/12 × 2 = 4×2 = 8 mole

= 22.4 × 8 L gases = 179.2 l gas.
Question 3
4 / -1

Volume of 4.4 g of CO₂ at NTP is:

A
22.4 L

B
44.8 L

C
2.24 L

D
4.48 L

Solution

M.wt of CO₂ = 12+16+16 = 44
Question 4
4 / -1

The maximum number of molecules is present in:

A
0.5 g of H₂ gas

B
10 g of O₂ gas

C
15 L of H₂ gas at STP

D
5 L of N₂ gas at STP

Solution

In 22.4l of H₂ maximum number of molecules
Question 5
4 / -1

At 0°C and one atm pressure, a gas occupies 100 cc. If the pressure is increased to one and a half-time and temperature is increased by one-third of absolute temperature, then final volume of the gas will be:

A
80 cc

B
88.9 cc

C
66.7 cc

D
100 cc

Solution
Question 6
4 / -1

Equal amounts of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is:

A
0.55 atm

B
0.11 atm

C
1 atm

D
0.12 atm

Solution

No. of moles of lighter gas = m/4

No. of moles of heavier gas = m/40
Question 7
4 / -1

There are 6.02 × 10²² molecules each of N₂ , O₂ and H₂ which are mixed together at 760 mm and 273 K. The mass of the mixture in grams is?

A
6.2

B
4.12

C
3.09

D
7

Solution

6.02 × 10²² molecules of each N₂ , O₂ and H₂
Question 8
4 / -1

At what pressure a quantity of gas will occupy a volume of 60 ml, if it occupies a volume of 100ml at a pressure of 720 mm? (while temperature is constant):

A
700 mm

B
800 mm

C
100 mm

D
1200 mm

Solution

According to Boyle's law P₁ V₁ = P₂ V₂

⇒ P₁ × 60 = 720 × 100
Question 9
4 / -1

A 10g of a gas at atmospheric pressure is cooled from o 273°C to o 0°C keeping the volume constant, its pressure would become:

A
1/2 atm

B
1/273 atm

C
2 atm

D
273 atm

Solution
Question 10
4 / -1

Which one of the following indicates the value of the gas constant R ?

A
1.987 cal K^-1 mol^-1

B
8.3 cal K^-1 mol^-1

C
0.0821 lit K^-1 mol^-1

D
1.987 joules K^-1 mol^-1

Solution

8.31 J.K^-1 mol^–1

1 cal = 4.2 J.