NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | CHEMISTRY IMPORTANT QUESTIONS PART-1 | CHAPTER -5 | STATES OF MATTER | EDUGROWN |

In This Post we are  providing Chapter-5 STATES OF MATTER NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON STATES OF MATTER

Question 1.
Ammonia and Sulphur dioxide gases are prepared in two comers of a laboratory. Which gas will be detected first by a student working in the middle of the laboratory and why?

Answer:
Molecular mass of NH₃ = 17 Molecular mass of SO₂ = 64
NH₃ is a lighter gas and diffuses at a faster speed than SO₂
∴ NH₃ gas will be detected first.
[∵ Acc. to Graham’s law of diffusion r1r2=d2d1−−√

Question 2.
What is the effect of hydrogen bonding on the viscosity of a liquid?

Answer:
Hydrogen bonding leads to an increase in the effective size of the moving unit in the liquid. Due to an increase in the size and mass of the molecule, there is greater interval resistance of the flow of the liquid. As a result, the viscosity of the liquid rises.

Question 3.
Which are the two faulty assumptions in the kinetic theory of gases.

Answer:

1. There is no force of attraction between the molecules of the gas.
2. The volume of the molecules of the gas is negligibly small as compared to the total space (volume) occupied by the gas.

Question 4.
What is the relationship between the density and molar mass of a gaseous substance? Derive it.

Answer:
Ideal gas equation is PV = nRT
or nV=pRT
Replacing n by mM, we get

Question 5.
What is meant by the term: Non-ideal or real gas?

Answer:
The gas which does not obey the Cas law:
Boyle’s law, Charles’ law, Avogadro^ law at all temperatures and pressures is a called-non-ideal or real gas. Most of the real gases show ideal behaviour at low pressure and high temperature.

Question 6.
Derive the ideal gas equation PV = nRT.

Answer:
According to Boyle’s law V ∝ 1P if n and T are constant.
According to Charles’ law V ∝ T at constant P.and n
According to Avogadro’s law V ∝ n at constant T and P
Combining the three laws
In
V ∝ TnP
or
PV ∝ nT
or
PV = nRT where R is a constant of proportionality called universal gas constant.

Question 7.
Why liquids have a definite volume, but no definite shape?

Answer:
It is due to the fact that in liquids intermolecular forces are strong enough to hold the molecules together, but these are strong enough to hold the molecules together, but these forces are not strong enough to fix them into definite or concrete positions as in solids. Hence they possess fluidity but no definite shape.

Question 8.
How do the real gases deviate from ideality above and below the Boyle point?

Answer:
Above their Boyle point, real gases show positive deviations from ideality and the values of Z are greater than one. The forces of attraction between the molecules are very feeble. Below the Boyle point, real gases first show a decrease, in Z value with increasing pressure, the value Of Z increases continuously.

Question 9.
Write down the van der Waals, equation for n moles of a real gas. What do the constants ‘a’ and ‘b’ stand for?

Answer:
(P + an2V2) (V – nb) = nRT
where p, Vindicate gas pressure and its volume. T is the Kelvin temperature of the gas. R is gas constant. Value of ‘a’ is a measure of the magnitude of intermolecular attractive forces within the molecule and is independent of temperature and pressure, ‘rib’ is approximately the total volume occupied by the molecules themselves. ‘a’ and ‘b’ are called van der Waals constants and their value depends upon the nature of the gas.

Question 10.
One mole of S02 gas occupied a volume of 350 mL at 27°C and 50 atm pressure. Calculate the compressibility factor of the gas. Comment on the type of deviation shown by the gas from ideal behaviour.

Answer:
Compressibility factor, Z =
Here n = 1, P = 50 atm, V = 350 × 10^-3 L = 0.35 L
R = 0.0821 L atm K^-1 mol^-1; T = 27 + 273 = 300 K
∴ Z = 50×0.351×0.0821×300 = 0.711
For an ideal gas Z = 1
As for the given gas Z < 1, it shows a negative deviation, i.e., it is more compressible than expected from ideal behaviour.

Question 11.
A mixture of CO and CO₂ is found to have a density of 1.5 g L^-1 at 20°C and 740 mm pressure. Calculate the composition of the mixture.

Answer:
1. Calculation of average molecular mass of the mixture
M = dRTP=1.50×0.0821×293740760 = 37.06

2. Calculation of percentage composition
Let mol% of CO in the mixture = x
Then mol% of CO? in the mixture = 100 – x

Average molecular mass

Question 12.
A 5-L vessel contains 14 g of nitrogen. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K.

Answer:
N₂ ⇌ 2N

Initial moles = 1.428 = 0.05
Moles left = 0.05 – 30100 × 0.05 = 2 × 0.015 = 0.03

∴ Total no. of moles = 0.035 + 0.030 = 0.065
i.e. n = 0.065 mol, V = 5 L, T = 1800 K; P =?
P = nRTV=0.065×0.0821×18005
= 1.92 atm.

Question 13.
A given mass of a gas occupies 919.0 mL in a dry state at STP. The same mass when collected over water at 15°C and 750 mm pressure occupies on litre volume. Calculate the vapor pressure of water at 15°C.

Answer:
If p is the vapor pressure of water at 15°C, then P₂ = 750 – p
From the gas equation P1V1T1=P2V2T2, we get
760×919273=(750−p)×1000288
or
p = 13.3 mm

∴ Vapour pr. of water = 13.3 mm.

Question 14.
A steel tank containing air at 15 atm pressure at 15°C ¡s provided with a safety valve that will yield at a pressure of 30 atm. To what minimum temperature must the air be heated to below the safety valve?

Answer:
P1P2=T1T2
i.e., 1530=288 T2
or
T₂ = 576 K
or
t₂°C = 576 – 273 = 303°C

Question 15.
Calculate the pressure exerted by 110 g of CO₂ in a vessel of 2 L capacity at 37°C. Given that the van der Waals constants are a = 3.59 L² atm mol^-2 and b = 0.0427 L mol^-1. Compare the value with the calculated value if the gas were considered ideal.

Answer:
According to van der Waals equation

110
Here, n = 11044 = 2.5 moles. Putting the given values, we get
P = 2.5×0.0821×310(2−2.5×0.0427)−3.59×2.52
= 33.61 atm – 5.61 atm = 28.0 atm

If the gas were considered an ideal gas. applying ideal gas equation
PV = nRT
or
P = nRTV
∴ P = 2.5×0.0821×3102 = 31.8 atm