# Class 11 Chemistry MCQ – States of Matter – Kinetic Energy and Molecular Speeds

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This set of Class 11 Chemistry Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “States of Matter – Kinetic Energy and Molecular Speeds”.

1. Calculate the root mean square speed of hydrogen in m/s at 27°C?
a) 2835.43 m/s
b) 2635.43 m/s
c) 2735.43 m/s
d) 2731.43 m/s

Explanation: The formula of root mean square speed is given by urms = $$\sqrt{3RT/M}$$. We have R = 8.314 kgm2/s2, M = 10-3 kg/mol and T = 300 k. So by substituting the formula we get, urms = $$\sqrt{3×8.314×300/10^{-3}}$$ = 2735.43 m/s.

2. What is the ratio of urms to ump in oxygen gas at 298k?
a) 1.124
b) 1.224
c) 1.228
d) 1.128

Explanation: The ratio of root mean square speed, represented as urms to the most probable speed, represented as ump is always the same for identical conditions and same gas. It is $$\sqrt{3RT/M}$$ divided by $$\sqrt{8RT/πM}$$ = 1.224.

3. The speed of three particles is recorded as 3 m/s, 4 m/s, and 5 m/s. What is a root mean square speed of these particles?
a) 4.082 m/s
b) 2.07 m/s
c) 3.87 m/s
d) 3.082 m/s

Explanation: The root means square speed of particles is nothing but the square root over the sum of squares of the particle’s speeds by a total number of particles. So by substituting, √32 + 42 + 52/3 = 4.082 m/s.
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4. What is the ratio of root mean square speed of 16 grams of Oxygen to 4 grams of hydrogen?
a) 2
b) 3
c) 4
d) 1

Explanation: The formula of root mean square speed of particles is given as $$\sqrt{3RT/M}$$. We know that the velocity of gas molecules is inversely proportional to the root over the mass of the gas here the mass of oxygen to the mass of hydrogen ratio is the answer. So $$\sqrt{16/4}$$ = 2.

5. Which of the following is greater for identical conditions and the same gas?
a) most probable speed
b) average speed
c) root mean square speed
d) most probable and average speed have the same value

Explanation: According to the formula, the root mean square speed is greater than the average speed and the average speed is greater than the most probable speed at given identical conditions and for the same gas.
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6. The root mean square speed of a gas at a certain condition is 1.128 times greater than the most probable speed.
a) true
b) false

Explanation: The ratio of root mean square speed to the mean probable speed is 1.224. So the above statement is considered to be wrong. The ratio between the main probable speed and the average speed and root mean square speed is 1: 1.128: 1.224.

7. What is the most probable speed of oxygen gas with the mass of 32 grams at 27-degree centigrade?
a) 33.74 m/s
b) 44.78 m/s
c) 57.94 m/s
d) 549.14 m/s

Explanation: The formula for the most probable speed of a gas is given as $$\sqrt{8RT/πM}$$. Here R is a universal gas constant which is always equal to 8.314 kgm2s-2, T=300 Kelvin and M = 0.032 kg So by substituting, we get an answer as 44.78 m/s.

8. Which among the following options do you think has the highest average speed?
a) chlorine
b) hydrogen
c) neon
d) oxygen

Explanation: The formula of average speed is given by $$\sqrt{2RT/M}$$, where R is universal gas constant, T is a temperature in Kelvin and M is the mass in kilograms. From the formula, we understand that the average speed is inversely proportional to the root over the mass. As hydrogen has the least mass among the options it has the highest average speed.

9. What is the ratio of the velocities of 2 moles of hydrogen to five moles of helium?
a) $$\sqrt{14}$$
b) $$\sqrt{10}$$
c) $$\sqrt{20}$$
d) $$\sqrt{50}$$

Explanation: The formula of average speed is given by $$\sqrt{2RT/M}$$. We know that the velocity of gas molecules is inversely proportional to the root over the mass of the gas. So here the ratio of velocities is $$\sqrt{5×4/2×1}$$ = $$\sqrt{10}$$.

10. What is the mean velocity of one Mole neon gas at a temperature of 400 Kelvin?
a) 11.533 m/s
b) 357.578 m/s
c) 367.79 m/s
d) 34 m/s

Explanation: The formula for mean velocity of a gas is given by the expression $$\sqrt{2RT/M}$$. for one mole of neon gas M is taken as 0.02 kg, temperature as 400 k, R as 8.314 kgm2s-2, so by substituting we get an answer as 11.533 m/s.

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