Atoms and Molecules Worksheet-4

Atoms and Molecules Worksheet-4

1. One ‘u’ stands for :

A. An atom of carbon (C-12)

B. 1/12^th of mass of carbon atom (C-12)

C. 1/12^th Of hydrogen atom.

D. One atom of all the elements.

2. How many times an atom of sulphur is heavier then an atom of carbon?

A. 32 times                                     B. 12 times

C. 8/3 times                                  D. 12/32 times

3. The number of oxygen atoms in 4.4 g of CO₂ is approx.-

A. 6 × 10²²         B. 6 × 10²³         C. 12 × 10²³       D. 1.2 × 10²³

4. A sample of CaCO₃ contains 3.01 × 10²³ ions if Ca²⁺ and CO₃^2–. The mass of the sample is :

A. 100 g              B. 50 g                C. 200 g             D. 5 g

5. Which of the following has maximum number of molecules?

A. 1g of CO₂                                   B. 1g of N₂

C. 1g of H₂                                      D. 1g of CH₄

6. The law of constant proportion was given by _______.

A. Dalton           B. Berzelius       C. Proust            D. Lavoisier.

7. The law of constant proportion is applied to ______.

A. Any element                             B. Any chemical compound

C. Pure chemical compound     D. None of these.

8. In carbon disulphide (CS₂), the mass of sulphur in combination with 3.0 g of carbon is :

A. 4.0 g              B. 6.0 g              C. 64.0 g            D. 16.0 g

9. What mass of carbon dioxide (CO₂) will contain 3.011 × 10²³ molecules?

A. 11.0 g             B. 22.0 g            C. 4.4 g               D. 44 g

10. Atom is smallest indivisible particle of matter was proposed by ________.

A. Rutherford   B. Dalton           C. Bohr               D. Einstein

Answer:

1. B

Explanation: Relative atomic mass is expressed in units known as a.m.u. It is simply represented as u (unified mass). An a.m.u is defined as the mass of one twelfth (1/12) of the mass of one atom of carbon taken as 12.

2. C

Explanation: Atomic mass of sulphur =

32/12 = 8/3 times.

An atom of sulphur is 8/3 times heavier than an atom of carbon.

3. D

Explanation: 44 g of CO₂ has oxygen atoms = 2 × 6.022 × 10²³

4.4 g of CO₂ has oxygen atoms

4. B

Explanation: 6.022 × 10²³ Ca²⁺ and CO₃^2– ions are present in CaCO₃ = 100 g

3.01 × 10²³ Ca²⁺ and CO₃^2– ions are present in CaCO₃ 

5. C

Explanation:

1 mole of a substance has 6.022 × 10²³ particles

1 g of H₂ (0.5 mol) has max. no. of molecules i.e. 3.011 × 10²³ molecules.

6. C

Explanation: Fact

7. C

Explanation: Law of constant proportion: A pure chemical compound always consists of the same elements that are combined together in a fixed (or definite) proportion by mass.

8. D

Explanation: In compound CS₂

12 g of carbon combine with 64 g sulphur

3 g of carbon will combine with  sulphur

9. B

Explanation: 6.022 × 10²³ molecules of CO₂ are present in 44g of CO₂

3.011 × 10²³ molecules of CO₂ will be present in =

10. B

Explanation: Fact