# Atoms and Molecules Worksheet-4

Atoms and Molecules Worksheet-4

1. One ‘u’ stands for :

A. An atom of carbon (C-12)

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

C. 1/12th Of hydrogen atom.

D. One atom of all the elements.

1. 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

1. 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

1. A sample of CaCO3 contains 3.01 × 1023 ions if Ca2+ and CO32–. The mass of the sample is :

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

1. 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

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

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

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

A. Any element                             B. Any chemical compound

C. Pure chemical compound     D. None of these.

1. 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

1. 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

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

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

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.

1. 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.

1. D

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

4.4 g of CO2 has oxygen atoms

1. B

Explanation: 6.022 × 1023 Ca2+ and CO32– ions are present in CaCO3 = 100 g

3.01 × 1023 Ca2+ and CO32– ions are present in CaCO3

1. 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.

1. C

Explanation: Fact

1. 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.

1. D

Explanation: In compound CS2

12 g of carbon combine with 64 g sulphur

3 g of carbon will combine with  sulphur

1. B

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

3.011 × 1023 molecules of CO2 will be present in =

1. B

Explanation: Fact