Question

The weight of $2.01 \times {10^{23}}$ molecules of $CO$ is:
A.$9.3g$
B.$7.2g$
C.$1.2g$
D.$3g$

Answer 1

Hint: In one mole of atom there are $6.02 \times {10^{23}}$molecules. If in the molecule we have to find the number of molecules then we have to multiply the Avogadro number by the molecule present in the compound.

Complete step by step solution:
First of all we will talk about the gram molecule, Avogadro number and atoms.
Atoms: It is the smallest unit of an element. The elements are made of the same kind of atoms. In the atoms there are three particles: electrons, protons and neutrons.
Electrons: They are the negatively charged particle having negligible mass. They move in the atoms outside the nucleus in a fixed orbit.
Protons: They are positively charged particles. They have mass. They are present in the nucleus (a special room in the atoms where neutrons and protons are present).
Neutrons: They have no charge but they have mass. They are also present in the nucleus along with protons.
Gram molecule: It is defined as the number of grams of the substance i.e. the molecular mass of the substance.
Atomic mass: It is defined as the total number of protons and neutrons present in the nucleus of an atom. It is represented by symbol A.
Atomic number: It is defined as the number of protons in the nucleus of an atom. It is represented by symbol Z.
In the given question we have to find the weight of $2.01 \times {10^{23}}$ molecules of $CO$. We know that the number of moles is equal to the ratio of given molecules of substance to the Avogadro number. We know that one gram molecule contains $6.02 \times {10^{23}}$ molecules. The quantity $6.02 \times {10^{23}}$ is known as Avogadro number. So number of moles is $\dfrac{{2.01 \times {{10}^{23}}}}{{6.02 \times {{10}^{23}}}} = 0.333$. We also know that one mole of carbon monoxide has mass equal to $28g$. So the weight of $0.333$ moles will be $28 \times 0.3333 = 9.3$grams.

So option A is correct.

Note: In one mole of compound the gram equivalent is $6.02 \times {10^{23}}$molecules. Then if we want to find the gram equivalent for the atoms separately then divide the Avogadro number by the number of atoms present in the compound.