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Question: The weight of \(2.01 \times {10^{23}}\) molecules of \(CO\) is: A.\(9.3g\) B.\(7.2g\) C.\(1.2g...

The weight of 2.01×10232.01 \times {10^{23}} molecules of COCO is:
A.9.3g9.3g
B.7.2g7.2g
C.1.2g1.2g
D.3g3g

Explanation

Solution

In one mole of atom there are 6.02×10236.02 \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 2.01×10232.01 \times {10^{23}} molecules of COCO. 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 6.02×10236.02 \times {10^{23}} molecules. The quantity 6.02×10236.02 \times {10^{23}} is known as Avogadro number. So number of moles is 2.01×10236.02×1023=0.333\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 28g28g. So the weight of 0.3330.333 moles will be 28×0.3333=9.328 \times 0.3333 = 9.3grams.

So option A is correct.

Note: In one mole of compound the gram equivalent is 6.02×10236.02 \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.