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Question: What is the mass of one atom of \(C-12\) in grams? a.) \(1.992\times {{10}^{-23}}gm\) b.) \(1.9...

What is the mass of one atom of C12C-12 in grams?
a.) 1.992×1023gm1.992\times {{10}^{-23}}gm
b.) 1.989×1023gm1.989\times {{10}^{-23}}gm
c.) 1.892×1023gm1.892\times {{10}^{-23}}gm
d.) 1.965×1023gm1.965\times {{10}^{-23}}gm

Explanation

Solution

. One mole of an element consists of Avogadro number of atoms that is 6.023×10236.023\times {{10}^{23}} also the number of moles is the ratio of the mass to the molecular mass of the element.

Complete step by step answer:
Given that
Number of atoms is equal to one, We know that One mole of an element consists of Avogadro number of atoms that is 6.023×10236.023\times {{10}^{23}}
\Rightarrow one atom of an element contains 1Avagadro number\dfrac{1}{\text{Avagadro number}} number of moles that is 6.023×10236.023\times {{10}^{-23}}
Therefore one atom of carbon = 6.023×10236.023\times {{10}^{-23}} moles

We also know that the number of moles is the ratio of the mass in grams and the molecular mass of the element or the mass of an element is the product of the number of moles and the molecular mass.
We have the number of moles of the carbon atom is = 6.023×10236.023\times {{10}^{-23}} moles
Given the molecular mass of the carbon is = 12
Therefore the mass of one mole of carbon is = number of moles × molecular weight\text{number of moles }\times \text{ molecular weight}
\Rightarrow mass of the carbon is =126.023×1023=1.992×1023g=\dfrac{12}{6.023\times {{10}^{23}}}=1.992\times {{10}^{-23}}g
Therefore the required mass of one atom of C12C-12 in grams is 1.992×1023g1.992\times {{10}^{-23}}g
So, the correct answer is “Option A”.

Note: Number of moles can be defined as the ratio of the mass of the element to the molecular mass of the element.
One mole of an element contains Avogadro number of atoms.
One mole of an element contains Avogadro number of molecules.
One mole of an element contains Avogadro number of ions.
One mole of an element contains Avogadro number of species.