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Question: The atomic mass of A and B are 20 g and 40 g respectively. If X g of A contains Y atoms, then how ma...

The atomic mass of A and B are 20 g and 40 g respectively. If X g of A contains Y atoms, then how many atoms does 2X g of B contains:
A. XX
B. YY
C. X+YX + Y
D. 2Y2Y

Explanation

Solution

Firstly we should be clear what exactly is the data given in ques which is the mass of the atoms independently. Now after then we have to use the given values to find the number of atoms in the mass of the 2X g of B and so in the terms of the given alphabets. Now after the proper solving we get that the atoms would be YY .

Complete step-by-step solution: Firstly we should be aware that the question is regarding the mathematical approach towards the relative masses and the atomic mass.
Now we have to apply a specific step by step approach towards finding the atomic mass of the obtained product, as it would be concluded by using the data mentioned in the question.
We know that to find the number of atoms first we have to find the number of moles because number of moles and number of atoms related to each other by avogadro number.
As per the data which is given in the question regarding the two atoms we get that the
The atomic weight of AA would be : 20g20g
The atomic weight of BB would be : 40g40g
And also that XgXg of AA contain YY atoms.
Now we have to use this data and derive the relation for the given problem.
For element A:
Number of moles =massatomic  mass = \dfrac{{mass}}{{atomic\;{\text{mass}}}}
Number of moles =X20 = \dfrac{X}{{20}}
Number of atoms Of A =NA×X20=Y = {N_A} \times \dfrac{X}{{20}} = Y……(1)
For element B:
Number of moles =2X40 = \dfrac{{2X}}{{40}}
Number of atoms of B =NA×2X40 = {N_A} \times \dfrac{{2X}}{{40}} =NA×X20=Y = {N_A} \times \dfrac{X}{{20}} = Y

Therefore the correct option would be option B, YY .

Note: The atomic mass of an isotope and the relative isotopic mass refers to a certain specific isotope of an element. Because substances are usually not isotopically pure, it is convenient to use the elemental atomic mass which is the average (mean) atomic mass of an element, weighted by the abundance of the isotopes.