Question

Which of the following contains the maximum number of molecules?
(A) 1 gm of $C{{O}_{2}}$
(B) 1 gm of ${{N}_{2}}$
(C) 1 gm of ${{H}_{2}}$
(D) 1 gm of $C{{H}_{4}}$

Answer 1

For a molecule, molecular weight in grams is equal to the Avogadro number. By calculating the molar mass of each given compound and then equating it with the above relation we would get the compound which is having the highest number of molecules.

Complete step by step answer:
- Let's start with the idea of Avogadro number and concept of moles. Avogadro’s number can be defined as a proportion that relates molar mass on an atomic scale to physical mass on a human scale. It is defined as the number of elementary particles such as molecules, atoms, compounds, etc. per mole of a substance.
- Or in other words, the number of units in one mole of any substance is called Avogadro’s number or Avogadro’s constant. Its value is given as $6.023\times {{10}^{23}}$.
- Avogadro number is often denoted as ${{N}_{A}}$.
- An additional property of Avogadro’s number is that the mass of one mole of a substance is equal to that substance’s molecular weight. Also, the gram atomic mass of an element is the mass of one mole of that element and the gram molecular mass of a compound refers to the mass of a single mole of the compound.

-Let's consider $C{{O}_{2}}$. By the above definition the gram atomic mass of $C{{O}_{2}}$ an be found.
For $C{{O}_{2}}$the molar mass
= 12+(16×2) ( since ,molar mass of Carbon =12 and molar mass of oxygen =16)
= 44 g
By the definition of Avogadro's number, we can write
Number of molecules of in 44 g of $C{{O}_{2}}$ =$6.023\times {{10}^{23}}$
Therefore in 1 g of $C{{O}_{2}}$= $\dfrac{6.023\times {{10}^{23}}}{44}$
= $1.37\times {{10}^{22}}$molecules

For ${{N}_{2}}$ the molar mass is 28 g.
Hence number of molecules in 28 g of ${{N}_{2}}$= $6.023\times {{10}^{23}}$
∴ In 1 g of ${{N}_{2}}$=$\dfrac{6.023\times {{10}^{23}}}{28}$
= $2.15\times {{10}^{22}}$ molecules

For ${{H}_{2}}$the molar mass is 2 g.
Hence number of molecules in 2 g of ${{H}_{2}}$= $6.023\times {{10}^{23}}$
∴ In 1 g of ${{H}_{2}}$ = $\dfrac{6.023\times {{10}^{23}}}{2}$
= $3.011\times {{10}^{23}}$ molecules

For $C{{H}_{4}}$ the molar mass is 16 g.
Hence number of molecules in 16 g of $C{{H}_{4}}$=$6.023\times {{10}^{23}}$
∴ In 1 g of $C{{H}_{4}}$ =$\dfrac{6.023\times {{10}^{23}}}{16}$
= $3.76\times {{10}^{22}}$ molecules
So, the correct answer is “Option C”.

Note: We could also find the answer in a simpler way. The given mass is the same for all the four gases.
Therefore, the gas having lowest molar mass will have maximum number of moles and maximum number of molecules. Among the given options ${{H}_{2}}$ is having the lowest molar mass (2 g). Hence ${{H}_{2}}$ will have a maximum number of molecules.