Question

in a vessel 2g O2, 2g H2, 2g N2 are present.Which has the highest number of atoms? ​

Answer 1

Answer== 2g H2

Explanation:

No.of atoms =

No.of moles× av. No.× atomicity

So, if we put the values for 2g O2

No.of atoms =0.75×10^23

If we put values for 2g H2

Then no. Of atoms =12×10^23

For 2gN2

No.of atoms =1.71×10^23

Answer 2

Answer:

The highest number of atoms present in $2g$ $H_{2}$, i.e., $6.022\times 10^{23}$ atoms.

Explanation:

Given,

The mass of $O_{2}$ = $2g$

The mass of $H_{2}$ = $2g$

The mass of $N_{2}$ = $2g$

The highest number of atoms present in =?

Case 1) $O_{2}$

• The given mass = $2g$

As we know,

• The molar mass of $O_{2}$ = $32g/mol$.

Thus,

• The number of moles = $\frac{Given mass}{Molar mass}$ = $\frac{2}{32}$ = $\frac{1}{16}$ mol.

Now,

• $1$ mole = $6.022\times 10^{23}$ atoms
• $\frac{1}{16}$ moles = $\frac{1}{16}\times 6.022\times 10^{23}$ atoms = $0.37\times 10^{23}$ atoms.

Case 2) $H_{2}$

• The given mass = $2g$

As we know,

• The molar mass of $H_{2}$ = $2g/mol$.

Thus,

• The number of moles = $\frac{Given mass}{Molar mass}$ = $\frac{2}{2}$ = $1mol$

Now,

• $1$ mole = $6.022\times 10^{23}$ atoms.

Case 3) $N_{2}$

• The given mass = $2g$

As we know,

• The molar mass of = $28g/mol$.

Thus,

• The number of moles = $\frac{Given mass}{Molar mass}$ = $\frac{2}{28}$ = $\frac{1}{14}$ mol.

Now,

• $1$ mole = $6.022\times 10^{23}$ atoms.
• $\frac{1}{14}$ moles = $\frac{1}{14} \times 6.022\times 10^{23}$ atoms = $0.43\times 10^{23}$ atoms.

Therefore,  

• The number of atoms in $2g$ $O_{2}$ = $0.37\times 10^{23}$ atoms
• The number of atoms in $2g$ $H_{2}$ = $6.022\times 10^{23}$ atoms
• The number of atoms in $2g$ $N_{2}$ = $0.43\times 10^{23}$ atoms.

As we can see, the highest number of atoms present in $2g$ $H_{2}$.