Question

The number of electrons in oxygen atom in 0.3 g-molecule of Mohr's salt ${FeSO}_4(N{H}_4{)}_2{SO}_{4.}{6H}_{2}O$ is:

• A. ${4.2N}_A$
• B. ${42N}_A$
• C. ${25.2N}_A$
• D. ${33.6N}_A$

Answer 1

The correct option is D ${33.6N}_A$

$FeS{O}_4(N{H}_4{)}_{2}S{O}_4.6H_{2}O$

Mohr salt - $0.3$ g-molecule

$1g$ molecule - $6.022\times {10}^{23}$

So, in $0.3g$ molecule - $0.3\times 6.022\times {10}^{23}$

The total oxygen atom in Mohr salt is - $14$

Total molecule - $14\times 0.3\times 6.022\times {10}^{23}=4.2\times 6.022\times {10}^{23}$

As we know that one atom of oxygen contains eight electrons.

Total electron - $4.2\times 8\times 6.022\times {10}^{23}=33.6\times 6.022\times {10}^{23}$

Avogadro number also represented by $N_A$

Hence, option D is the correct answer.