Question

The number of positively and negatively charged ions present in $1L$ of $0.1M$ aluminium sulphate solution are:

• A. $1.2\times {10}^{22}+ve$ ions and $1.8\times {10}^{22}-ve$ ions
• B. $1.2\times {10}^{23}+ve$ ions and $1.8\times {10}^{22}-ve$ ions
• C. $1.2\times {10}^{23}+ve$ ions and $1.8\times {10}^{23}-ve$ ions
• D. $1.2\times {10}^{21}+ve$ ions and $1.8\times {10}^{21}-ve$ ions

Answer 1

The correct option is C $1.2\times {10}^{23}+ve$ ions and $1.8\times {10}^{23}-ve$ ions

Reaction represented as: $A{l}_2{\left(S{O}_4\right)}_3\to 2A{l}^{3+}+3{S{O}_4}^{2-}$
Number of moles in $1l$ of $0.1M$ aluminium sulphate solution are= $1\times 0.1=0.1mole$.
1 mole of $A{l}_2{\left(S{O}_4\right)}_3$ produce $2$ moles of $A{l}^{3+}$ and $3$ moles of ${S{O}_4}^{2-}$ therefore 0.1mole will produce $0.2$ mole and $0.3$ mole respectively.
The number of positively ions present in 0.1mole $A{l}_2{\left(S{O}_4\right)}_3$ are $0.2\times 6.023\times {10}^{23}=1.2\times {10}^{23}+ve$ ions and The number of positively ions present in 0.1mole $A{l}_2{\left(S{O}_4\right)}_3$ are $0.3\times 6.023\times {10}^{23}=1.8\times {10}^{23}-ve$ ions