Question

Calculate number of electrons present in $9.5g$ of $P{O}_4^{3-}$.

• A. $6N_A$
• B. $4.7N_A$
• C. $5N_A$
• D. $0.1N_A$

Answer 1

Answer: (C)

$5N_A$

$MassofP{O}_4^{3-}=9.5$ $g$

$Numberofmoles=\frac{WeightofP{O}_4^{3-}}{Molecularweight}=\frac{9.5}{95}=0.1$

$1mole=6.023\times {10}^{23}ions$

$\Rightarrow 0.1moles=0.1\times 6.023\times {10}^{23}=6.023\times {10}^{22}ions$

Each ion of $P{O}_4^{3-}$ has $50$ electrons

$\therefore$ Total number of electrons

$=6.023\times {10}^{22}\times 50$

$=3\times {10}^{24}$ electrons

$=6.023\times {10}^{22}\times \frac{100}{2}$

$=6.023\times {10}^{23}\times \frac{10}{2}$

$=\left(N_A\right)5$

$=5N_A$