Question

How many milligram of iron ($F{e}^{2+}$) are equal to $1$ mL of $0.1055$ N $K_{2}C{r}_2{O}_7$ equivalent?

• A. $5.9mg$
• B. $0.59mg$
• C. $59mg$
• D. $5.9\times {10}^{-3}mg$

Answer 1

The correct option is A $5.9mg$

No. of ions on ${Fe}^{+2}=$no. of ions on $K_2{Cr}_2{O}_7$

Normality X volume $=$ no. of ions on $K_2{Cr}_2{O}_7$=moles X charge

$0.1055\times 1=\mathrm{0.1.55}=mol\times 1$ (charge$=1$ on ${Fe}^{+2}$, ${Fe}^{+2}⟶{Fe}^{+3}+e^{-}$ )

$\frac{x}{56}=0.1055mg$

$x=5.9mg$