91Ó°ÊÓ

A mole is a common scientific unit for measuring significant quantities of very small entities such as atoms, molecules, or other designated particles in chemistry.

$$\begin{gathered} \mathrm{The}\mathrm{reaction}\mathrm{is}:{\mathrm{Hg}}_{2}2++2{\mathrm{l}}^{-}→{\mathrm{Hg}}_2{\mathrm{i}}_2\left(\mathrm{s}\right) \\ \mathrm{Concentration}\mathrm{of}\mathrm{Kl}=0.1\mathrm{M} \\ \mathrm{Concentration}\mathrm{of}{\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2=40\mathrm{mL}=0.04\mathrm{M} \\ \mathrm{volume}\mathrm{of}{\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2=40\mathrm{mL}=0.04\mathrm{L} \\ \mathrm{To}\mathrm{begin},\mathrm{determine}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2 \\ \mathbf{Moles}\mathbf{=}\mathbf{Concentration}\mathbf{}\mathbf{×}\mathbf{Volume}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=0.04\mathrm{M}×0.04\mathrm{L} \\ =0.0016\mathrm{mol} \end{gathered}$$

And since the mole ration between ${\mathrm{Hg}}_2\left({\mathrm{NO}}_3\right)$and ${{\mathrm{Hg}}_2}^{2+}$is$1.1$, the number of moles of${{\mathrm{Hg}}_2}^{2+}$will also equal$0.0016$mol.

So can now compute the moles of ${{\mathrm{Hg}}_2}^{2+}$reacted using the mole ratio between $l^{-}$and $l^{-}$

Because the mole ratio of $l^{-}$to $\mathrm{Kl}$is $1.1$, the number of moles of $\mathrm{Kl}$will also be equal to $0.0032$mol.

Finally, calculate the required volume because we know the concentration and number of moles of $\mathrm{Kl}$

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{Kl}=\frac{\mathrm{Moles}}{\mathrm{Concentration}} \\ =\frac{0.0032\mathrm{mol}}{0.1\mathrm{M}} \\ =0.032\mathrm{L} \\ =32\mathrm{mL} \end{gathered}$$

Therefore, the volume of needed is$32\mathrm{mL}$.