91Ó°ÊÓ

• The moles of a solute per litre of a solution are known as molarity.
• The molar concentration of a solution is also known as molarity.

$8.33×{10}^{-5}\mathrm{M}.$As illustrated below, the concentration of Mn in the supplied unknown combination is estimated,

$(13.5\mathrm{μ²\mu }/\mathrm{mL})\frac{1.00}{6.00}=2.25\mathrm{μ²\mu }/\mathrm{mL}$

For a standard combination, combine:

role="math" localid="1663655475613" $$\begin{gathered} \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\left(\frac{{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{S}\right]}\right) \\ \frac{1.05}{\left[2.50\mathrm{μ²\mu }/\mathrm{mL}\right]}=\mathrm{F}\left(\frac{1.00}{\left[2.00\mathrm{μ²\mu }/\mathrm{mL}\right]}\right) \\ \mathrm{F}=\frac{1.05}{\left[2.50\mathrm{μ²\mu }/\mathrm{mL}\right]}×\frac{\left[2.00\mathrm{μ²\mu }/\mathrm{mL}\right]}{1.00} \\ =\frac{\left[2.1\mathrm{μ²\mu }/\mathrm{mL}\right]}{\left[2.50\mathrm{μ²\mu }/\mathrm{mL}\right]} \\ =0.840 \end{gathered}$$

In the case of an unknown mixture:

$$\begin{gathered} \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\left(\frac{{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{S}\right]}\right) \\ \frac{0.185}{\left[Fe\right]}=0.840\left(\frac{0.128}{\left[2.25\mathrm{μ²\mu }/\mathrm{mL}\right]}\right) \\ \frac{0.185}{\left[Fe\right]}=\frac{0.10752}{\left[2.25\mathrm{μ²\mu }/\mathrm{mL}\right]} \\ \left[Fe\right]=\frac{0.185×\left[2.25\mathrm{μ²\mu }/\mathrm{mL}\right]}{0.10752} \\ =\frac{\left[0.41625\mathrm{μ²\mu }/\mathrm{mL}\right]}{0.10752} \\ =3.87\mathrm{μ²\mu }/mL \end{gathered}$$

The Fe concentration at the beginning has to be,

$6.00/5.00(3.87\mathrm{μ}g/mL)=4.65\mathrm{μ}g/mL$

The molarity of this can be stated as