Question

An alloy contains$\mathbf{~}\mathbf{2}\mathbf{.}\mathbf{0}$ wt% Ni. What volume of 0.83 wt% DMG should be used to provide a 50%excess of DMG for the analysis of 1.8 g of steel? What mass of$\mathbf{Ni}\mathbf{(}\mathbf{DMG}{\mathbf{)}}_{\mathbf{2}}$ precipitate is expected?

Short answer

In the reaction, the needed volume of alcoholic DMG to raise it by50% excess is33mL The Ni-DMGcomplex red precipitate has a mass of 0.18 g

Step-by-step solution

Concept used

The concentration of a solution is measured by the number of moles of solute per litre of solution, which is known as molarity (M). The formula for calculating a solution's molarity is:

$\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{\left(}\mathit{M}\mathit{\right)}\mathit{=}\frac{\mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{}\mathit{o}\mathit{f}\mathit{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathit{\left(}\mathit{i}\mathit{n}\mathit{m}\mathit{o}\mathit{l}\mathit{\right)}}{\mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{\hspace{0.33em}}\mathit{o}\mathit{f}\mathit{\hspace{0.33em}}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{(}\mathit{i}\mathit{n}\mathit{\hspace{0.33em}}\mathit{L}\mathit{)}}$

To measure the nickel content in steel:

In the given reaction, the needed volume of alcoholic DMG to raise it by 50% excess.

The reaction is,

The mass of 2.00Wt% nickel in 1.8g of steel is calculated as

$$\begin{gathered} =\frac{1.8\mathrm{g}}{100\%}\times 200\% \\ =0.036\mathrm{g} \end{gathered}$$

DMG and nickel have a 2 :1 ratio. DMG's mass is estimated as follows:

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}=\frac{\mathrm{Mass}}{\mathrm{Mole}} \\ \mathrm{molc}=\frac{0.036\mathrm{gNi}}{58.69\mathrm{g}/\mathrm{molNi}}=6.134\times {10}^{-4}\mathrm{molNi} \\ 2(6.134\times {10}^{-4}\mathrm{molNi})(116.12\mathrm{g}/\mathrm{molDMG}/\mathrm{molNi})=0.142\mathrm{gDMG} \end{gathered}$$

The excess of 50% of DMG is computed as

$(1.5)(0.142\mathrm{g}\mathrm{DMG})=0.214\mathrm{g}\mathrm{DMG}$

The quantity of DMG in 2.15% is calculated as follows:

$\frac{0.214\mathrm{g}\mathrm{DMG}}{0.0083\mathrm{g}\mathrm{DMG}/\mathrm{g}\mathrm{solution}}=25.78\mathrm{g}\mathrm{solution}$

$\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$

$\frac{25.78\mathrm{g}\mathrm{solution}}{0.790\mathrm{g}/\mathrm{mL}}=33\mathrm{mL}$

Hence, the required volume of alcoholic DMG to raise by 50% more is 33mL

The quantity of Ni-DMGcomplex red precipitate is calculated as

$\mathrm{Molarmass}=\frac{\mathrm{Mass}}{\mathrm{Mole}}$

$\mathrm{mole}=\frac{0.036\mathrm{gNi}}{58.69\mathrm{g}/\mathrm{molNi}}=6.134\times {10}^{-4}\mathrm{molNi}$

$$\begin{gathered} 6.134\times {10}^{-4}\mathrm{molNi}=6.134\times {10}^{-4}\mathrm{molNi}-\mathrm{DMG} \\ 6.134\times {10}^{-4}\mathrm{molNi}-\mathrm{DMG}\times 288.91\mathrm{g}/\mathrm{molNi}-\mathrm{DMG}=0.18\mathrm{g} \end{gathered}$$

Therefore, the Ni-DMG complex red precipitate mass is 0.18 g