91Ó°ÊÓ

(a)

According to the above chart,a standard addition graph of final pb(II) have to be prepared ,contributed by addition of standard pb(II) solution.

The below graph represents their respective observance

So,the trendline equation for the Abs vs cone graph is $\mathrm{y}=49.52\mathrm{x}+1.094$,which is in the form of $\mathrm{y}=\mathrm{mx}+\mathrm{c}$

In the graph, $\mathrm{y}-\mathrm{axis}\mathrm{is}\mathrm{signal},\mathrm{x}-\mathrm{axis}\mathrm{is}\mathrm{pb}\left(\mathrm{ll}\right)\mathrm{in}\mathrm{ppm},\mathrm{y}-\mathrm{intercept}\mathrm{is}\mathrm{c},\mathrm{slope}\mathrm{is}\mathrm{m}.$

Here,the signal is zero at $\mathrm{y}-\mathrm{intercept}=1.094\mathrm{and}\mathrm{m}=49.52$

$\mathrm{pb}\left(\mathrm{ll}\right)\mathrm{is}\frac{\mathrm{intercept}}{\mathrm{slope}}=\frac{1.094}{49.52}=0.0221\mathrm{ppm}$

Hence,$\mathrm{pb}\left(\mathrm{ll}\right)$inunskippedaliquent$=0.0221\mathrm{ppm}$

(b)

The unskipped aliquant is consist of

$0.5\mathrm{mL}$of original solution

$\mathrm{A}$and diluted to a final volume of

$50\mathrm{ml}.$

$$\begin{gathered} {\mathrm{C}}_1{\mathrm{V}}_1\left(\mathrm{unskipped}\mathrm{aliquent}\right)={\mathrm{C}}_2{\mathrm{V}}_2\left(\mathrm{original}\mathrm{solution}\right) \\ {\mathrm{C}}_2=\frac{0.0221\mathrm{ppm}×5.0\mathrm{ml}}{0.50\mathrm{ml}} \\ {\mathrm{C}}_2=0.0221\mathrm{ppm} \end{gathered}$$

So.pb(II) is original solution $=0.0221\mathrm{ppm}$

$\mathrm{ppm}=\frac{\mathrm{mg}\mathrm{of}\mathrm{solute}}{\mathrm{lit}\mathrm{of}\mathrm{solution}}$

So,the pb(II) in original solution is

$$\begin{gathered} \mathrm{A}=0.221\mathrm{ppm} \\ \mathrm{pb}\left(\mathrm{ll}\right)=0.221\mathrm{mg}/\mathrm{L} \\ \mathrm{pb}\left(\mathrm{ll}\right)=221\frac{\mathrm{μ²\mu }}{\mathrm{Lit}} \end{gathered}$$

So,The pb(II) in original solution is$221\frac{\mathrm{μ²\mu }}{\mathrm{Lit}}$