Answer:
The right answer is "[tex][Na^+]=5.57 \ ppm[/tex]".
Explanation:
The given values are:
[tex]\frac{A_s}{A_{coffee}} =\frac{0.840}{1.000}[/tex]
According to the question,
The concentration of standard will be:
= [tex]\frac{7.80}{50.00}\times 4.80[/tex]
= [tex]0.156\times 4.80[/tex]
= [tex]0.748 \ ppm[/tex]
Coffee after dilution,
⇒ [tex]\frac{A_{coffee}}{C_{coffee}} =F\times \frac{A_s}{C_s}[/tex]
or,
⇒ [tex]C_{coffee}=\frac{A_{coffee}}{A_s}\times \frac{C_s}{F}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{1.000}{0.840}\times \frac{0.748}{1.68}[/tex]
⇒ [tex]=1.191\times 1.3104[/tex]
⇒ [tex]=1.561 \ ppm[/tex]
hence,
In unknown sample, the concentration of coffee will be:
= [tex]1.561\times \frac{50}{14}[/tex]
= [tex]\frac{78.05}{14}[/tex]
= [tex]5.57 \ ppm[/tex]
