Question

A sample was then prepared containing 14.00 mL of coffee and 8.00 mL of 4.80 ppm Li , and diluted to a 50.00 mL total volume. The ratio of the internal standard signal to the analyte signal in the unknown sample was found to be 0.840/1.000 (Li signal/Na signal). Determine the unknown concentration of Na in the coffee sample.

Answer 1

Answer:

The right answer is "$[Na^+]=5.57 \ ppm$".

Explanation:

The given values are:

$\frac{A_s}{A_{coffee}} =\frac{0.840}{1.000}$

According to the question,

The concentration of standard will be:

=  $\frac{7.80}{50.00}\times 4.80$

=  $0.156\times 4.80$

=  $0.748 \ ppm$

Coffee after dilution,

⇒  $\frac{A_{coffee}}{C_{coffee}} =F\times \frac{A_s}{C_s}$

or,

⇒  $C_{coffee}=\frac{A_{coffee}}{A_s}\times \frac{C_s}{F}$

On substituting the values, we get

⇒               $=\frac{1.000}{0.840}\times \frac{0.748}{1.68}$

⇒               $=1.191\times 1.3104$

⇒               $=1.561 \ ppm$

hence,

In unknown sample, the concentration of coffee will be:

=  $1.561\times \frac{50}{14}$

=  $\frac{78.05}{14}$

=  $5.57 \ ppm$