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Pie
3 years ago
8

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. Th

e 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.
Chemistry
1 answer:
mafiozo [28]3 years ago
7 0

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

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