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3 years ago

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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.

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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$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

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So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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