Question

Calculate and report the precise concentration of undiluted stock standard solution #1 for AR in micromoles per liter from ppm by mass. Assume that the density of water is 1.00g/ml. This is your most concentrated undiluted standard solution for which you measured the absorbance.

Answer 1

The question is incomplete, complete question is ;

Allura Red (AR) has a concentration of 21.22 ppm. What is this is micro moles per liter? Report the precise concentration of the undiluted stock solution #1 of AR in micromoles per liter. This is your most concentrated (undiluted) standard solution for which you measured the absorbance. Use 3 significant figures. Molarity (micro mol/L) =

Answer:

The molarity of the solution of allura red is 42.75 micro moles per Liter.

Explanation:

The ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

The ppm concentration of allura red = 21.22 ppm

This means that 21.22 mg of allura red was present 1 kg of solution.

Mass of Allura red = 21.22 mg = $21.22\times 0.001 g$

1 mg = 0.001 g

Mass of solution = 1 kg = 1000 g

Density of the solution = Density of water = d = 1.00 g/mL

( since solution has very small amount of solute)

Volume of the solution :

$=\frac{1000 g}{1.00 g/mL}=1000 mL$

1000 mL = 1 L

Volume of the solution, V = 1 L

Moles of Allura red = $\frac{21.22\times 0.001 g}{496.42 g/mol}=4.275\times 10^{-5} mol=4.275\times 10^{-5}\times 10^{6} \mu mole$

Molarity of the solution ;

$=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

$M=\frac{4.275\times 10^{-5}\times 10^6 \mu mol}{1 L}=42.75 \mu mol/L$

The molarity of the solution of allura red is 42.75 micro moles per Liter.