So my chem experiment asks us: What is the molarity of a 6.0 ppm solution (it's a 6.0 ppm chlorophyll solution, hexane is the solvent)

Now, I've found a few ways to solve for it but they both give me different values. If it's possible, could you just tell me which is right.

#1 Way: 6parts/1million --> 0.006g/L --> take 0.006g convert w/ molar mass chlorophyll (1 mole/ 893.49 g) --> 6.71524 10^(-6) moles

put 6.71524 10^(-6) over 1L --> 6.71524 10^(-6) M

0R 6parts/1million --> 6g/1000,000g --> take 6g convert w/ molar mass chlorophyll (1 mole/ 893.49 g) --> 6.71524 10^(-3) moles

use hexane's density to convert 1000,000g to (mL first then) L (density: 1mL/0.6548g) --> 1527.18388 L

put 6.71524 10^(-3) moles over 1527.18388 L --> 4.39710^(-6) M

OR 6parts/1million -->6g/1000,000g take 6g convert w/ molar mass chlorophyll (1 mole/ 893.49 g) --> 6.71524 10^(-3) moles

put 6.71524 10^(-3) moles over 1000,000 --> 6.71524 *10^(-9) M

I think the second one is the right one but I'm not sure. Could you please help me? Thanks!

The second approach is correct. The other two approaches are not correct because they are incomplete; first approach would have been right IF the 6.0 ppm was MEASURED in hexane. Third approach cannot be right since it calculates moles and grams but not L.

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Tringa0 Tringa0 · Answer 2 · 2020-01-16T06:02:27-05:00

Answer:

The first option is correct.

Explanation:

Concentration of chlorophyll is the ppm = 6.0 ppm

(1 ppm = 1 mg/kg)

6.0 ppm = 6.0 mg/kg

Mass of chlorophyll = 6.0 mg= 0.006 g (1 mg = 0.001 g)

Moles of chlorophyll =\(\frac{0.006 g}{893.49 g/mol}=6.71524\times 10^{-6} mol\)

volume of the solution = 1 L

Concentration(c) = \(\frac{n}{V(L)}\)

Where : n = moles of the compound

V = Volume of the solution

Concentration of the chlorophyll :

\(\frac{6.71524\times 10^{-6} mol}{1 L}=6.71524\times 10^{-6} M\)