Question

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 --> 4.397*10^(-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!

Answer 1

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. 

Answer 2

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 :

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