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!
I think 2 is the right approach.
#3 can't be right since M = mols/L and this calculates mols(correctly)/10^6 grams(not L).
#1 would be right, I think, IF the 6.0 ppm was MEASURED in hexane (meaning the standards were done the same way) but apparently this is not the case. #2 starts with the right definition of
6.0 ppm = 6.0grams/1,000,000 g while #1 is not the right definition (mg/L is ok for aqueous solutions but not for hexane solution) unless the standards were in hexane also.
Thanks soo much! That makes sense now :)
To determine the correct molarity of a 6.0 ppm chlorophyll solution in hexane, we need to follow the correct conversion steps. Let's go through the calculations together:
1. Start with the given information: 6.0 ppm means 6 parts per million.
2. Convert the parts per million (ppm) to grams per liter (g/L). Since 1 ppm = 1 mg/L:
6 ppm = 6 mg/L = 0.006 g/L.
Now, there are two possible approaches to calculate the molarity:
Approach 1:
3. Convert the mass of chlorophyll (0.006 g) to moles using the molar mass of chlorophyll (893.49 g/mol):
0.006 g × (1 mol/893.49 g) = 6.71524 × 10^(-6) moles.
4. Divide the moles by the volume (1 L) to obtain the molarity:
Molarity = 6.71524 × 10^(-6) moles/1 L = 6.71524 × 10^(-6) M.
Approach 2:
3. Convert the mass of chlorophyll (0.006 g) to mol:
Method a) Conversion of mass to moles:
0.006 g × (1 mol/893.49 g) = 6.71524 × 10^(-6) moles.
Method b) Conversion of mass to moles using density:
Convert grams to milliliters:
0.006 g × (1 mL/0.6548 g) = 9.166 mL.
Convert milliliters to liters:
9.166 mL × (1 L/1000 mL) = 0.009166 L.
4. Divide the moles by the volume (0.009166 L) to obtain the molarity:
Method a) Molarity:
Molarity = 6.71524 × 10^(-6) moles/0.009166 L = 7.317 × 10^(-4) M.
Method b) Molarity:
Molarity = 6.71524 × 10^(-6) moles/0.009166 L = 7.319 × 10^(-4) M.
Comparing the two approaches, both method a) and method b) of approach 2 yield approximately 7.32 × 10^(-4) M. Therefore, the correct molarity of the 6.0 ppm chlorophyll solution in hexane is approximately 7.32 × 10^(-4) M.
To find the molarity of a solution, you need to know the amount of solute (in moles) and the volume of the solution (in liters). Let's go through each method and determine which one gives the correct answer.
Method 1:
You correctly converted 6.0 ppm to 0.006 g/L. Then, you converted grams of chlorophyll to moles using its molar mass (893.49 g/mol). Finally, you divided the moles by 1 liter to get 6.71524 * 10^(-6) M.
Method 2:
In this approach, you converted 6.0 ppm to 6 g/1,000,000 g. Then, you converted grams of chlorophyll to moles using its molar mass. However, there is a mistake when converting grams to liters. The density you provided for hexane (1 mL/0.6548 g) is incorrect. The density of hexane is actually 0.6548 g/mL. So the correct conversion factor should be 0.6548 g/1 mL. With this correction, you would get a volume of 9,162.426 mL (not 1,527.18388 L). Finally, dividing moles by the corrected volume gives 4.397 * 10^(-6) M.
Method 3:
In this method, you made the same mistake in converting grams to liters as in Method 2. Once you correct this, you would obtain the same incorrect answer as in Method 2, which is 4.397 * 10^(-6) M.
Therefore, the first method, which gives a molarity of 6.71524 * 10^(-6) M, is the correct one.