if the ration of water drops to EDTA is 6.7 for the known sample and 10 for the unknown sample what is the concentration of the unknown sample? (The concentration of the known sample is 315 ppm)
a. 150mg/L
b. 210mg/L
c. 280mg?L
d. 320mg/L
I would set up a proportion.
315ppm/6.7 = xppm/10
and solve for x.
that proportion solving for X doesnt give me a number listed as an option..I got 470
I never liked the drops way of doing things. Having said that, I must take the blame and say I made a mistake.
If it takes 3 drops of the EDTA for the known of 315 ppm and it takes only 2 drops for the unknown, the unknown must be weaker. You can do it by drops of EDTA (which makes more sense to me)
315 x 2/3 = ??
or you can use the ratio set up to give a smaller number; i.e.,
315 x (6.7/10) = ?? ( you don't get the same number because the 6.7 was rounded off from 6.66667 to 6.7).
or you can set up a ratio/proportion but the reverse of what I have above.
I suppose it's obvious but I didn't even look at the options for answers; perhaps I should have looked.
To determine the concentration of the unknown sample, we can use the ratio of water drops to EDTA for both the known and unknown samples, along with the concentration of the known sample.
Let's calculate the concentration of the unknown sample step by step:
1. Start by comparing the ratios of water drops to EDTA for the known and unknown samples.
- The known sample has a ratio of water drops to EDTA of 6.7.
- The unknown sample has a ratio of water drops to EDTA of 10.
2. We can establish a proportion between the two ratios:
- 6.7 (known sample ratio) / 10 (unknown sample ratio) = 315 ppm (known sample concentration) / x (unknown sample concentration).
3. Cross-multiply the values:
- 6.7 * x = 10 * 315.
4. Solve the equation for x:
- 6.7x = 3150.
- Divide both sides of the equation by 6.7:
- x = 3150 / 6.7.
- x ≈ 469.925.
5. Convert parts per million (ppm) to milligrams per liter (mg/L):
- x (unknown sample concentration) ≈ 469.925 mg/L.
Therefore, the concentration of the unknown sample is approximately 469.925 mg/L. None of the given answer choices match exactly, so the closest option would be 470 mg/L, which is not listed.