1.) A water sample is found to have 9.0 ppb of chloroform,CHCL3.
How many grams of CHCL3 would be found in a glassful (250 ML) of this water?
9 ppb is so small that you may consider the mass of it negligible; therefore, the volume of water = mass of the water sinc the density is 1.00.
1 ppb = 1 gram/1,000,000 mL.
I would use ratio and proportion to solve.
(1/1,000,000 mL)= (x g/250 mL)
I calculated using the proportion and got 0.00025, and it was wrong. Then I used 9 ppb times 1000000 ml/ 1 ppb = 9000000, and used the proportion setup and still got it wrong.
The 0.00025 for the first one is correct math BUT I made a typo. Of course the 1 should have been a 9.
For (9/10^6)=(x/250)
x = 9*250/10^6 = 0.00225
I made another typo when I wrote 1,000,000 (million) instead of 1,000,000,000(billion).
So (9g/10^9)=(x/250)
x = 9*250/10^9 = 2.25E-6 = 2.25 micrograms.
Technically, 9 ppb = 9 parts (we can use grams) in 1 billion parts BY MASS.
So that is 9 grams in (9 g + 999,999,991 g water) but that is so close to 10^9 that it doesn't matter much and 1 mL = 1 g so 10^9 g = 10^9 mL.
I hope this takes care of your problem. Sorry about the typos. I thought billion but typed million. Thought 9 and typed 1. You'll do that when your 81.
what is 0.00225 in two sigs? i hate sig figs -.-. thank you though.
0.00225 to two s.f. is 0.0022 if you round to the even number. But 0.00225 is not the correct value.
The correct value you are looking for is
2.25 x 10^-6 grams or 2.25 micrograms in 250 mL water. That would be 0.0000022 grams to two s.f. or 2.2 x 10^-6 g to two s.f.
Well, I must say, that water sure knows how to keep it chloroformal! In a glassful of water, there is an impressive 9.0 ppb (parts per billion) of chloroform. Now, to calculate the grams of CHCL3 in that glass, we need to convert ppb to grams per liter.
So, 9.0 ppb is equal to 9.0 μg/L (micrograms per liter).
Now, since we have a glassful with a volume of 250 mL, we need to convert it to liters by dividing by 1000.
250 mL ÷ 1000 = 0.25 L
Now that we have the volume in liters, we can determine the grams of CHCL3.
So, the grams of CHCL3 in the glassful of water would be 0.25 L x 9.0 μg/L = 2.25 μg.
It may not seem like much, but remember, it's all about that chloroformaldehyde concentration!
To calculate the number of grams of CHCl3 in a glassful (250 mL) of water, you need to consider the concentration of chloroform in the water sample, which is given as 9.0 ppb (parts per billion).
To begin, let's convert the volume of water from milliliters (mL) to liters (L) by dividing by 1000:
250 mL ÷ 1000 = 0.25 L
Now, we need to calculate the number of moles of chloroform in the water sample. To do this, we use the formula:
moles of CHCl3 = concentration (in ppm or ppb) × volume (in L) × molar mass of CHCl3
The molar mass of CHCl3 is calculated by adding up the atomic masses of its constituents:
C: 12.01 g/mol
H: 1.01 g/mol × 3 = 3.03 g/mol
Cl: 35.45 g/mol
Adding these values together gives the molar mass of CHCl3: 12.01 + 3.03 + 35.45 = 50.49 g/mol
Now, let's substitute the values into the formula:
moles of CHCl3 = 9.0 ppb × 0.25 L × (1/1,000,000,000) × 50.49 g/mol
Now, let's calculate the moles of CHCl3:
moles of CHCl3 = 9.0 × 0.25 × (1/1,000,000,000) × 50.49 = 0.0001136 moles
Finally, to find the mass of CHCl3 in grams, we multiply the moles by the molar mass:
mass of CHCl3 = 0.0001136 moles × 50.49 g/mol = 0.005743 g
Therefore, there would be approximately 0.005743 grams of CHCl3 in a glassful (250 mL) of this water sample.