A public water supply was found to have 0.8ppb by mass of chloroform, CHCl3. a) How many CHCl3 molecules would be present in a 350mL glass of this water? b)If the CHCl3 in part a could be isolated, would this quantity be detectable on an ordinary analytical balance that measures mass with a precision of +or- 0.0001g?
Given Conc CHCl₃ = 0.80ppb => [0.80 g CHCl₃ / 1 x 10⁹ g Water]∙100% = 8 x 10¯⁸ % w/w … Assuming 350 ml water = 350 grams...
=> 8 x 10¯⁸ % of 350-g = 2.8 x 10¯⁷ g CHCl₃ *
=>2.8 x 10¯⁷ g CHCl₃ = 2.8 x 10¯⁷ g CHCl₃/119 g/mole = 2.353 x 10¯⁹ mole CHCl₃
=> 2.353 x 10¯⁹ mole CHCl₃ = 2.353 x 10¯⁹ mole CHCl₃(6.022 x 10²³ molecules CHCl₃/moleCHCl₃)
=> 1.42 x 10¹⁵ molecules CHCl₃.
*If mass is in order of 10¯⁷ gram => not detectable by balance that accurate to only 10¯⁴ gram.
the mass of 350 mL of water is 350 g
0.8 ppb is ... .8 * .35 μg = .28 micrograms
not detectable on the balance in question
0.8ppb=0.8ng/mL
0.8ng/mL*(350mL)=280ng
Since ng is 10^-9g and the scale only measures to the 4th decimal place. I would say no.
a. 0.8 ppb CHCl3 = 0.8 ug/L CHCl3 so in 350 mL there is
0.8 ug/L x 0.350 L = approx 0.3 ug. How many mols is that?
1 mol CHCl3 = 119.5 so 0.3E-6 g/119.5 = approx 2.5E-8 mols. There are 6.022E23 molecules in a mol. # molecules = 6.022E23 molecules/mol x numb of mols = ?
b. you have 0.3E-6 grams.or 3E-7 or 0.0000003 grams. Your balance will weigh to 0.0001 g.
To answer both parts of the question, we need to use some basic chemistry and mathematical calculations.
a) To find the number of CHCl3 molecules in the given volume of water, we need to know the concentration of CHCl3 in parts per billion (ppb) and convert it to moles.
1. First, let's convert the volume of water from mL to liters:
350 mL = 350/1000 = 0.35 L
2. Next, we need to convert the concentration from ppb to moles per liter. Since 1 ppb represents 1 part of CHCl3 per billion parts, we can use the following conversion factors:
1 ppb = 1 molecule/1 billion molecules = 1/1,000,000,000 moles/L
Therefore, the concentration of CHCl3 in moles per liter would be:
0.8 ppb = 0.8/1,000,000,000 moles/L
3. Now, we can calculate the number of moles of CHCl3 in the given volume of water:
Moles of CHCl3 = concentration (moles/L) x volume (L)
Moles of CHCl3 = (0.8/1,000,000,000) x 0.35
4. Finally, we can calculate the number of CHCl3 molecules using Avogadro's number (6.022 x 10^23 molecules/mol):
Number of CHCl3 molecules = Moles of CHCl3 x Avogadro's number
b) To determine if the quantity of CHCl3 can be detected on an ordinary analytical balance, we need to compare the mass to the precision of the balance.
1. First, we need to calculate the mass of CHCl3 in grams:
Mass of CHCl3 = Moles of CHCl3 x molar mass of CHCl3
The molar mass of CHCl3 is:
Carbon (C): 12.01 g/mol
Hydrogen (H): 1.01 g/mol (3 H atoms)
Chlorine (Cl): 35.45 g/mol (1 Cl atom)
Molar mass of CHCl3 = 12.01 + (1.01 x 3) + 35.45 = 119.38 g/mol
Mass of CHCl3 = Moles of CHCl3 x 119.38
2. Now, we can compare the mass to the precision of the analytical balance:
Precision of the balance = +/- 0.0001 g
If the mass of CHCl3 falls within the range of +/- 0.0001 g, it would be detectable on the balance. Otherwise, it would not be detectable.
By following these steps, you can calculate the number of CHCl3 molecules in the water and determine whether its quantity would be detectable on an ordinary analytical balance.