Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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.