Question

22.
14. A certain public water supply contains 0.10 ppb (part per
billion) of chloroform (CHCI). How many molecules of
CHCI, would be obtained in 0.478 mL drop of this water?​

Answer 1

Answer:

The molecules obtained = 2.41 × 10^11

Explanation:

Amount of chloroform (CHCl₃) = 0.1 ppb = 0.1 μg/l

in 1000 mL, amount of chloroform = 0.1 μg

In 0.478 mL, amount of chloroform = (0.1/1000) × 0.478 μg

                                                          = 0.0478 × 10^(-9) gm

                                                         = 4.78 × 10^(-11) gm

Moles of chloroform = wt/molecular weight

                                 = 4.78 × 10^(-11) /119.38

                                 = 4 × 10^(-13) moles

In one mole, number of molecules = 6.022 × 10^23

Therefore in 4 × 10^(-13) moles no. of molecules = 6.022 × 10^23 × 4 × 10^(-13)

                                                                               = 2.41 × 10^11

Hope this answer helps.