Question

A certain public water supply contains 0.10 ppb of chloroform . How many molecules of CHCL3 would be obtained in 0.478 ml drop of this water

Answer 1

Dear Student,

◆ Answer -

N(chloroform) = 2.4088×10^14 molecules

◆ Explanation -

Mass of water is -

W(Water) = 0.478 × 1

W(Water) = 0.478 g

Amount of chloroform in this water is -

W(Chloroform) = W(Water) × 0.1 ppb

W(Chloroform) = 0.478 × 0.1×10^-6

W(Chloroform) = 4.78×10^-8 g

No of chloroform molecules in this water is -

N(chloroform) = NA × W(chloroform)/M(chloroform)

N(chloroform) = 6.022×10^23 × 4.78×10^-8 / 119.5

N(chloroform) = 2.4088×10^14 molecules

Therefore, 2.4088×10^14 molecules will be present in 0.478 ml of water.

Hope this helps you...

Answer 2

Answer:

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