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Q.

A certain public water supply contains 0.10 ppb (part per billion) of chloroform (CHCl3). How many molecules of CHCI3 would be obtained in 0.478 mL drop of this water ?(assumed d = 7 g/mL)

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a

4×10−13×NA

b

10−3×NA

c

4×10−10×NA

d

None of these

answer is A.

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Detailed Solution

0.47a mL =0.478 g of water; 109 g water contain 0.10 gCHCI3∴0.478 g water contain0.1109×0.478gCHCl3∴nCHCl3=0.1109×0.478119.5∴No. of molecules =0.1109×0.478119.5×NA=4×10−13×NA
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