Question

A certain public water supply contains $0.10$ ppb (part per billion) of chloroform ($CHC{l}_3$). How many molecules of $CHC{l}_3$ would be obtained in $0.478$mL drop of this water?

• A. $4\times {10}^{-13}\times N_A$
• B. ${10}^{-3}\times N_A$
• C. $4\times {10}^{10}\times N_A$
• D. None of these

Answer 1

The correct option is A $4\times {10}^{-13}\times N_A$

$0.478mL\equiv 0.478g$ of water ; ${10}^{9}g$ water contain $0.10gCHC{l}_3$
$\therefore 0.478g$ water contain
$\frac{0.1}{{10}^9}\times 0.478gCHC{l}_3$
$\therefore$ $n_{CHC{l}_3}=\frac{0.1}{{10}^9}\times \frac{0.478}{119.5}$
$\therefore$ No. of molecules $=\frac{0.1}{{10}^9}\times \frac{0.478}{119.5}\times N_A$
$=4\times {10}^{-3}\times N_A$