Question

A public water supply contains 0.10 ppb (part per billion) of chloroform $\left(CHC{l}_3\right)$. How many molecules of $CHC{l}_3$ would be obtained from 0.478 mL 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 $4\times {10}^{-13}\times N_A$

Density of water = 1 g/mL
Weight of 0.478 mL of water = 0.478 g
Parts per billion $=\frac{\text{weight of}CHC{l}_3\left(g\right)}{\text{total weight of water (g)}}\times {10}^9\Rightarrow 0.10=\frac{\text{weight of}CHC{l}_3}{0.478}\times {10}^9$

Weight of $CHC{l}_3=0.10\times 0.478\times {10}^{-9}=4.78\times {10}^{-11}g$
Molar mass of $CHC{l}_3=119.5$ g/mol
119.5 g of $CHC{l}_3$ contains $N_{A}molecules$ .
$4.78\times {10}^{-11}$ g will contain = $\frac{N_A}{119.5}\times 4.78\times {10}^{-11}g=4\times {10}^{-13}\times N_A$
No. of molecules = $4\times {10}^{-13}\times N_A$