Question

What will be the molality of chloroform in the water sample which contains 15 ppm chloroform by mass?

• A. $1.25\times {10}^{-4}m$
• B. $2.5\times {10}^{-4}m$
• C. $1.5\times {10}^{-3}m$
• D. $1.25\times {10}^{-5}m$

Answer 1

Answer: (A)

$1.25\times {10}^{-4}m$

The concentration of chloroform=15 ppm

$\therefore$ 10^6 g of solution contains 15 g chloroform

Thus, 1000g of solution contains $\frac{15}{{10}^6}\times 1000=15\times {10}^{-3}g$ chloroform

Molality=$\frac{numberofmolesofsolute}{massofsolvent\left(kg\right)}$

and number of moles = $\frac{mass}{molarmass}$

For chloroform, $CHC{l}_3$

mass$=15\times {10}^{-3}g$,

molar mass=$12+1+3\times 35.5=119.5g/mol$

Weight of water = 1000g = 1 kg

Molality = $\frac{15\times {10}^{-3}}{119.5\times 1}=1.25\times {10}^{-4}m$

Thus, option $A$ is correct.