Question

7. A sample of drinking water is contaminated with chloroform. If the level of contamination was 20
ppm (by mass) then this contamination is equivalent to *
(4/4 Points)
(1) 0.7 percentage by mass
13
12) 0.7 percentage by volume
(B) 0.000167 molal
(4) 0.0167 molal​

Answer 1

Answer:

Contamination is equivalent to $1.255 \times 10^{-4} \mathrm{~m}$.

Explanation:

20 ppm corresponds to $20 \mathrm{~g}$ chloroform in $1000,000 \mathrm{~g}$ of solution.

(1) The given amount of chloroform is $20 \mathrm{ppm}$ (by mass). So, 20 gram chloroform is present in $10^6$ gram of solution

The formula of percent by mass is as follows:

$\text { percent mass }=\frac{\text { mass of chloroform }}{\text { mass of solution }} \times 100$

Substitute 20 gram for mass of chloroform and $10^6 \mathrm{gram}$ for mass of solution

$\begin{aligned}& \text { percent mass }=\frac{20}{10^6} \times 100 \\& \text { percent mass }=1.5 \times 10^{-3}\end{aligned}$

Therefore, the percent by mass is $1.5 \times 10^{-3} \%$.

(2) We will use the mole formula to determine the mole of chloroform, present in 20 gram.

The mole formula is as follows:

$\text { Mole }=\frac{\text { Mass }}{\text { Molar mass }}$

The molar mass of the chloroform is $119.5 \mathrm{~g} / \mathrm{mol}$.

Substitute $119.5 \mathrm{~g} / \mathrm{mol}$ for molar mass and 20 gram for mass of iron.

$\begin{aligned}& \text { mole }=\frac{20 \mathrm{~g}}{119.5 \mathrm{~g} / \mathrm{mol}} \\& \text { mole }=0.1255\end{aligned}$

So, 20 grams of chloroform have 0.1255 mole chloroform.

We will determine the mass of solvent by subtracting the mass of solute (chloroform) from mass of solution.

$\begin{aligned}& \text { mass of solvent }=10^6 \mathrm{~g}-20 \mathrm{~g} \\& \Rightarrow \text { mass of solvent }=999985 \mathrm{~g}\end{aligned}$

Convert the mass of solvent form gram to $\mathrm{kg}$ as follows:

$\begin{aligned}& 1000 \mathrm{~g}=1 \mathrm{~kg} \\& \Rightarrow 999985 \mathrm{~g}=999.985 \mathrm{~kg}\end{aligned}$

Use the molality formula to determine the molality of chloroform solution as follows:

$Molality$$=\frac{\text { Moles of solute }}{\text { kgof solvent }}$

Substitute $999.985 \mathrm{~kg}$ for $\mathrm{kg}$ of solvent and 0.1255 for mole of solute.

$\begin{aligned}& \text { Molality }=\frac{0.1255 \mathrm{~mol}}{999.985 \mathrm{~kg}} \\& \Rightarrow \text { Molality }=1.255 \times 10^{-4} \mathrm{~m}\end{aligned}$

Therefore, the molality of chloroform in the water sample $1.255 \times 10^{-4} \mathrm{~m}$.

To learn more about similar question visit:

https://brainly.in/question/10387230

https://brainly.in/question/10726231

#SPJ1