Question

A sample of drinking water was found to be severely contaminated with chloroform. The level of contamination was 15 ppm (by mass) i) express this in percent by mass. Ii) determine the molality of chloroform in this water sample.

Answer 1

Answer :

• Mass percent of chloroform = 1.5 × 10⁻³ %.
• Molality of chloroform in sample = 1.255 × 10⁻⁴

Step-by-step explanation:

Given that,

• Level of contamination = 15 ppm (by mass).

Also,

• Molar mass of chloroform = 119.5 g/mol.

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(i) 15 ppm means 15 parts in million (10⁶) parts.

Therefore, Mass % of chloroform =

$\implies\rm{\dfrac{15}{10^6}\times{}100}$

$\implies\rm{15\times{}10^{-4}=}\:\bold{1.5\times{}10^{-3}{\%}.}$

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(ii) Molar mass of chloroform (given) = 119.5 g/mol

Now, 100 g of the sample contain = 1.5 × 10⁻³ g chloroform.

Then, 1000 g of the sample will contain =1.5 × 10⁻² g chloroform.

We know,

$\rm{Molality=\dfrac{W_B\times{}1000}{M_B\times{}W_A}}$

Putting the values, we get,

$\implies\rm{\dfrac{1.5\times{}10^{-2}\times{}1000}{119.5\times{}1000}}$

$\implies\bold{1.255\times{}10^{-4}\:m.}$