Q.A sample of drinking water was found to be severely contaminated with chloroform, $CHCl_3$, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

The problem involves converting concentration units: from parts per million (ppm) to mass percentage, and then to molality.

Part (i): Express 15 ppm in per cent by mass.

1. Understand ppm: 15 ppm by mass means 15 grams of chloroform ($CHCl_3$) are present in $10^6$ grams of the water sample (solution).
2. Apply mass percentage formula: Per cent by mass is calculated as $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$.
3. Calculate:

Part (ii): Determine the molality of chloroform in the water sample.

1. Identify solute and solvent masses: From 15 ppm, we have 15 g of $CHCl_3$ (solute) in $10^6$ g of solution.
   • Mass of solvent (water) = Mass of solution - Mass of solute $= 10^6 \text{ g} - 15 \text{ g} = 999985 \text{ g} = 999.985 \text{ kg}$.
2. Calculate moles of solute:
   • Molar mass of $CHCl_3 = 12.011 (\text{C}) + 1.008 (\text{H}) + 3 \times 35.453 (\text{Cl}) = 119.378 \text{ g/mol}$. …

When dealing with very dilute solutions, like contaminants in drinking water, expressing concentration as a percentage can lead to extremely small, inconvenient numbers. This is why units like parts per million (ppm) or parts per billion (ppb) are used. These units essentially scale up the percentage idea to make the numbers more manageable.

Parts per million (ppm), when specified "by mass," means there are a certain number of mass units of solute for every million mass units of solution. For example, 15 ppm by mass means 15 grams of chloroform in $10^6$ grams of the water sample.

Mass percentage is simply the mass of the solute divided by the total mass of the solution, multiplied by 100. It's a direct way to express the proportion of a component in a mixture.

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which uses the volume of the solution, molality uses the mass of the solvent. This makes molality independent of temperature, as mass does not change with temperature, whereas volume does. This is particularly useful in colligative properties calculations.

Let's break down the problem into two parts.

Part (i): Express 15 ppm (by mass) in per cent by mass.

1. Understand ppm by mass:

   The statement "15 ppm (by mass)" means that there are 15 parts by mass of chloroform for every $10^6$ parts by mass of the total solution. We can interpret "parts" as any consistent mass unit, such as grams.

   So, we have 15 g of $CHCl_3$ in $10^6$ g of the water sample.

2. Recall the formula for mass percentage:

   Mass percentage of a component $= \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100\%$

3. Apply the formula:

   Using the interpretation from step 1:

   Mass of $CHCl_3 = 15 \text{ g}$

   Mass of solution $= 10^6 \text{ g}$

   Mass percentage of $CHCl_3 = \frac{15 \text{ g}}{10^6 \text{ g}} \times 100\%$

   Mass percentage of $CHCl_3 = 15 \times 10^{-6} \times 100\%$

   Mass percentage of $CHCl_3 = 15 \times 10^{-4}\%$

   Mass percentage of $CHCl_3 = \mathbf{0.0015\%}$

Part (ii): Determine the molality of chloroform in the water sample.

To determine molality, we need two pieces of information:

• Moles of solute ($CHCl_3$)
• Mass of solvent (water) in kilograms

We will continue to use the information that the contamination level is 15 ppm by mass.

1. Assume a convenient mass of solution:

   Let's assume we have $10^6 \text{ g}$ (or 1000 kg) of the water sample. This makes it easy to work with the ppm definition.

2. Calculate the mass of chloroform ($CHCl_3$) in the assumed solution:

   From the 15 ppm (by mass) information, if we have $10^6 \text{ g}$ of solution, then the mass of chloroform is:

   Mass of $CHCl_3 = 15 \text{ g}$

3. Calculate the molar mass of chloroform ($CHCl_3$):

   We need the atomic masses of Carbon (C), Hydrogen (H), and Chlorine (Cl).

   Atomic mass of C $\approx 12.01 \text{ g/mol}$

   Atomic mass of H $\approx 1.008 \text{ g/mol}$

   Atomic mass of Cl $\approx 35.45 \text{ g/mol}$

   Molar mass of $CHCl_3 = (1 \times 12.01) + (1 \times 1.008) + (3 \times 35.45) \text{ g/mol}$

   Molar mass of $CHCl_3 = 12.01 + 1.008 + 106.35 \text{ g/mol}$

   Molar mass of $CHCl_3 = 119.368 \text{ g/mol}$

   We can round this to $119.37 \text{ g/mol}$ for calculations.

4. Calculate the moles of chloroform:

   Moles of $CHCl_3 = \frac{\text{Mass of } CHCl_3}{\text{Molar mass of } CHCl_3}$ …

Method: Unit Conversion & Molality Calculation

This problem uses ppm to mass percent conversion followed by molality formula.

(i) Express 15 ppm as per cent by mass

Concept:

ppm (parts per million) means parts of solute per million parts of solution.

Percent means parts per hundred.

Steps:

1. Recall the relation:

   $1 \text{ ppm} = \frac{1 \text{ g solute}}{10^6 \text{ g solution}}$

2. Convert ppm to fraction:

   $15 \text{ ppm} = \frac{15}{10^6}$

3. Convert fraction to percent (multiply by 100):

   $\frac{15}{10^6} \times 100 = \frac{15}{10^4} = 1.5 \times 10^{-3}$

Answer:

Percent by mass = $1.5 \times 10^{-3}\%$

(ii) Determine the molality of chloroform

Concept:

Molality ($m$) = moles of solute per kg of solvent.

Steps:

1. Assume 1000 g (1 kg) of solution for easy calculation.

   • Mass of chloroform = $15 \text{ ppm} \times 1 \text{ kg} = 15 \times 10^{-6} \times 1000 \text{ g} = 0.015 \text{ g}$

2. Find molar mass of $CHCl_3$:

   $C = 12,\ H = 1,\ Cl = 35.5 \times 3 = 106.5$

   Molar mass = $12 + 1 + 106.5 = 119.5 \text{ g/mol}$

3. Calculate moles of chloroform:

   $\text{Moles} = \frac{0.015}{119.5} \approx 1.255 \times 10^{-4} \text{ mol}$ …

Here are the most common mistakes students make when solving this problem, along with the conceptual fixes to avoid them.

Mistake 1: Confusing ppm with a direct percentage

The Error: Students often think 15 ppm means 15% by mass, or they try to convert by simply dividing by 100.

Why it happens: They don't internalize that ppm (parts per million) is a ratio of masses, not a direct count. The "million" refers to the total mass of the solution, not the number of particles.

How to avoid it:

• Remember the definition: $1 \text{ ppm} = \frac{1 \text{ g solute}}{10^6 \text{ g solution}}$.
• To convert ppm to percent, divide by $10^4$ (because percent is parts per hundred, and $10^6 / 10^4 = 100$).
• Correct calculation:

So the answer is $1.5 \times 10^{-3} \%$ (or 0.0015%).

Mistake 2: Using the wrong molar mass for chloroform ($CHCl_3$)

The Error: Students forget to add the mass of all three chlorine atoms, or they use the atomic mass of Carbon as 12.01 but round Chlorine incorrectly.

Why it happens: Rushing through the formula or not double-checking the periodic table values.

How to avoid it:

• Write out the atomic masses clearly:
  • C = 12.01 g/mol
  • H = 1.008 g/mol
  • Cl = 35.45 g/mol (three atoms)
• Correct molar mass:

Round to 119.37 g/mol for calculations.

Mistake 3: Forgetting that ppm is based on the total solution mass

The Error: Students treat the 15 ppm as if it refers to 15 g of solute in 1,000,000 g of solvent (water), not the solution.

Why it happens: They confuse ppm with molality (which uses kg of solvent) or they don't read the problem carefully ("by mass" means solution mass).

How to avoid it:

• For ppm, the denominator is total mass of solution (solute + solvent).
• For molality, the denominator is mass of solvent only (in kg).
• Correct approach for molality:
  • In 15 ppm, there are 15 g of $CHCl_3$ in $10^6$ g of solution.
  • Mass of solvent (water) = $10^6 - 15 \approx 10^6$ g (since 15 is negligible).
  • Convert to kg: $10^6 \text{ g} = 1000 \text{ kg}$.
  • Moles of $CHCl_3$ = $\frac{15}{119.37} \approx 0.1257 \text{ mol}$.
  • Molality:

Mistake 4: Not simplifying the approximation for solvent mass

The Error: Students try to subtract 15 from 1,000,000 exactly, then get confused about significant figures or waste time. …