Question

A sample of drinking water has 15 ppm (by mass) of a carcinogen (molar mass 120 g mol\(^{-1}\)). The molality of carcinogen in water sample is:

• A. \(2.50 \times 10^{-4}\)
• B. \(2.50 \times 10^{-3}\)
• C. \(1.25 \times 10^{-4}\)
• D. \(1.25 \times 10^{-3}\)

Hint:

1 ppm = 1 mg per kg of solution. Always convert mg → g before calculating moles.

Answer 1

The Correct Option is C

Step 1: Understanding ppm definition.
15 ppm means 15 mg of solute per 1 kg of solution.

Step 2: Converting mass of solute.
\[ 15 \text{ mg} = 0.015 \text{ g} \]
So in 1 kg water, solute = 0.015 g.

Step 3: Calculating moles of carcinogen.
Molar mass = 120 g/mol: \[ n = \frac{0.015}{120} = 1.25 \times 10^{-4} \text{ mol} \]

Step 4: Using definition of molality.
Molality \(m = \frac{\text{moles of solute}}{\text{kg of solvent}}\): \[ m = \frac{1.25 \times 10^{-4}}{1} \]

Step 5: Final result.
Thus molality = \(1.25 \times 10^{-4} \, \text{mol kg}^{-1}\).
Final Answer: \[ \boxed{1.25 \times 10^{-4}} \]