Question

The concentration of 1 L of CaCO\(_3\) solution is \(10^{-5}\) M. Its concentration in ppm is. (Ca=40u; C=12u; O=16u)

• A. \(10\)
• B. \(1000\)
• C. \(100\)
• D. \(1\)

Hint:

When calculating ppm, ensure you convert the molar concentration to mass, and then use the formula to calculate ppm. Remember, ppm represents milligrams of solute per liter of solution.

Answer 1

The Correct Option is D

We know that the concentration in ppm (parts per million) is calculated as: \[ \text{ppm} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (L)}} \times 10^6 \] Given: - The molar mass of CaCO\(_3\) (Ca = 40 u, C = 12 u, O = 16 u) = 40 + 12 + 3(16) = 100 u
- Molar concentration = \(10^{-5}\) M
Thus, the mass of 1 mole of CaCO\(_3\) is 100 g. The mass of CaCO\(_3\) in 1 L of solution will be: \[ \text{Mass of CaCO\(_3\)} = 100 \times 10^{-5} = 0.001 \, \text{g} \] Now, calculating ppm: \[ \text{ppm} = \frac{0.001}{1} \times 10^6 = 1 \, \text{ppm} \] Thus, the concentration of CaCO\(_3\) in ppm is \(1\).