Some amount of dichloromethane (CH₂ Cl₂) is added to 671.141 mL of chloroform (CHCl₃) to prepare 2.6 × 10⁻³ M solution of CH₂ Cl₂(DCM). The concentration of DCM is ___________ ppm (by mass).
Given :
atomic mass : C = 12
H = 1
Cl = 35.5
density of CHCl₃=1.49 g cm⁻³

Solution

Chemistry ma’am · 4 steps

The solution is very dilute, so its volume is the chloroform volume, and that chloroform weighs exactly 1 kg.

1. Moles of DCM. At 2.6 × 10⁻³ M the solution is very dilute, so take its volume as the chloroform volume, 0.671141 L. Moles of DCM = 2.6 × 10⁻³ × 0.671141 = 1.745 × 10⁻³ mol.
2. Mass of DCM. Molar mass of CH₂Cl₂ = 12 + 2 + 71 = 85 g mol⁻¹. Mass of DCM = 1.745 × 10⁻³ × 85 = 0.1483 g.
3. Mass of chloroform. Mass = volume × density = 671.141 × 1.49 = 1000 g. That odd volume was chosen so the solvent is exactly 1 kg.

   ppm = (mass of solute)/(mass of solution) × 10⁶ = 0.1483/1000 × 10⁶ ≈ 148.3

4. So. The DCM is a tiny part of the total mass, so adding it to the 1000 g barely changes the answer. The concentration is 148 ppm by mass.

The common slip: 221. Taking 671.141 mL as 671.141 g of solvent skips the density and gives 221, not 148.

The idea to keep. ppm by mass is grams of solute per million grams of solution. Turn every volume into a mass with the density first. For a very dilute solution, solution volume ≈ solvent volume.

Answer from the NTA final answer key. Solution by PrepWiser.