Some amount of dichloromethane `(CH_(2)Cl_(2))` is added to `671.141mL` of chloroform `(CHCl_(3))` to prepare `2.6times10^(-3)M` solution of `CH_(2)Cl_(2) (DCM)`.The concentration of `DCM` is........ Given : atomic mass : `C=12,H=1,Cl=35.5` density of `CHCl_(3)=1.49gcm^(-3)`

Answer

To solve the problem, we need to determine the concentration of dichloromethane (DCM) in parts per million (ppm) after it is added to chloroform (CHCl₃). Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the mass of chloroform (CHCl₃)
We know the density of chloroform and the volume provided in the question. The formula to calculate mass from density and volume is:

\[ \text{Mass} = \text{Density} \times \text{Volume} \]

Given:
- Density of CHCl₃ = 1.49 g/cm³
- Volume of CHCl₃ = 671.141 mL = 671.141 cm³ (since 1 mL = 1 cm³)

Now, substituting the values:

\[ \text{Mass of CHCl}_3 = 1.49 \, \text{g/cm}^3 \times 671.141 \, \text{cm}^3 = 1000 \, \text{g} \]

### Step 2: Calculate the number of moles of dichloromethane (CH₂Cl₂)
We are given the molarity of the DCM solution and the volume of chloroform. The formula for molarity (M) is:

\[ M = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Rearranging for the number of moles gives us:

\[ \text{Number of moles} = M \times \text{Volume in liters} \]

Given:
- Molarity of DCM = \(2.6 \times 10^{-3} \, \text{M}\)
- Volume of CHCl₃ = 671.141 mL = 0.671141 L

Now substituting the values:

\[ \text{Number of moles of CH}_2\text{Cl}_2 = 2.6 \times 10^{-3} \, \text{mol/L} \times 0.671141 \, \text{L} = 1.74 \times 10^{-3} \, \text{mol} \]

### Step 3: Calculate the molar mass of dichloromethane (CH₂Cl₂)
To find the molar mass, we sum the atomic masses of all the atoms in the molecule:

\[ \text{Molar mass of CH}_2\text{Cl}_2 = (1 \times 12) + (2 \times 1) + (2 \times 35.5) = 12 + 2 + 71 = 85 \, \text{g/mol} \]

### Step 4: Calculate the mass of dichloromethane (CH₂Cl₂)
Using the number of moles and the molar mass, we can find the mass:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

Substituting the values:

\[ \text{Mass of CH}_2\text{Cl}_2 = 1.74 \times 10^{-3} \, \text{mol} \times 85 \, \text{g/mol} = 0.1479 \, \text{g} = 147.9 \, \text{mg} \]

### Step 5: Calculate the concentration of DCM in ppm
To convert the mass of DCM to ppm, we use the formula:

\[ \text{ppm} = \left( \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \right) \times 10^6 \]

The mass of the solution is the mass of chloroform (1000 g) plus the mass of DCM (0.1479 g):

\[ \text{Mass of solution} = 1000 \, \text{g} + 0.1479 \, \text{g} = 1000.1479 \, \text{g} \]

Now substituting the values into the ppm formula:

\[ \text{ppm} = \left( \frac{0.1479 \, \text{g}}{1000.1479 \, \text{g}} \right) \times 10^6 \approx 148.32 \, \text{ppm} \]

### Final Answer:
The concentration of DCM is approximately **148.32 ppm**.

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