Some amount of dichloromethane CH$$_2$$Cl$$_2$$ is added to 671.141 mL of chloroform CHCl$$_3$$ to prepare $$2.6 \times 10^{-3}$$ M solution of CH$$_2$$Cl$$_2$$ DCM. The concentration of DCM is ______ ppm (by mass).
Given: Atomic mass: C = 12; H: 1; Cl = 35.5 density of CHCl$$_3$$ = 1.49 g cm$$^{-3}$$

The concentration in ppm by mass is given by

$$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times10^6$$

The mass of chloroform is

$$\text{Mass of }CHCl_3=671.141\times1.49\approx1000\ \text{g}$$

Thus, the mass of the solvent is approximately $1000\ \text{g}$.

The molar mass of $CH_2Cl_2$ is

$$12+(2\times1)+(2\times35.5)=85\ \text{g mol}^{-1}$$

Since the solution is very dilute, its volume is approximately equal to the solvent volume.

$$V=671.141\ \text{mL}=0.671141\ \text{L}$$

The moles of $CH_2Cl_2$ are

$$n=MV=(2.6\times10^{-3})\times0.671141\approx1.745\times10^{-3}\ \text{mol}$$

Hence, the mass of $CH_2Cl_2$ is

$$m=n\times M=(1.745\times10^{-3})\times85\approx0.1483\ \text{g}$$

Substituting into the ppm expression,

$$\text{ppm}=\frac{0.1483}{1000}\times10^6=148.3$$

Therefore, the concentration of the solution is

$$\boxed{148\ \text{ppm}}$$