Question

Some amount of dichloromethane $\left( CH _2 Cl _2\right)$ is added to $671.141 \,mL$ of chloroform $\left( CHCl _3\right)$ 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 $CHCl _3=1.49\, g \,cm ^{-3}$

Answer 1

Correct Answer: 221

Step 1: Calculate the Molar Mass of \(\text{CH}_2\text{Cl}_2\)
\[\text{Molar Mass of } \text{CH}_2\text{Cl}_2 = (12) + 2(1) + 2(35.5) = 12 + 2 + 71 = 85 \, \text{g mol}^{-1}.\]
Step 2: Calculate the Mass of \(\text{CH}_2\text{Cl}_2\)
The molarity (\(M\)) is given by:
\[M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}}.\]
Substituting \(M = 2.6 \times 10^{-3}\) and \(\text{Volume of solution} = 671.141 \, \text{mL} = 0.671141 \, \text{L}\):
\[\text{Moles of solute} = M \times \text{Volume (in L)} = 2.6 \times 10^{-3} \times 0.671141 = 1.745 \times 10^{-3} \, \text{moles}.\]
Mass of \(\text{CH}_2\text{Cl}_2\) is:
\[\text{Mass} = \text{Moles} \times \text{Molar Mass} = 1.745 \times 10^{-3} \times 85 = 0.148 \, \text{g}.\]
Step 3: Calculate the Total Mass of the Solution
Density of \(\text{CHCl}_3\) is given as \(1.49 \, \text{g cm}^{-3}\). Volume of \(\text{CHCl}_3 = 671.141 \, \text{mL}\):
\[\text{Mass of } \text{CHCl}_3 = \text{Density} \times \text{Volume} = 1.49 \times 671.141 = 1000.0 \, \text{g}.\]
Total mass of the solution:
\[\text{Total Mass} = \text{Mass of solute} + \text{Mass of solvent} = 0.148 + 1000.0 = 1000.148 \, \text{g}.\]
Step 4: Calculate Concentration in ppm
\[\text{Concentration (ppm)} = \frac{\text{Mass of solute (g)}}{\text{Total Mass of solution (g)}} \times 10^6 = \frac{0.148}{1000.148} \times 10^6.\]
Simplifying:
\[\text{Concentration (ppm)} = 0.148 \times 10^3 = 221 \, \text{ppm}.\]