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QJEEM23Chdab3: JEE Main 2023 Chemistry - Class 11 JEEM23Chdab3 Solution: Some amount of dichloromethane is added to of chloroform to…

Some amount of dichloromethane (CH2Cl2)\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right) is added to 671.141 mL671.141 \mathrm{~mL} of chloroform (CHCl3)\left(\mathrm{CHCl}_{3}\right) to prepare 2.6×10−3M2.6 \times 10^{-3} \mathrm{M} solution of CH2Cl2(DCM)\mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{DCM}) . The concentration of DCM\mathrm{DCM} is ___________ ppm (by mass).

Given :

atomic mass : C = 12

H = 1

Cl = 35.5

density of CHCl3=1.49 g cm−3\mathrm{CHCl}_{3}=1.49 \mathrm{~g} \mathrm{~cm}^{-3}

Step-by-Step Verified Solution

Mass of CHCl3=671.141×1.49=1000\mathrm{CHCl_3=671.141\times1.49=1000} gm

2.6×10−3=moles of CH2Cl20.6711412.6 \times {10^{ - 3}} = {{moles\,of\,C{H_2}C{l_2}} \over {0.671141}}

⇒\Rightarrow moles of CH2Cl2=1.74496×10−3C{H_2}C{l_2} = 1.74496 \times {10^{ - 3}}

mass of CH2Cl2=148.32×10−3C{H_2}C{l_2} = 148.32 \times {10^{ - 3}} gm

Composition of CH2Cl2=148.32×10−31000×106C{H_2}C{l_2} = {{148.32 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}

= 148.32 ppm

≈\approx 148

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