Virat Batch 2027
Champion Batch 2028
Divine JEE 1-on-1
Some amount of dichloromethane (CH2Cl2)\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)(CH2Cl2) is added to 671.141 mL671.141 \mathrm{~mL}671.141 mL of chloroform (CHCl3)\left(\mathrm{CHCl}_{3}\right)(CHCl3) to prepare 2.6×10−3M2.6 \times 10^{-3} \mathrm{M}2.6×10−3M solution of CH2Cl2(DCM)\mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{DCM})CH2Cl2(DCM). The concentration of DCM\mathrm{DCM}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}CHCl3=1.49 g cm−3
Mass of CHCl3=671.141×1.49=1000\mathrm{CHCl_3=671.141\times1.49=1000}CHCl3=671.141×1.49=1000 gm
2.6 × 10 − 3 = moles of CH2Cl2 0.671141\\2.6 \times {10^{ - 3}} = {{moles\,of\,C{H_2}C{l_2}} \over {0.671141}}2.6 × 10 − 3 = 0.671141molesofCH2Cl2
⇒\\ \Rightarrow ⇒ moles of CH2Cl2 = 1.74496 × 10 − 3C{H_2}C{l_2} = 1.74496 \times {10^{ - 3}}\\CH2Cl2 = 1.74496 × 10 − 3
mass of CH2Cl2 = 148.32 × 10 − 3C{H_2}C{l_2} = 148.32 \times {10^{ - 3}}CH2Cl2 = 148.32 × 10 − 3 gm
Composition of CH2Cl2 = 148.32 × 10 − 3 1000 × 106C{H_2}C{l_2} = {{148.32 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}CH2Cl2 = 1000148.32 × 10 − 3 × 106
= 148.32 ppm
≈\approx≈ 148
Select an option to instantly check whether it is correct or wrong.
Learn on the go with our mobile app. Access courses, live classes, and study materials anytime, anywhere.
Empowering students to achieve their IIT dreams through personalized learning and expert guidance.
Copyright © 2026 Divine JEE. All Rights Reserved.
