Some amount of dichloromethane ( CH 2 Cl 2) is added to 671.141 mL of chloroform ( CHCl 3) to prepare 2.6 × 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

Question

Some amount of dichloromethane ( CH 2 Cl 2) is added to 671.141 mL of chloroform ( CHCl 3) to prepare 2.6 × 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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Molarity =( text mole / text volume ) 2.6 × 10-3 =( x / 85/0.67141) x =0.148 g conc. Fo DCM in ppm =(0.148/1.49 × 671.141) × 106 =148 ppm

Q. Some amount of dichloromethane (CH2​Cl2​) is added to 671.141mL of chloroform (CHCl3​) to prepare 2.6×10−3M solution of CH2​Cl2​(DCM). The concentration of DCM is____ppm(by mass). Given: Atomic mass : C=12;H:1;Cl=35.5 density of CHCl3​=1.49gcm−3

Molarity = volume mole ​ 2.6×10−3=0.67141x/85​ x=0.148g conc. Fo DCM in ppm =1.49×671.1410.148​×106 =148ppm

Molarity $= \frac{mole}{volume}$
$2.6 \times 1 0^{- 3} = \frac{x /85}{0.67141}$
$x = 0.148 g$
conc. Fo DCM in ppm $= \frac{0.148}{1.49 \times 671.141} \times 1 0^{6}$
$= 148 pp m$