Some amount of dichloromethane $\left(\mathrm{CH_2Cl_2}\right)$ is added to 671.141 mL of chloroform $\left(\mathrm{CHCl_3}\right)$ to prepare a $2.6 \times 10^{-3}\ m$ (molal) solution of $\mathrm{CH_2Cl_2}$ (DCM). The concentration of DCM is ______ ppm (by mass). Given: Atomic masses: $\mathrm{C} = 12$, $\mathrm{H} = 1$, $\mathrm{Cl} = 35.5$; density of $\mathrm{CHCl_3} = 1.49\ \mathrm{g\ cm^{-3}}$

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Some Basic Concepts of Chemistry

Some amount of dichloromethane

(C H_{2} C l_{2})

is added to 671.141 mL of chloroform

(CHC l_{3})

to prepare a

2.6 \times 1 0^{- 3} m

(molal) solution of

C H_{2} C l_{2}

(DCM). The concentration of DCM is ______ ppm (by mass).

Given: Atomic masses:

C = 12

H = 1

Cl = 35.5

; density of

CHC l_{3} = 1.49 g c m^{- 3}

Answer: 221

The mass of chloroform solvent is

1.49 \times 671.141 = 1000 g

, that is exactly

1 kg

. A molality of

2.6 \times 1 0^{- 3}

then means

2.6 \times 1 0^{- 3} mol

of dichloromethane. With a molar mass of

12 + 2 + 71 = 85 g mo l^{- 1}

, this is

2.6 \times 1 0^{- 3} \times 85 = 0.221 g

. The mass fraction in parts per million is

\frac{0.221}{1000.221} \times 1 0^{6} \approx 221

. The concentration of dichloromethane is

221 ppm

by mass.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer