Question

The henry's law constant of gaseous methyl bromide CH₃Br is k=0.159 mol/L -atm at 25^oC. what mass of gaseous CH₃Br willl dissolve in 225 mL of water at 25^oC and at a CH₃Br partial pressure of 250 mm Hg? ( Molar Mass CH₃Br=94.93)

please explain

A. 5.23 g

B. 3.40 g

C. 8.49 g

D. 22.1 g

E. 1.12 g

Answer 1

According to henry’s laW ,the amount of gas dissolved in any solvent is proportional to its partial pressure in the gaseous phase.

Mathematically,C=K*P

Where k=0.159 mol/L -atm =henry’s constant of gaseous methyl bromide

C=concentration of methyl bromide in water

P=partial pressure of methyl bromide=250 mm Hg=250mmHg/760 mmHg atm-1=0.329 atm

C=(0.159 mol/L –atm)*0.329 atm=0.0523 mol/L

The concentration of methyl bromide=0.0523 mol/L

Moles of methyl br in 225ml(0.225L)=0.0523 mol/L*0.225L=0.0118 mol

Mass=moles*molar mass=0.0118 mol*94.93 g/mol=1.120 g(option E)