Henry's law constant of oxygen is 1.4 × 10-3 mol 1-1 atm -1 at 298 K. How much of oxygen is dissolved in 100 ml at 298 K when the partial pressure of oxygen is 0.5 atm?

Question

Henry's law constant of oxygen is 1.4 × 10-3 mol 1-1 atm -1 at 298 K. How much of oxygen is dissolved in 100 ml at 298 K when the partial pressure of oxygen is 0.5 atm? (A) 1.4 g (B) 3.2 g (C) 2.24 mg (D) 22.4 mg

Tardigrade Admin · Accepted Answer

C = K H ⋅ P O 2 =1.4 × 10-3 × 0.5=7 × 10-4 mol / L ∴ Number of moles of oxygen dissolved in 1000 ml =7 × 10-4 ∴ Number of moles of oxygen dissolved in 100 ml =(7 × 10-4 × 100/1000)=7 × 10-5 ∴ Weight of oxygen in 100 ml =7 × 10-5 × 32 =2.24 × 10-3 g =2.24 mg

Q. Henry's law constant of oxygen is 1.4×10−3mol1−1atm−1 at 298K. How much of oxygen is dissolved in 100ml at 298K when the partial pressure of oxygen is 0.5 atm?

1.4 g

3.2 g

2.24 mg

22.4 mg

C=KH​⋅PO2​​ =1.4×10−3×0.5=7×10−4mol/L ∴ Number of moles of oxygen dissolved in 1000ml =7×10−4 ∴ Number of moles of oxygen dissolved in 100ml =10007×10−4×100​=7×10−5 ∴ Weight of oxygen in 100ml =7×10−5×32 =2.24×10−3g=2.24mg

Q. Henry's law constant of oxygen is $1.4 \times 1 0^{- 3} m o l 1^{- 1} a t m^{- 1}$ at $298 K$ . How much of oxygen is dissolved in $100 m l$ at $298 K$ when the partial pressure of oxygen is 0.5 atm?