1.

At 20°C and 0.02 atm pressure, what is the solubility of oxygen in 1L of water. Given KH for oxygen is 4.6 x 10^4 atm

Answer» ACCORDING to Henry's law, P=K H ×x∴x O 2 = K H P O 2 = 4.34×10 4 0.2 =4.6×10 −6 Moles of WATER= 181000 =55.5mol∴x O 2 = N H 2 O +n O 2 n O 2 ≈ n H 2 O n O 2 ⇒n O 2 =4.6×10 −6 ×55.5=2.55×10 −4 mole∴ Molarity = 2.55×10 −4 M


Discussion

No Comment Found

Related InterviewSolutions