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
Answers
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=
18
1000
=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
To find : the solubility of oxygen in 1 L of water at 20°C and 0.02 atm pressure.
where Henry's constant, KH = 4.6 × 10⁴ mol/Litre atm
Solution : solubility of gas in a liquid at a particular temperature is given by Henry's law.
i.e., P = x × KH
here , KH = 4.6 × 10⁴ atm , P = 0.02 atm
so, x = 0.02/4.6 × 10⁴ = 0.00435 × 10¯⁴ = 4.35 × 10^-7
now amount of water = 1L = 1000 ml
so, mass of water = density of water × volume = 1 g/ml × 1000 ml = 1000g
∴ no of moles of water = 1000/18 = 55.5 moles [ we know, molar mass of water is 18 g/mol ]
x = n₁/(n₁ + n₂)
here n₁ is no of moles of oxygen gas
n₂ is no of moles of water
x << 1 it means n₁ << n₂ so, x ≈ n₁/n₂
now, n₁/55.5 = 4.35 × 10^-7
⇒n₁ = 241.425 × 10^-7 = 2.41425 × 10^-5 mol of oxygen gas is present in 1000 ml of water so, solubility of water is 2.41425 × 10^-5 M ≈ 2.4 × 10^-5 M.
Therefore the solubility of oxygen in water is 2.4 × 10^-8 M.
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