If O2 gas is bubbled through water at 303 K, the number of millimoles of O2 gas that dissolve in 1 litre of water is_______. (Nearest integer)
(Given : Henry’s Law constant for O2 at 303 K is 46.82 k bar and partial pressure of O2 = 0.920 bar)
(Assume solubility of O2 in water is too small, nearly negligible)
(Given : Henry’s Law constant for O2 at 303 K is 46.82 k bar and partial pressure of O2 = 0.920 bar)
(Assume solubility of O2 in water is too small, nearly negligible)
Correct Answer: 1
Approach Solution - 1
To find the number of millimoles of O2 that dissolve in 1 litre of water, we employ Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically, it is expressed as:
P = kH × x
where:
- P is the partial pressure of the gas (0.920 bar).
- kH is the Henry's Law constant (46.82 k bar = 46820 bar, since 1 k bar = 1000 bar).
- x is the mole fraction of the gas in solution.
We rearrange the formula to solve for x:
x = P / kH = 0.920 / 46820
Calculating x gives us:
x ≈ 1.965 × 10-5
For ideal dilute solutions, the molality is approximately equal to the mole fraction when the solubility is negligible. Thus, we can find the number of moles of O2 by multiplying x by the number of moles of water in 1 litre. The molar mass of water is approximately 18 g/mol, and 1 litre of water is about 1000 g, equivalent to (1000 g) / (18 g/mol) ≈ 55.56 moles.
Therefore, the number of moles of O2 is approximately:
moles of O2 = x × 55.56 ≈ 1.092 × 10-3
To convert to millimoles, multiply by 1000:
millimoles = 1.092 × 10-3 × 1000 = 1.092
Rounding to the nearest integer, we find that the number of millimoles of O2 is:
1
Approach Solution -2
According to Henry’s law,
\(X(\text{oxygen}) = \frac{p(\text{oxygen})}{K_H} = \frac{0.920}{46.82 \times 10^3} = 1.96 \times 10^{-5}\)
Since, 1 litre of water contains 55.5 mol of it,
therefore,\(→ n \) represents moles of O2 in solution.
\(X(\text{oxygen}) = \frac{n}{n + 55.5} \approx \frac{n}{55.5}\)
\(\frac{n}{55.5} = 1.96 \times 10^{-5}\)
\(n = 108.8 \times 10^{-5} = 1.08 \times 10^{-3} \, \text{moles}\)
m moles of oxygen = 1.08 × 10–3 × 103= 1 m mole
So, the answer is 1.
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.