Calculate the concentration of dissolved gas in water at $25^\circ\text{C}$ if partial pressure of gas at same temperature is 0.15 atm . $[\text{K}_\text{H} = 0.15 \text{ moldm}^{-3} \text{atm}^{-1}]$}
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$S = K_H P$. It's a simple multiplication of the constant and the pressure.
Step 1: Understanding the Concept:
Henry's Law states that the solubility (concentration) of a gas in a liquid is directly proportional to the partial pressure of that specific gas located above the surface of the liquid. Step 2: Key Formula or Approach:
The mathematical form of Henry's Law utilized based on the given units is:
\[ C = \text{K}_\text{H} \times P \]
where:
$C$ = concentration of the dissolved gas (solubility)
$\text{K}_\text{H}$ = Henry's law constant
$P$ = partial pressure of the gas Step 3: Detailed Explanation:
Given values from the problem statement:
Partial pressure ($P$) = $0.15 \text{ atm}$
Henry's law constant ($\text{K}_\text{H}$) = $0.15 \text{ moldm}^{-3} \text{atm}^{-1}$
Substitute the numerical values directly into the law's equation:
\[ C = \left(0.15 \text{ moldm}^{-3} \text{atm}^{-1}\right) \times (0.15 \text{ atm}) \]
\[ C = 0.0225 \text{ moldm}^{-3} \]
Since standard volume conversions state that $1 \text{ dm}^3 = 1 \text{ L}$, the concentration unit $\text{moldm}^{-3}$ is exactly equivalent to standard Molarity ($\text{M}$).
\[ C = 0.0225 \text{ M} \]
Step 4: Final Answer:
The concentration of the dissolved gas is $0.0225 \text{ M}$.