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Step-by-Step Solution
Step 1: Write Henry’s Law
Henry’s law states that the partial pressure of a gas ( P_{\text{gas}} ) in a solution is proportional to its mole fraction ( X_{\text{gas}} ), according to
P_{\text{gas}} = K_H \times X_{\text{gas}},
where K_H is the Henry’s law constant.
Step 2: Express the Mole Fraction of the Dissolved Gas
The mole fraction X_{\text{gas}} of CO2 can be written approximately as
X_{\text{CO}_2} = \frac{n(\text{CO}_2)}{n(\text{H}_2\text{O}) + n(\text{CO}_2)},
Since n(\text{CO}_2) is usually very small compared to n(\text{H}_2\text{O}) , we consider:
X_{\text{CO}_2} \approx \frac{n(\text{CO}_2)}{n(\text{H}_2\text{O})}.
Step 3: Calculate the Moles of Water
We are given 0.9 L of water. Assuming the density of water is about 1 g/mL:
\text{Mass of water} \approx 0.9 \times 1000 \text{ g} = 900 \text{ g}.
The molar mass of water is 18 g/mol, so
n(\text{H}_2\text{O}) = \frac{900 \text{ g}}{18 \text{ g/mol}} = 50 \text{ mol}.
Step 4: Substitute into Henry’s Law
Given:
Partial pressure of CO2, P_{\text{gas}} = 0.835 \text{ bar}
Henry’s law constant for CO2 at 298 K, K_H = 1.67 \times 10^3 \text{ bar}
According to Henry’s law:
0.835 = (1.67 \times 10^3) \times \frac{n(\text{CO}_2)}{50},
where we used n(\text{H}_2\text{O}) = 50 \text{ mol} in the denominator for simplicity.
Step 5: Solve for the Moles of CO₂
Rearranging to find n(\text{CO}_2) :
n(\text{CO}_2) = \frac{0.835 \times 50}{1.67 \times 10^3}.
Calculate this value:
n(\text{CO}_2) = \frac{41.75}{1.67 \times 10^3} = 0.025 \text{ mol (approximately)}.
Step 6: Convert to Millimoles and Final Answer
1 mole = 1000 millimoles, therefore:
\text{Millimoles of CO}_2 = 0.025 \times 1000 = 25 \text{ mmol}.
The question asks for the value of x in millimoles (nearest integer), so:
x = 25
