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Chapter 11: Problem 57

The solubility of nitrogen in water is $8.21 \times 10^{-4} \mathrm{mol} / \mathrm{L}\( at \)0^{\circ} \mathrm{C}\( when the \)\mathrm{N}_{2}$ pressure above water is 0.790 \(\mathrm{atm}\) . Calculate the Henry's law constant for \(\mathrm{N}_{2}\) in units of \(\mathrm{mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P,\) where \(C\) is the gas concentration in mol/L. Calculate the solubility of \(\mathrm{N}_{2}\) in water when the partial pressure of nitrogen above water is 1.10 atm at \(0^{\circ} \mathrm{C} .\)

Short Answer

Expert verified
The Henry's law constant for N₂ is approximately \(1.04 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}\cdot\mathrm{atm}}\), and the solubility of N₂ in water when the partial pressure of nitrogen is 1.10 atm at \(0^{\circ} \mathrm{C}\) is approximately \(1.14 \times 10^{-3} \mathrm{mol}/\mathrm{L}\).

Step by step solution

01

(Step 1: Extract the given data)

(We are given the solubility of nitrogen (\(\mathrm{mol}/\mathrm{L}\)) and the partial pressure above water (\(\mathrm{atm}\)). Let's extract: Solubility of N₂ in water: \(8.21 \times 10^{-4} \mathrm{mol} / \mathrm{L}\) Partial pressure of N₂ above water: \(0.790 \mathrm{atm}\))
02

(Step 2: Calculate the Henry's law constant)

(Using the equation \(C=kP\), we can solve for the Henry's law constant \(k\) by rearranging the equation to \(k=C/P\). Plugging in the given values, we have: \(k = \frac{8.21 \times 10^{-4} \mathrm{mol} / \mathrm{L}}{0.790 \mathrm{atm}}\)) Next, perform the calculation to find the value of \(k\): \(k = \frac{8.21 \times 10^{-4} \mathrm{mol} / \mathrm{L}}{0.790 \mathrm{atm}} \approx 1.04 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}\cdot\mathrm{atm}}\)
03

(Step 3: Calculate the new solubility of N₂)

(We are now asked to find the solubility of N₂ when the partial pressure is 1.10 atm. Using the same equation as the previous step, rearrange to find the concentration: \(C = kP\)). We have \(k = 1.04 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}\cdot\mathrm{atm}}\) and \(P = 1.10 \mathrm{atm}\). Multiply these values to find the new solubility of N₂ in water: \(C = \left(1.04 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}\cdot\mathrm{atm}}\right) (1.10 \mathrm {atm}) \approx 1.14 \times 10^{-3} \mathrm{mol}/\mathrm{L}\) The solubility of N₂ in water when the partial pressure of nitrogen is 1.10 atm at \(0^{\circ} \mathrm{C}\) is approximately \(1.14 \times 10^{-3} \mathrm{mol}/\mathrm{L}\).

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