The solubility of N_2 gas in water at 25 text | vedclass

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(A) According to Henry's Law, the solubility $(S)$ of a gas is directly proportional to its partial pressure $(P)$: $S = k_H 	imes P$.Step $1$: Calcu

Vedclass · Accepted Answer

$4.80 	imes 10^{-4} 	ext{ mol L}^{-1}$

Vedclass · Answer

(A) According to Henry's Law, the solubility $(S)$ of a gas is directly proportional to its partial pressure $(P)$: $S = k_H 	imes P$.Step $1$: Calculate Henry's constant $(k_H)$ using the initial conditions:$k_H = \frac{S_1}{P_1} = \frac{6.85 	imes 10^{-4} 	ext{ mol L}^{-1}}{1 	ext{ bar}} = 6.85 	imes 10^{-4} 	ext{ mol L}^{-1} 	ext{ bar}^{-1}$.Step $2$: Calculate the new solubility $(S_2)$ at $P_2 = 0.70 	ext{ bar}$:$S_2 = k_H 	imes P_2 = (6.85 	imes 10^{-4} 	ext{ mol L}^{-1} 	ext{ bar}^{-1}) 	imes (0.70 	ext{ bar}) = 4.795 	imes 10^{-4} 	ext{ mol L}^{-1} \approx 4.80 	imes 10^{-4} 	ext{ mol L}^{-1}$.

The solubility of $N_2$ gas in water at $25 \text{ }^\circ\text{C}$ and $1 \text{ bar}$ is $6.85 \times 10^{-4} \text{ mol L}^{-1}$. Calculate the solubility of $N_2$ gas in water at the same temperature when the partial pressure of $N_2$ is $0.70 \text{ bar}$.

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