The Henry's law constant for the solubility of $N_2$ gas in water at $298 ext{ K}$ is $1.0 imes 10^5 ext

Question

The Henry's law constant for the solubility of $N_2$ gas in water at $298 	ext{ K}$ is $1.0 	imes 10^5 	ext{ atm}$. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ formed from air dissolved in $10 	ext{ moles}$ of water at $298 	ext{ K}$ and $5 	ext{ atm}$ pressure is:

Accepted Answer

1.  **Calculate Partial Pressure of Nitrogen ($p_{N_2}$):**
    According to Dalton's Law, the partial pressure of a component in a gas mixture is the product of its mole fraction in the gas phase and the total pressure.
    $$p_{N_2} = 	ext{Mole fraction in air} 	imes P_{	ext{total}}$$
    $$p_{N_2} = 0.8 	imes 5 	ext{ atm} = 4.0 	ext{ atm}$$

2.  **Calculate Mole Fraction of Nitrogen in Solution ($x_{N_2}$):**
    Using Henry's Law formula: $p = K_H 	imes x$
    $$4.0 	ext{ atm} = (1.0 	imes 10^5 	ext{ atm}) 	imes x_{N_2}$$
    $$x_{N_2} = \frac{4.0}{1.0 	imes 10^5} = 4.0 	imes 10^{-5}$$

3.  **Calculate Moles of Nitrogen ($n_{N_2}$):**
    The mole fraction $x_{N_2}$ is defined as $\frac{n_{N_2}}{n_{N_2} + n_{H_2O}}$. For dilute solutions, $n_{N_2} \ll n_{H_2O}$, so we approximate $x_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}}$.
    $$4.0 	imes 10^{-5} = \frac{n_{N_2}}{10}$$
    $$n_{N_2} = 4.0 	imes 10^{-5} 	imes 10 = 4.0 	imes 10^{-4} 	ext{ mol}$$

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