Liquid Solution

To determine the number of moles of methane dissolved in 1 L of water at 293 K with a partial pressure of 1 bar, we employ Henry's law and the concept of mole fraction.Given:Partial pressure of methane, $p_0=1$ barHenry's constant for methane, $K_{H}=0.4K$ barStep 1: Apply Henry's LawHenry's law can be expressed as:$p_0=K_{H}\chi$Solving for the mole fraction $(\chi)$ of methane, we have:$\;1=0.4\times 10^3\times\chi$$\;\chi=2.5\times 10^{-3}$Step 2: Calculate Mole FractionThe mole fraction of methane $(x)$ in the solution is given by:$x = \;\frac{n(\text{ methane })}{n(\text{ methane })+n(\text{ water })}$With 1 L of water, the number of moles of water is:$n(\text{ water }) = \;\frac{1000\ \text{g}}{18\ \text{g/mol}} = 55.5\ \text{mol}$Assuming $n$ (methane) is significantly smaller than 55.5 mol, substituting into Equation 2 simplifies to:$2.5\times 10^{-3} = \;\frac{n}{55.5}$Solving for $n$, the moles of methane:$\;n = 2.5\times 10^{-3}\times 55.5$$\;n = 1.38\times 10^{-1}$Thus, the number of moles of methane dissolved in 1 L of water is $1.38\times 10^{-1}$.