A container with a volume of 20.0L holds (g) and (l) at 300 K. The pressure is found to be 1.0 atm. — States of Matter and Gaseous State Chemistry Question
Question
A container with a volume of 20.0L holds $N_2$(g) and $H_2O$(l) at 300 K. The pressure is found to be 1.0 atm. The water is then split into hydrogen and oxygen by electrolysis. After the reaction is complete, the pressure is 1.86 atm. What mass (in g) of water was present initially in the container? The aqueous tension of water at 300 K is 0.04 atm. ($R = 0.08 \text{ L-atm/K-mol}$)
💡 Solution & Explanation
Initial pressure of $N_2 = 1.0 - 0.04 = 0.96\text{ atm}$. Assuming complete electrolysis, no liquid water remains, so aqueous tension vanishes. Total final pressure $= 1.86\text{ atm}$. Pressure from electrolysed gases ($H_2$ and $O_2$) $= 1.86 - 0.96 = 0.90\text{ atm}$. Moles of gas generated $n = \frac{PV}{RT} = \frac{0.90 \times 20.0}{0.08 \times 300} = 0.75\text{ mol}$. The reaction $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$ yields 3 moles of gas per 2 moles of water. Moles of water $= \frac{2}{3} \times 0.75 = 0.50\text{ mol}$. Mass of water $= 0.50\text{ mol} \times 18\text{ g/mol} = 9\text{ g}$. Therefore, correct answer is 9.