A quantity of 4 ml of a gas at 1 atm and 300 K is dissolved in 1 L of water. The volume (in ml) of g — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

According to Henry's law, the mass (or number of moles, $n$) of gas dissolved is proportional to the pressure ($P$) and the volume of the solvent ($V_{solvent}$). Thus, $n \propto P \times V_{solvent}$. Let $n_1$ be the initial moles dissolved. For the second state, $P_2 = 4 P_1$ and $V_{solvent2} = 0.5 V_{solvent1}$. So, $n_2 = n_1 \times (4) \times (0.5) = 2 n_1$. According to the ideal gas law, the volume of the gas at the new pressure is $V_2 = (n_2 R T) / P_2$. Substituting $n_2 = 2n_1$ and $P_2 = 4P_1$, we get $V_2 = (2n_1 R T) / (4 P_1) = 0.5 \times (n_1 R T / P_1) = 0.5 V_1$. Since $V_1 = 4 \text{ ml}$, $V_2 = 0.5 \times 4 \text{ ml} = 2 \text{ ml}$. Therefore, correct answer is 2.