The solubility of gas in water exposed to the atmosphere, when its partial pressure is , is determin — Solutions and Colligative Properties Chemistry Question
Question
The solubility of $N_2$ gas in water exposed to the atmosphere, when its partial pressure is $593\text{ mm Hg}$, is determined to be $5.3 \times 10^{-4}\text{ M}$. Calculate its solubility (expressed in $10^{-4}\text{ M}$) at $760\text{ mm Hg}$ and at the same temperature. (Round off your final numerical answer to one decimal place, e.g., if you calculate 1.23, write 1.2)
Answer: 6.8
💡 Solution & Explanation
Because solubility $S \propto P$ at constant temperature, we can write $\frac{S_1}{P_1} = \frac{S_2}{P_2}$. Substituting the known values: $\frac{5.3 \times 10^{-4}}{593} = \frac{S_2}{760}$. Solving for $S_2$ gives $S_2 = \frac{5.3 \times 10^{-4} \times 760}{593} = 6.79 \times 10^{-4}\text{ M}$. Rounded to one decimal place, the required value is $6.8$.
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