MHTCET 14 May 2023 Shift 2 Solved Paper

According to Henry's Law, the solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas above the liquid. The law can be mathematically expressed as:$S = k_H \cdot P$Where:$S$ is the solubility of the gas (in $\mathrm{mol} \ \mathrm{dm}^{-3}$ ).$k_H$ is Henry's law constant (in $\mathrm{mol} \ \mathrm{dm}^{-3} \ \mathrm{bar}^{-1}$ ).$P$ is the partial pressure of the gas (in bar). To find Henry's law constant for the given conditions, we need to rearrange the equation to solve for $k_H$ :$k_H = \; \frac{S}{P}$Given that the solubility $S = 0.028 \ \mathrm{mol} \ \mathrm{dm}^{-3}$ and the partial pressure $P = 0.346$ bar, we can plug these values into the equation:$k_H = \; \frac{0.028 \ \mathrm{mol} \ \mathrm{dm}^{-3}}{0.346 \ \mathrm{bar}}$To calculate $k_H$$\;k_H = \; \frac{0.028}{0.346} \ \mathrm{mol} \ \mathrm{dm}^{-3} \ \mathrm{bar}^{-1}$$\;k_H \approx 0.08092 \ \mathrm{mol} \ \mathrm{dm}^{-3} \ \mathrm{bar}^{-1}$ If you review the options provided, the closest value to 0.08092 is option A, $0.081 \ \mathrm{mol} \ \mathrm{dm}^{-3} \ \mathrm{bar}^{-1}$. Therefore, the correct option is: