Given, $K_H = 1.67 \times 10^{8} \, \text{Pa},$ $P_{\mathrm{CO}_2} = 2.5 \, \text{atm} = 2.5 \times 101325 \, \text{Pa}$ Applying Henry's law, $P_{\mathrm{CO}_2} = K_H \times x_{\mathrm{CO}_2}$ $\therefore \;\; x_{\mathrm{CO}_2} = \frac{P_{\mathrm{CO}_2}}{K_H}$ $= \frac{2.5 \times 101325 \, \text{Pa}}{1.67 \times 10^{8} \, \text{Pa}}$ $= 1.517 \times 10^{-3}$ i.e., $x_{\mathrm{CO}_2} = \frac{n_{\mathrm{CO}_2}}{n_{\mathrm{H}_2\mathrm{O}} + n_{\mathrm{CO}_2}}$ $= \frac{n_{\mathrm{CO}_2}}{n_{\mathrm{H}_2\mathrm{O}}} = 1.517 \times 10^{-3}$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; [n_{\mathrm{CO}_2}$ is negligible w.r.t. $n_{\mathrm{H}_2\mathrm{O}} \ ]$ For 500 mL of soda water, Volume of water = 500 mL; mass of water = 500 g, moles of water $= \frac{500}{18} = 27.78$ i.e., $n_{\mathrm{H}_2\mathrm{O}} = 27.78$ $\therefore \; \frac{n_{\mathrm{CO}_2}}{27.78} = 1.517 \times 10^{-3}$ or $n_{\mathrm{CO}_2} = 42.14 \times 10^{-3} \, \text{mole},$Mass of $\mathrm{CO}_2 = 42.14 \times 10^{-3} \times 44 \, \text{g}$$= 1.854 \, \text{g}$