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$\text{Hint:}$ $\Delta\text{S}_{vap} = \frac{\Delta\text{H}_{vap}}{\text{T}}$ $\text{To calculate the entropy change during the vaporization of water at its boiling point, we can use the formula:}$ $\Delta\text{S}_{vap} = \frac{\Delta\text{H}_{vap}}{\text{T}}$ $\text{Where:}$ $\Delta\text{S}_{vap} = \text{entropy change of vaporization}$ $\Delta\text{H}_{vap} = \text{enthalpy change of vaporization (latent heat)}$ $\text{T} = \text{temperature in Kelvin}$ $\text{Given information:}$ $\Delta\text{H}_{vap} = 2.257 \text{ kJ/g}$ $\text{T} = 373 \text{ K}$ $\text{Step 1: Calculate the entropy change per gram of water.}$ $\Delta\text{S}_{vap} = \frac{\Delta\text{H}_{vap}}{\text{T}}$ $\Delta\text{S}_{vap} = \frac{2.257 \text{ kJ/g}}{373 \text{ K}}$ $\text{Converting kJ to J:}$ $\Delta\text{S}_{vap} = \frac{2.257 \times 10^3 \text{ J/g}}{373 \text{ K}}$ $\Delta\text{S}_{vap} = 6.05 \text{ JK}^{-1}\text{g}^{-1}$ $\text{Step 2: Calculate the entropy change for one mole of water.}$ $\text{The molar mass of water (H}_2\text{O) is 18 g/mol.}$ $\text{So, for one mole of water:}$ $\Delta\text{S}_{vap} (\text{per mole}) = 6.05 \text{ JK}^{-1}\text{g}^{-1} \times 18 \text{ g/mol}$ $\Delta\text{S}_{vap} (\text{per mole}) = 108.9 \text{ JK}^{-1}\text{mol}^{-1}$ $\text{Therefore, the entropy change in the conversion of one mole of liquid water at 373 K to vapor is 108.9 JK}^{-1}\text{mol}^{-1}\text{, which corresponds to option 3.}$