**(i)** For vaporization of diethyl ether,

∆_{vap}H^{Θ} = 26 kJ mol^{−1} = 26 × 10^{3} J mol^{−1}

T = 35 + 273= 308 K

\(\therefore\) Δ_{vap}S^{Θ} = \(\frac{\Delta_{vap}H^{\ominus}}{T}\)

= \(\frac{26\times10^3\,J\,mol^{-1}}{35K}\)

= 84.4 JK^{-1} mol^{-1}

**(ii)** The conversion of vapour into liquid is condensation. The enthalpy of condensation is negative of enthalpy of vaporization.

\(\therefore\) ∆_{cond}H^{Θ }= −26 kJ mol^{−1}

= −26 × 10^{3} J mol^{−1}

^{\(\therefore\) }∆_{cond}S^{Θ} = \(\frac{\Delta_{cond}H^{\ominus}}{T}\)

= \(\frac{-26\times10^3\,J\,mol^{-1}}{308K}\)

= -84.4 J K^{-1} mol^{-1}