$18.0 \ g$ of water completely vaporises at $100^{\circ}C$ and $1 \ bar$ pressure and the enthalpy change in the process is $40.79 \ kJ \ mol^{-1}$. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?
(N/A) Given: Mass of water $= 18.0 \ g$,Pressure $= 1 \ bar$.
Since the molar mass of $H_2O$ is $18.0 \ g \ mol^{-1}$,$18.0 \ g$ of $H_2O$ corresponds to $1 \ mole$.
Enthalpy change for vaporising $1 \ mole$ of $H_2O$ is given as $\Delta_{vap}H = 40.79 \ kJ \ mol^{-1}$.
Therefore,the enthalpy change for vaporising $2 \ moles$ of $H_2O = 2 \times 40.79 \ kJ = 81.58 \ kJ$.
The standard enthalpy of vaporisation $(\Delta_{vap}H^{\circ})$ at $100^{\circ}C$ and $1 \ bar$ pressure is $40.79 \ kJ \ mol^{-1}$.
Since the molar mass of $H_2O$ is $18.0 \ g \ mol^{-1}$,$18.0 \ g$ of $H_2O$ corresponds to $1 \ mole$.
Enthalpy change for vaporising $1 \ mole$ of $H_2O$ is given as $\Delta_{vap}H = 40.79 \ kJ \ mol^{-1}$.
Therefore,the enthalpy change for vaporising $2 \ moles$ of $H_2O = 2 \times 40.79 \ kJ = 81.58 \ kJ$.
The standard enthalpy of vaporisation $(\Delta_{vap}H^{\circ})$ at $100^{\circ}C$ and $1 \ bar$ pressure is $40.79 \ kJ \ mol^{-1}$.
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