Ncert Exercises & Solutions

18.0 g of water completely vaporises at $100 ° C$ 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?

Given that, quantity of water

$= 18.0 g$ , pressure = 1 bar

As we know that,

$18.0 g H_{2} O = 1 mole H_{2} O$

Enthalpy change for vaporising 1 mole of

$H_{2} O = 40.79 kJ {mol}^{- 1}$

$∴$ Enthalpy change for vaporising 2 moles of

$H_{2} O = 2 \times 40.79 kJ = 81.358 kJ$

Standard enthalpy of vaporisation at

$100 ° C$ and 1 bar pressure,

$∆_{vap} H ° = + 40.79 kJ {mol}^{- 1}$

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