18.0 g of water completely vapourises 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 vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

Question

18.0 g of water completely vapourises 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 vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?, Enthalpy change for vapourising two moles of water under the same conditions is +81.58 kJ and ∆ vap H = +40.79 kJ mol –1, Enthalpy change for vapourising two moles of water under the same conditions is +69.58 kJ and ∆ vap H = +40.79 kJ mol –1, Enthalpy change for vapourising two moles of water under the same conditions is +81.58 kJ and ∆ vap H = +45.79 kJ mol –1, Enthalpy change for vapourising two moles of water under the same conditions is +69.58 kJ and ∆ vap H = +45.79 kJ mol –1

MARKS App · Accepted Answer

Correct Option is : (A) Enthalpy change for vapourising two moles of water under the same conditions is +81.58 kJ and ∆ vap H = +40.79 kJ mol –1

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