18.0.g water completely vaporises at 100°C and 1 bar pressure and the enthalpy change in the process 40.79 kJ moll. What will be the enthalpy change for vaporising two moles of water under the same condition

Options

A. `81.58 kJ, 81.58 kJ `
B. `40. 79. kJ , 40. 79 kJ`
C. `40.79 kJ , 81.58 kJ`
D. `81.58 kJ , 40. 79 kJ`

Explanation

`18.0 g H _(2) O = 1 mol H _(2) O`
Enthalpy change for vaprising 1 mole of `H _(2) O = 40. 79 kJ `
`therefore ` Enthalpy change for vaporising 2 moles of `H _(2) O = 2 xx 40. 79 kJ = 81. 58 kJ`
Standard enthalpy of vaporisation at `100 ^(@) C` and 1 bar pressure, `Delta _(vap) H ^(@) =+ 40. 79 kJ mol ^(-1)`