`A (l) overset(1atn)hArr A(g), Delta_(vap)H = 460.6 cal mol^(-1)`, boiling point `-50K`. What is the boiling point at `10` atm

Boiling point of water is 373k then the boiling point of H_2S is

Boiling point of water is 373k then the boiling point of H_2S is

DeltaH_("vap") of the process, A_((l)) hArr A_(("vap")) is 460.6 Cal mol^(-1) . The normal boiling point of the A is 50K. When pressure is increased to 10 atm. Then the boiling point of the A is (50)x K. The value of x is

DeltaH_("vap") of the process, A_((l)) hArr A_(("vap")) is 460.6 Cal mol^(-1) . The normal boiling point of the A is 50K. When pressure is increased to 10 atm. Then the boiling point of the A is (50)x K. The value of x is

H_(2)O (1) hArr H_(2) O(g) at 373K,DeltaH^(@)=8.31kcal mol^(-1) Thus, boiling point of 0.1 molal sucrose solution is

Boiling point of a liquid is 50 K at 1 atm and triangleH_vap=460.6 cal mol^-1 What will be its boiling point at 10 atm.

A liquid is in equilibrium with its vapour at a certain temperature. If DeltaH_(vap) = 60.24 kJ mol^-1 and DeltaS_(vap) = 150.6 J mol^-1 then the boiling point of the liquid will be

A liquid is in equilibrium with its vapour at a certain temperature. If deltaH_(vap) = 60.24 kJ mol^(-1) and deltaS_(vap) = 150.6 J mol^(-1) then the boiling point of the liquid will be

Calculate the entropy changes of fusion and vapourisation for chlorine from the following data: Delta_(fus)H = 6.40 kJ mol^(-1) , melting point =- 100^(@)C Delta_(vap)H = 20.4 kJ mol^(-1) , boiling point =- 30^(@)C