Raghuvir
Last Activity: 6 Years ago
In 1st ... 0.25 mol HCl will react wilth 0.25 mol NaOH, so the reaction will be NaOH + HCl ~ H2O + NaCl, so ∆H1=∆H .In 2nd reaction also there is reaction between an strong acid and strong base but 0.5 mol will not have any impact.so ∆H2=∆H. In 3rd also there is same reaction but 10 milimoles remain in react that will only change pH of Sol. So ∆H1=