Question

When a 3.11 g sample of solid sodium hydroxide was dissolved in a calorimeter in 130.0 g of water, the temperature rose from 18.9 oC to 31.3 o C . Calculate Δ H (in kJ/mol NaOH ) for the following solution process: NaOH(s)→Na+(aq)+OH−(aq).  Assume that it’s a perfect calorimeter and that the specific heat of the solution is the same as that of pure water.

Answer 1

3.11g sample of solid sodium hydroxide was dissolved in de calorimeter in 130.0 g of water. rature rose from 18.goc to 31.3°c. NaOH as - Nacaal + oHelaas BH dissolution. → Heat Heat absorbed by the solution = a solution given off by the dissolution of salt = Shaiss Adiss a solution minus sign here because - Heat lost carries a negative To find the Heat absorbed by the solutions a - You can use the equation. a solution - a=mc.ot where - q= Heat gained by the water m= 130g m=mass of the water DT= 31.3°C - 18.goc= 12.4°C (2 specific heat of water c = Specific Heat of solution ~ water = 4.186] Dit = change in temperature goc q= 1308 x 4.186 J x 12.40 a = 6747.832 J which implies Soo You know that solution absorbed 6747.832). that the dissolution of NaOH(s) gave off 674708325 AH dise =-67470832] Moles of Naos in 3.11 g = 3.11g Mass nolar Max = 0.0777mol 39.997 Amol You know that. Sediss when 0-0777 mol of NaOH(s) are dissolved in water a so use this info to find the enthalpy of Dissolution when I mole of Naoles) dissolves 1 mol NaOH X - 6747.832 KJ 86-845 KJ. 1000x 0.0777 moles Naosh landiss = - 86.848 kJ motal] Page No.