Calculate the expected change in temperature for 3.5 g ofNH 4 NO 3 dissolved in

Question

Calculate the expected change in temperature for 3.5 g ofNH4NO3 dissolved in 100.0 mL ofH2O. Assume that the calorimeter absorbs no heat.(ie Ccal=0)(Remember to use dimensional analysis) NH4NO3(s) -->NH4+(aq) + NO3- (aq)    H=+25.7 kJ/mol weight of water is 1.0g/mL and has a specific heat of 4.18J/gx C Calculate the expected change in temperature for 3.5 g ofNH4NO3 dissolved in 100.0 mL ofH2O. Assume that the calorimeter absorbs no heat.(ie Ccal=0)(Remember to use dimensional analysis) NH4NO3(s) -->NH4+(aq) + NO3- (aq)    H=+25.7 kJ/mol weight of water is 1.0g/mL and has a specific heat of 4.18J/gx C

Explanation / Answer

   Fromula : q = mst Given :                    NH4NO3(s) -->NH4+(aq) + NO3- (aq)    H = +25.7 kJ/mol           When one mole of NH4NO3(s) disoolovedin water 25.7 kJ/mol energy is absorbed from water , asresult there was a change in temperture of water .           3.5 gof NH4NO3 = 3.5g/ 80.04g.mol-1                                        = 0.0437mol Therefore, amount of heat absorbed from waterwhen  3.5 g of NH4NO3 =  0.0437moldissolved is                                         0.0437mol *25kJ.mol-1 = 1.093kJ                                                                             =1093J    change in temperature ,t =q / ms                                                = 1093J / (100g*4.18J.g-1 .0C-1 )                                                = 2.610C Here , 100mLwater = 100 g of water Here , 100mLwater = 100 g of water

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.