Question

Calculate the enthalpy change for the dissolution of 5.00 g of sodium hydroxide (NaOH) in 100.0 g of water (H2O). The temperature change was measured to be -3.45°C. Given that the specific heat of water is 4.18 J/g°C and the density of water is 1.00 g/mL. (Assume that the heat evolved from NaOH is absorbed by the water.)

Answer 1

To calculate the enthalpy change for the dissolution of NaOH in water, we can use the formula:q = mcTwhere q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, and T is the change in temperature.In this case, we are given the mass of water  100.0 g , the specific heat capacity of water  4.18 J/gC , and the change in temperature  -3.45C . We can plug these values into the formula:q =  100.0 g  4.18 J/gC  -3.45C q = -1441.7 JSince the temperature change is negative, the dissolution of NaOH in water is exothermic, meaning heat is released. Therefore, the enthalpy change for the dissolution of 5.00 g of NaOH in 100.0 g of water is -1441.7 J.If you want to express the enthalpy change per mole of NaOH, you can divide the enthalpy change by the moles of NaOH dissolved:Molar mass of NaOH = 22.99 g/mol  Na  + 15.999 g/mol  O  + 1.007 g/mol  H  = 39.996 g/molMoles of NaOH = 5.00 g / 39.996 g/mol = 0.125 molEnthalpy change per mole of NaOH = -1441.7 J / 0.125 mol = -11533.6 J/mol or -11.53 kJ/mol