1. When 14.01 mL. of HCl of unknown concentration (but less than that of the bas

Question

1. When 14.01 mL. of HCl of unknown concentration (but less than that of the base) are reacted with 15 mL. of 3.068 M NaOH, 1.56 kJ of heat are released. What is the molarity of the HCl solution? 2. 17.18 g of a substance, initially at 98.82 °C, was poured into 9.94 g of water in a calorimeter The initial temperature of the water and the calorimeter was 23.5 1C. The calorimeter has a heat capacity of 8.04 J/K. The temperature after mixing, TP was determined to be 36.2 1C. Calculate the specific heat capacity of the substance

Explanation / Answer

1 . HCl+NaOH ---> NaCl+ H2O

1 mol HCl = 1 mol NaOH

No of mol of NaOH taken = M*V

                      = 3.068*15

           = 46.02 mmol

standard Enthalpy of neutralisation of NaOH vs HCl = 57.9 kj/mol

No of mol of HCl , NaOH reacted = 1.56/57.9 = 0.027 mol

No of mol of HCl reacted = 27 mmol

molarity of HCl reacted = n/v

                        = 27/14.01

           = 1.928 M

2. heat lost by the substance = heat gained by the water + calorimeter

     m*s*DT1    = mass of water*specific heat of water*DT2 + heat capacity*DT2

    17.18*s*(98.82-36.21) = 9.94*4.184*(36.21-23.51)+ 8.04*(36.21-23.51)

s = specific heat of substance = 0.586 j/g.c

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