1. When 14.01 mL. of HCl of unknown concentration (but less than that of the bas
ID: 593578 • Letter: 1
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 substanceExplanation / 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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