A) 11.56 kJ
B) 5.73 kJ
C) 22.92 kJ
D) 17.19 kJ
Correct Answer: B
Solution :
The molarity and normality are same in case of \[NaOH\] and \[HCl\] because the acidity and basicity of these are one and one respectively. Since, number of moles = molarity \[\times \]volume \[(l)\] \[\therefore \] No. of moles of \[NaOH\] solution \[=\frac{100\times 1}{1000}=0.1\] and number of moles of \[HCl\] solution \[=\frac{300\times 1}{1000}=0.3\] Hence, 0.1 mole \[NaOH\] is neutralised by mole \[HCl\]. We know that heat of neutralisation of 1 mole \[HCl\] and 1 mole \[NaOH\] is 57.3 kJ. Hence, heat of neutralisation of 0.1 mol \[HCl\] and 0.1 mole \[NaOH\] will be \[=57.3\times 0.1=5.73\,kJ\]
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