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CBSE Class 11-science Answered

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Asked by lovemaan5500 | 21 Jan, 2019, 06:37: AM
answered-by-expert Expert Answer
Given:
 
 500 ml, N/10 H2SO4 
 
 400 ml, N/15 NaOH
 
Reaction:
 
2NaOH + H2SO4  → Na2SO4 + 2H2O   
 
500 ml of N/10 H2SO4 = 0.5 L × 0.1 N = 0.05 equivalent moles of H2SO4
 
400 ml of N/15 NaOH = 0.4 L × 0.66 N = 0.026 equivalent moles of NaOH
 
Enthalpy of neutralisation of strong acid and a strong base is 57.1 kJ
 
So for 0.026 equivalent moles of NaOH = 0.026×57.1
 
                                                         = 1.522 kJ
 
Amount of heat liberated is 1.522 kJ.
 
 
 
Answered by Varsha | 22 Jan, 2019, 02:43: PM

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