CBSE Class 11-science Answered
6.63 sum
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Asked by lovemaan5500 | 21 Jan, 2019, 06:37: AM
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, 14:43: PM
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