30g of NaOH was dissolved in 80ml of water.The temperature increased from 27°C to 70°C.Estimate the enthalpy change for the dissolution of NaOH
ΔH=mcΔT\Delta H= mc\Delta TΔH=mcΔT
Assuming density of solution=1g/ml= 1g/ml=1g/ml
and taking c=4.2kJ/(kg.K)c=4.2kJ/(kg.K)c=4.2kJ/(kg.K)
ΔH=801000×4.2×(70−27)\Delta H = \frac{80}{1000}×4.2×(70-27)ΔH=100080×4.2×(70−27)
=14.448kJ=14.448kJ=14.448kJ
Molar enthalpy of dissolution:
=4030×14.448=19.264kJ=\frac{40}{30}×14.448=19.264kJ=3040×14.448=19.264kJ
∴ΔHdiss=−19.264kJ mol−1\therefore \Delta H_{diss}=-19.264kJ\>mol^{-1}∴ΔHdiss=−19.264kJmol−1
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