Asked Dec 8, 2019

In a calorimeter, 100.0 g of water, initially at 21.8 *C, is mixed with 0.300 moles of LiOH. After dissolving, the temperature of the water is 38.7 *C. What was the heat of dissolution (dHdiss) of LiOH in kJ/mole (3 sig figs)


Expert Answer

Step 1


Mass of water=100 g

Moles of LiOH=0.300 moles

Initial temperature=21.8oC

Final temperature=38.7oC

Step 2

To calculate heat of dissolution:

Heat of dissolution also called enthalpy of dissolution, the equation is given as


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q=mcAT where, q-amount of heat transfered m=mass of water c=specificheat capacity of water=4.184J/g°C AT=change in temperature,T,-T,

Step 3

First we have to calculate the amount of heat transfer...


Image Transcriptionclose

q=mcAT =100 gx4.184J/g°C*(38.7-21.8)°C =7070,96J


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