Hi I\'m having trouble solving this problem: Consider the dissolving of ammonium

Question

Hi I'm having trouble solving this problem: Consider the dissolving of ammonium nitrate,NH4NO3(s) forwhich H= +25.3kJ/mol. If 133 grams of ammonium nitrate aredissolved in 367 grams of water, and the initial temperature is25.O CELCIUS . what will be the final temperature of thesystem ? you should assume that the specific heat of the entiresystem (the ammonium nitrate and water combined)is 4.18J/g C. Any help will be much appreciated Hi I'm having trouble solving this problem: Consider the dissolving of ammonium nitrate,NH4NO3(s) forwhich H= +25.3kJ/mol. If 133 grams of ammonium nitrate aredissolved in 367 grams of water, and the initial temperature is25.O CELCIUS . what will be the final temperature of thesystem ? you should assume that the specific heat of the entiresystem (the ammonium nitrate and water combined)is 4.18J/g C. Any help will be much appreciated

Explanation / Answer

When we dissolve NH4NO3 in water , water cools becasuethe process is endothermic. We know, moles = mass / molar mass                           =133g / 80.04g/mol = 1.66mol Given that when one mole of NH4NO3 in water , itrequires 25.3kJ so 1.66mol of NH4NO3 requires     = (25.3kJ/1mol)* 1.66mol = 41.99kJ We know that q = mst 41.99kJ (1000J/1kJ) =367g*4.18J/g.0C*(tf - 25) 41990J = 1534.06J/0C *(tf -25) (tf - 25) = 41990J /1534.06J/0C               = 27.370C tf =  27.370C   -25.00C   =2.370C

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