When a 5.00-g sample of KSCN is dissolved in water in a calorimeter that has a total heat capacity of 2.594 kJ·K–1, the temperature decreases by 0.480 K. Calculate the molar heat of solution of KSCN.
How do you solve this?
How do you solve this?
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Add the atomic wts in the formula KSCN to get a total of 97, so one mole of KSCN weighs 97 gms.
so 5.00 gms will be 5.00 over 97 gms/mole = .052 moles KSCN
Energy involved in lowering the temp of the calorimeter .48 degrees K would be
.48 degrees times 2.594 Kj/degree K = l.25 Kj heat energy lost by the calorimeter.
now we can scale up to the heat energy lost when one mole of KSCN dissolves.
l.25 Kj lost over .052 moles KSCN = X Kj lost over one mole KSCN
cross multiply and solve for X Kg/mole heat of solution. I got + 24 Kj per mole heat of solution.
so 5.00 gms will be 5.00 over 97 gms/mole = .052 moles KSCN
Energy involved in lowering the temp of the calorimeter .48 degrees K would be
.48 degrees times 2.594 Kj/degree K = l.25 Kj heat energy lost by the calorimeter.
now we can scale up to the heat energy lost when one mole of KSCN dissolves.
l.25 Kj lost over .052 moles KSCN = X Kj lost over one mole KSCN
cross multiply and solve for X Kg/mole heat of solution. I got + 24 Kj per mole heat of solution.