Solution

a) The calibration factor is calculated by the expression below.

b) Obtain the amount of energy absorbed from 2.45 grams of ammonium nitrate

According to the equation above every mole of ammonium nitrate that dissolves absorbs 25.2Kj of energy.

SO!

Step 1) Mole of ammonium nitrate added = 2.456 /80 = 0.031 mole

Step 2) If one mole of ammonium nitrate gives off 25.2Kj of energy then.

0.031 X 25.2Kj/mol = 0.772Kj = 772joules

c) The amount of degrees drop that 2.45g of ammonium nitrate will achieve in the calorimeter is given by the expression

772 joules / 311.4 joules/C =2.48C
So the temperature of the contents of the calorimeter will be at
20.34 - 2.48 = 17.86C

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