In problems I, II, and I Calorimeter constant is 0.00 j/°GC Specific Heat of wat

Question

In problems I, II, and I Calorimeter constant is 0.00 j/°GC Specific Heat of water or aqueous solutions is 4·18 j/g °C I. 112.2280 g of metal were heated to 101.3 °C and poured into a calorimeter containing 44.20 g of water at 261 5 °C After stirring, the temperature of the water rose rapidly to 33.00 °C before slowly starting to fall Calculate: a) The Specific Heat of the metal b) The Atomie Weight Calculated using the rule of Dulong and Petit II. To a calorimeter containing 43.70 g of water at 25.10 °C,4.4174 g of a salt was added with stirring. As the salt dissolved, the temperature rapidly changed to 18.15 °C before slowly returning to room temperature. Calculate: c) The Heat of Solution

Explanation / Answer

Heat of water =

4.184 J/g C × 43.70g × (18.15-25.10 C)

= -1270.74 J or 1.270 KJ

Heat of formation of solution:

Heat / amount of salt= -1270.74 /4.4174 g

= -287.67 J/g = -0.287 KJ/ g

Now calculate the moles of salt:

4.4174 g/ 113.64 g/ mol

=0.0389 mol

Heat of formation of solution per mol:

Heat / moles of salt= 1270.74 J / 0.0389 mol

=32666.8 J/mol

=32.67 KJ/mol

with the help of the heat of formation for the solution & the given heat of formation for the solid salt & molecular mass of the solid salt, we can find the Heat of Formation of the Solution:

32.67Kj/mol – 76.49kJ/mol = - 43.82 Kj/mol

reaction:

            AAA +BBB ---- > CCC

Volume of A= 13.0 ml, B= 8.0 and water = 8.0

Total volume = 29 ml

Density = 1.00 g/ ml so mass = 29 g

Moles of A: 13.0/1000*1.56 = 0.020 mol

Moles of B: 8.0/1000*4.05 = 0.0324 mol

Here A is limiting agent thus the moles of C = 0.020 mol

Total energy = mCdT

= 4.184 J/g C × 29.00g × (30.45-28.30 C)

= 260.8724 J

=0.261 kJ

Now calculate the heat of reaction in KJ/ mol divide this values by number of moles:

=0.261 KJ/0.020 mol

=13.0kj/mol

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