Question

A pair of students found the temperature of 100. g of water to be 26.5°C. They then dissolved 8.89 g of AgNO3 in the water. When the salt had dissolved, the temperature of the water was 23.7°C.
(a) Calculate ?T for the water.


°C

(b) The dissolution was ?


(f) When 8.89 grams of AgNO3 is dissolved, how many moles of cation are produced?



How many moles of anion are produced?

Answer 1

Explanation:

Initial temperature of the water = $T_i=26.5^oC$

Final temperature of the water = $T_f=23.7^oC$

a) Change in temperature of the water,ΔT = $T_f-T_i$

$\Delta T=23.7^oC-26.5^oC=-2.8^oC$

Change in temperature of the water,ΔT is -2.8°°C.

b) Endothermic reaction : Reaction in which heat absorbed and the temperature if the surrounding is decreased.

Exothermic reaction : Reaction in which heat released and the temperature if the surrounding is increased.

On dissolving silver nitrate in water the temperature of the water decreased 2.8 degree Celsius  which means that energy water was absorbed by solid silver nitrate to get dissolve in water. Hence, endothermic reaction.

(f)  Mass of silver nitrate = 8.89 g

Moles of silver nitrate = $\frac{8.89 g}{170 g/mol}=0.05229 mol$

$AgNO_3(aq)\rightarrow Ag^+(aq)+NO_3^-(aq)$

1 mole of silver nitrate gives 1 mole of silver ion i.e. cation.

Then 0.05229 moles of silver nitrate will give:

$\frac{1}{1}\times 0.05229 mol=0.05229 mol$ silver ions.

0.05229  moles of cations are produced.

1 mole of silver nitrate gives 1 mole of nitrate ion i.e. anion .

Then 0.05229 moles of silver nitrate will give:

$\frac{1}{1}\times 0.05229 mol=0.05229 mol$ nitrate ions.

0.05229  moles of anion are produced.