Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
=
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.
B) they have no charges and are inside an atom.*
Answer: The molar enthalpy change is 73.04 kJ/mol
Explanation:

moles of HCl= 
As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.
volume of water = 100.0 ml + 50.0 ml = 150.0 ml
density of water = 1.0 g/ml
mass of water = 

q = heat released
m = mass = 150.0 g
c = specific heat = 
= change in temperature = 


Thus 0.0415 mol of HCl produces heat = 3031.3 J
1 mol of HCL produces heat = 
Thus molar enthalpy change is 73.04 kJ/mol