Explanation:
Molarity of solution = 1.00 M = 1.00 mol/L
In 1 L of solution 1.00 moles of calcium chloride is present.
Mass of solute or calcium chloride = m

Mass of solution = M
Volume of solution = V = 1L = 1000 mL
Density of solution , d= 1.07 g/mL

1) The value of %(m/M):

2) The value of %(m/V):



n = Equivalent mass
n = 
3) Normality of calcium ions:
Moles of calcium ion = 1 mol (1
mole has 1 mole of calcium ion)


4) Normality of chlorine ions:
Moles of chlorine ion = 2 mol (1
mole has 2 mole of chlorine ion)


Moles of calcium chloride = 1.00 mol
Mass of solvent = Mass of solution - mass of solute
= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)
5) Molality of the solution :

Moles of calcium chloride = 
Mass of solvent = 959 g
Moles of water = 
Mass of solvent = 959 g
6) Mole fraction of calcium chloride =

7) Mole fraction of water =

8) Mass of solution = m'
Volume of the solution= v = 100 mL
Density of solution = d = 1.07 g/mL

Mass of 100 mL of this solution 107 grams of solution.
9) Volume of solution = V = 100 mL
Mass of solution = M'' = 107 g
Mass of solute = m
The value of %(m/V) of solution = 11.1%

m = 11.1 g
Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g
95.9 grams of water was present in 100 mL of given solution.