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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.