Respuesta :
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
[tex]m = 1 mol\times 111 g/mol = 111 g[/tex]
Mass of solution = M
Volume of solution = V = 1L = 1000 mL
Density of solution , d= 1.07 g/mL
[tex]M=d\times V=1.07 g/mL\times 1000 mL=1,070 g[/tex]
1) The value of %(m/M):
[tex]\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%[/tex]
2) The value of %(m/V):
[tex]\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%[/tex]
[tex]Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}[/tex]
[tex]Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}[/tex]
n = Equivalent mass
n = [tex]\frac{\text{molar mass of ion}}{\text{charge on an ion}}[/tex]
3) Normality of calcium ions:
Moles of calcium ion = 1 mol (1 [tex]CaCl_2[/tex] mole has 1 mole of calcium ion)
[tex]n=\frac{40 g/mol}{2}=20 [/tex]
[tex]=\frac{1 mol}{20 g/mol\times 1L}=0.050 N[/tex]
4) Normality of chlorine ions:
Moles of chlorine ion = 2 mol (1 [tex]CaCl_2[/tex] mole has 2 mole of chlorine ion)
[tex]n=\frac{35.5 g/mol}{1}=35.5[/tex]
[tex]=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N[/tex]
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 :
[tex]\frac{1 mol}{0.959 kg}=1.043 mol/kg[/tex]
Moles of calcium chloride = [tex]n_1=1mol[/tex]
Mass of solvent = 959 g
Moles of water = [tex]n_2=\frac{959 g}{18 g/mol}=53.28 mol[/tex]
Mass of solvent = 959 g
6) Mole fraction of calcium chloride =
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842[/tex]
7) Mole fraction of water =
[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816[/tex]
8) Mass of solution = m'
Volume of the solution= v = 100 mL
Density of solution = d = 1.07 g/mL
[tex]m'=d\times v=1.07 g/ml\times 100 g= 107 g[/tex]
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%
[tex]11.1\%=\frac{m}{100 mL}\times 100[/tex]
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.
