Transcript Bendroji chemija

CONCENTRATION
OF
SOLUTIONS
Solution = Solute + Solvent
The amount of solution can be expressed by:
- mass m (g, kg) or
- volume V (cm3, mL, dm3, L, m3)
m = Vx
- density  (g/cm3, kg/m3)
PERCENT COMPOSITION (by mass)
(mass percentage)
C - parts of solute per 100 parts of solution
grams of solute per 100 grams of solution
Example: 25  aqua NaCl solution. How to prepare solution - 400 g?
100 g solution - 25 g NaCl - 75 g H2O
400 g solution - x g NaCl - y g H2O
x = 400 . 25 / 100 = 100 g NaCl
y = 400 . 75 / 100 = 300 g H2O
V = m /  = 400 / 1.25 = 320 cm3 (mL)
MOLAR CONCENTRATION
(molarity)
CM - tells us the number of moles of solute in exactly
one liter of a solution (mol/L, mol/1000 mL (cm3))
Example: 0.2 M H2SO4 aqua solution. How to prepare 4 Litres?
1 L solution - 0.2 moles
4L
“
- x “
1 mol (H2SO4) - 98 g/mol
0.8 mol
- yg
x = 4 . 0.2 / 1 = 0.8 moles H2SO4
y = 0,8 . 98 / 1 = 78.4 g H2SO4
DILUTION OF SOLUTIONS
Is used to prepare diluted solutions from concentrated or
standard (stock) solutions
c1 V1 = c2 V2
(CM)
c1 m1 = c2 m2
c1 V1 d1= c2 V2 d2
(C)
(C)
1 – concentrated (standard) solution, 2 - diluted solution
1 EXAMPLE.:
200 mL of a solution was prepared by dissolving 17.4 g of K2SO4
in water. What is mass percentage (C) and molarity (CM) of this
solution (d=1.1 g/cm3) ?
C - ?
mt = V . d = 200 . 1.1 = 220 g
220 g solution - 17.4 g K2SO4
100 g
“
x g “
x = 7.9 g / 100 g solution = 7.9 
1 EXAMPLE.:
200 mL of a solution was prepared by dissolving 17.4 g of K2SO4
in water. What is mass percentage (C) and molarity (CM) of this
solution (d=1.1 g/cm3) ?
CM - ?
200 mL solution 1000 mL “
-
1 mol
y mol
-
17.4 g K2SO4
x g “
174 g
87 g
x = 87 g/L
y = 0.5 mol/L
2 EXAMPLE.:
If you were going to make 5 L of a 12  (d2=1.08 g/cm3) solution of
sulfuric acid, how much of concentrated solution (60 , d1=1.5
g/cm3) would you need to use?
c1 m1 = c2 m2
c1 V1 d1= c2 V2 d2
(C)
(C)
V1 = c2 V2 d2 / c1 d1 = 12 . 5000 . 1.08 / 60 . 1.5 =
= 720 cm3 60  H2SO4