Molar and molal concentration, despite the similar names, the values are different.The main difference is that when determining the molal concentration calculation is not on the volume of the solution as the detection molarity, and weight of the solvent.
Overview of solutions and solubility
true solution is called a homogeneous system, which includes in its membership a number of components that are independent of each other.One of them is the solvent, and the rest are substances dissolved in it.The solvent is considered to be the substance that is the solution most.
Solubility - a substance with other substances to form a homogeneous system - solutions in which it is in the form of individual atoms, ions, molecules or particles.A concentration - is a measure of solubility.
Consequently, solubility is the ability of the substance are uniformly distributed in the form of elementary particles throughout the volume of the solvent.
True solutions are classified as follows:
- by type of solvent - non-aqueous and water;
- solute in appearance - solutions of gases, acids, alkalis, salts, etc .;
- for interaction with electric shock - electrolytes (substances that conduct electricity) and non-electrolytes (substances that are not capable of electrical conductivity);
- to concentrate - dilute and concentrated.
concentration and means of expression
Concentration called content (by weight) of the substance dissolved in a certain amount (weight or volume) of the solvent, or in a certain volume of the total solution.It happens the following types:
1. The concentration of interest (expressed in%) - she talks about how many grams of solute contained in 100 grams of solution.
2. molar concentration - is the number of gram moles per 1 liter of solution.It indicates how many gram-molecules contained in 1 liter of solution of the substance.
3. The concentration of normal - it is the number of gram-equivalents per 1 liter of solution.Indicates how many shows gram-equivalents of solute in 1 liter of solution.
4. molal concentration shows how much solute in moles accounted for 1 kilogram of solvent.
5. Titer determines the content (in grams) of solute dissolved in 1 milliliter of solution.
Molar and molal concentration is different from each other.Consider their individual characteristics.
molar concentration
formula for its determination:
Cv = (v / V), where
v - the amount of dissolved substance, mol;
V - total volume, liter or m3.
example, "0.1 M H2SO4 solution" indicates that 1 liter of the solution is the presence of 0.1 mol (9.8 g) of sulfuric acid.
molal concentration
should always take into account the fact that the molal and molar concentrations have completely different meanings.
What is the molal concentration of the solution?The formula for determining it is:
Cm = (v / m), where
v - the amount of dissolved substance, mol;
m - mass of solvent kg.
example, a record of 0.2 M NaOH solution means that in 1 kilogram of water (in this case, it is the solvent), solution of 0.2 moles NaOH.
more formulas needed for calculations
lot of guidance, it may be necessary to molal concentration was calculated.The formulas, which can be useful to address the main problems are presented below.
Under the amount of material ν understand a certain number of atoms, electrons, molecules, ions or other of its particles.
v = m / M = N / NA = V / Vm, where:
- m - weight of the compound, g or kg;
- M - molar mass, g (or kg) / mol;
- N - number of structural units;
- NA - the number of structural units in 1 mole of substance, the Avogadro constant: 6.02.1023 mole-1;
- V - total volume, l or m3;
- Vm - molar volume, l / mol or m3 / mol.
Last calculated by the formula:
Vm = RT / P, where
- R - constant, 8.314 J / (mol. K);
- T - gas temperature, K;
- P - the gas pressure, Pa.
Examples tasks for molarity and molality.Task №1
determine the molar concentration of potassium hydroxide solution in 500 ml.Weight KOH solution is equal to 20 grams.
Determination
molar mass of potassium hydroxide is:
ICOS = 39 + 16 + 1 = 56 g / mol.
expect much potassium hydroxide solution contained in:
ν (KOH) = m / M = 20/56 = 0,36 mol.
take into account that the volume of solution must be expressed in liters:
500 ml = 500/1000 = 0.5 liters.
Determine the molar concentration of potassium hydroxide:
Cv (KOH) = v (KOH) / V (KOH) = 0,36 / 0,5 = 0,72 mol / liter.
task №2
What sulfur oxide (IV) under normal conditions (i.e., when P = 101,325 Pa, and T = 273 K) should take in order to prepare a solution of sulfurous acid at a concentration of 2.5 mol /liter volume of 5 liters?
Determination
determine how much sulfurous acid contained in the solution:
ν (H2SO3) = Cv (H2SO3) ∙ V (solution) = 2,5 ∙ 5 = 12.5 mol.
equation obtain sulphurous acid has the following form:
SO2 + H2O = H2SO3
According to this:
ν (SO2) = ν (H2SO3);
ν (SO2) = 12,5 mol.
Bearing in mind that, under normal conditions, 1 mole of gas has a volume of 22.4 liters, count the amount of sulfur oxides:
V (SO2) = ν (SO2) ∙ 22,4 = 12,5 ∙ 22,4 = 280liters.
task №3
determine the molar concentration of NaOH in the solution when it is a mass fraction equal to 25.5%, and a density of 1.25 g / ml.
Determination
accepted as a sample solution volume of 1 liter and determine its mass:
m (solution) = V (solution) ∙ p (solution) = 1000 ∙ 1,25 = 1250 grams.
expect much in the sample of alkali by weight:
m (NaOH) = (w ∙ m (solution)) / 100% = (25,5 ∙ 1250) / 100 = 319 grams.
molar mass of sodium hydroxide is:
MNaOH = 23 + 16 + 1 = 40 g / mol.
expect much sodium hydroxide contained in the sample:
v (NaOH) = m / M = 319/40 = 8 mol.
Determine the molar concentration of alkali:
Cv (NaOH) = v / V = 8/1 = 8 mol / liter.
Problem №4
in water (100 g) was dissolved 10 grams of salt NaCl.Set the concentration of the solution (molality).
Determination
Molar mass of NaCl is:
MNaCl = 23 + 35 = 58 g / mol.
Number NaCl, contained in the solution:
ν (NaCl) = m / M = 10/58 = 0.17 mole.
In this case, the solvent is water:
100 grams of water = 100/1000 = 0.1 Kg H2O in the solution.
molal concentration of the solution is equal to:
Cm (NaCl) = v (NaCl) / m (water) = 0.17 / 0.1 = 1.7 mol / kg.
Problem №5
determine the molal concentration of 15% solution of alkali NaOH.
Determination
15% alkali solution means that every 100 grams of solution contains 15 grams of NaOH and 85 grams of water.Or that every 100 kilograms of the solution has 15 kilograms of NaOH and 85 kg of water.In order to prepare it, it is necessary in 85 grams (kilograms) H2O dissolved 15 grams (kg) of alkali.
molar mass of sodium hydroxide is:
MNaOH = 23 + 16 + 1 = 40 g / mol.
now find the amount of sodium hydroxide solution:
ν = m / M = 15/40 = 0.375 mol.
weight of solvent (water) in kg:
85 grams H2O = 85/1000 = 0.085 kg H2O in the solution.
then defined molal concentration:
Cm = (ν / m) = 0,375 / 0,085 = 4,41 mol / kg.
In accordance with these typical problem can be solved, and most of the others on the definition of molality and molar.
