Often, concentration of solution is described qualitatively, through the use of such words as dilute and concentrated. These are relative terms.
Suppose we have 3 solutions: A,B, and C with 20 ml,30 ml, and 50 ml alcohol in 100 ml of water each, respectively. In this case, B is more dilute than C while it is more concentrated than A. A solution can be diluted by adding solvent or removing solute. On the other hand, a solution can be concentrated by adding solute or removing solvent.
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Quantitative Description
Molarity is defined as the number of moles of solute per litre of solution:
(Molarity)=(Number of litres of solution)(Number of moles of solute).
It is generally represented by M. □
3.7 gCa(OH)X2 is dissolved in 250 ml of solution. What is its molarity?
The molar mass of Ca(OH)X2 is 40+32+2=74 g. So the number of moles is obtained as follows:
(Number of moles)=(Molar mass)(Given weight)=743.7=0.05.
Now, 250 ml=0.25 L. Hence the molarity is 0.250.05=0.2 M.□
We have mixed two glucose solution: 500 ml of 0.4 M and 800 ml of 0.25 M. What is the molarity of the final solution formed after mixing these two solutions?
We know that (Number of moles of solute)=(Molarity)×(Volume of solution).
In the first solution, we have 1000500×0.4=0.2 mol.
In the second solution, we have 1000800×0.25=0.2 mol.
When two solutions are mixed, they will have a total of 0.2+0.2=0.4 mol of glucose, and the final volume is 1.3 L. So the final molarity is 1.30.4=134 M.□
A 0.750 M solution of sulfuric acid HX2SOX4 has a density of 1.046 grams per milliliter at 20∘C. What is the molality of the solution?
To simplify the problem, assume you have exactly 1 L of solution. Then you have
(Moles of HX2SOX4)(Mass of solution)(Mass of solute)(Mass of solvent)(Molality)=(Molarity)×(Volume)=0.75×1=0.75=(Density)×(Volume)=1.046×1000=1046 (g)=(Molar Mass)×(Number of moles)=98×0.75=73.5 (g)=(Mass of solution)−(Mass of solute)=1046−73.5=972.5 (g)=0.9725 (kg)=(Mass of solvent)(Moles of solute)=0.97250.75=0.771 (M).□
Find the molality of 4 molar aqueous pure nitric acid solution. Consider that the density of the solution is 1 gm/ml for the sake of this question.
Normality is defined as the number of gram equivalents per liter of solution:
(Normality)=(Number of litres of solution)(Number of gram equivalents).
It is generally represented by N.
Molarity can be converted to normality by multiplying by valency factor:
(Normality)=(Valency Factor)×(Molarity).□
What is the normality of a 0.5 MHX2SOX4 solution?
The n-factor of HX2SOX4 is 2 since sulphuric acid has two replaceable HX+ ions.
Since (Normality)=(n-factor)×(Molarity), the normality is
0.5×2=1 N.□
In a beaker containing 2 liters of 3 normal solution of pure sodium hydroxide, how much hydrochloric acid of the same volume must be added to neutralize all of sodium hydroxide contained in it?
Submit your answer in grams (g).
Mole fraction for a component is defined as the number of moles of that component per mole of mixture. In a mixture of p components, the mole fraction of the kth component is given by xk=n1+n2+n3+⋯+npnk, where na represents the number of moles of component a. □
An aqueous solution weighing 100 gram has 60 gram of pure chalk. Then find the mole fraction of the chalk.
If the answer can be expressed as ba, where a and b are coprime positive integers, then submit the value of a+b.
Formality is defined as the number of moles of formula units in one liter of solution. It is used for ionic compounds only: (Formality)=(GFM)×[Volume of solution (ml)]W×1000, where GFM stands for gram formula mass. □
Percentage weight by weight is defined as (Mass of solution)(Mass of solute)×100.□
Percentage weight by volume is defined as [Volume of solution (ml)][Mass of solute (g)]×100.□
Percentage volume by volume is defined as (Volume of solution)(Volume of solute)×100.□
Find the molarity of aqueous solution of pure hydrochloric acid which is 15% volume-by-volume.
Details and Assumptions:
Density of solution=2 gm/ml.
Density of solute=1.5 gm/ml.
Parts per million (PPM) is defined as
(Mass of solution)(Mass of solute)×106.
As this concentration unit is generally used for solutions in which the mass of solute is very small as compared to the mass of solution, we can say that this concentration unit is the same as
(Mass of solvent)(Mass of solute)×106.□
Relation between the Concentration Terms
We will take mass of solute to be ma, mass of solvent mb, density of solution d, molecular mass of solute Ma, and molecular mass of solvent Mb
So we know that
mb g of solute is present in ma g of solvent, and
mamb g of solute is present in 1 g of solvent,
which implies
ma1000mb g of solute in 1000 g of solvent.
So, Molality(m)=Mbma1000mb⟹ma×Mb1000mb.
Now, say, assume
mb g of solute in ma+mb g of solution
mb g of solute in dma+mb ml of solution.
Then we have ma+mb1000mb×d in 1 L of solution, which implies Molarity(M)=Mbma+mb1000mb×d.
So MolarityMolality=d×mama+mb.
Here we get the relation guys with derivation. For instance, take the relation between normality (N) and molarity (M) to be
Molarity×n-factor=Normality.□
Dependence on Volume/Temperature
Some of the concentration terms depend on the volume of solution/solvent like molarity and normality. The volume changes as the temperature is varied. Thus, such concentration terms are temperature dependent. On the other hand, terms like molality and mole fraction do not depend on volume, thus being temperature independent. The density is often required to convert temperature independent terms into temperature dependent terms or vice versa.