# Concentration of Solution Problems and Definition

## Concentration of solution

The concentration of solution is the amount of solute present in a unit volume of solution.

### Mass Percentage

It is the mass of the component per 100g of the solution

$=\frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100$

### Volume Percentage

It is the volume of the component per 100 parts by the volume of the solution.

$=\frac{\text{Volume of component}}{\text{Total volume of solution}} \times 100$

### Mass by volume percentage

It is the mass of the solute dissolved in 100 mL of the solution.

### Parts per million

It is used when the concentration of the solute is in the trace. The concentration of pollutants in water or atmosphere is often expressed in terms of μg/mL or ppm.

$=\frac{\text{Volume of component}}{\text{Total number of parts of all} \\ \text{ Components of the solution}} \times 100$

### Parts per billion

It is the amount of solute dissolved in 109 parts of a solution.

$=\frac{\text{Mass of a solute particle}}{\text{Total mass of solution}} \times 10^9$

### Concentration or strength in g/L

It represents the number of grams of solute present in 1L of the solution.

$=\frac{\text{Mass of a solute (in g)}}{\text{volume of solution(in L)}}$

### Mole fraction (χ)

it is the ration of the number of moles of solute or solvent to the total number of moles of solute and solvent.

n = Number of moles

$\chi_{\text{ solute}}=\frac{\text{n}_{\text{ solute}}}{\text{n}_{\text{ solute}}+\text{n}_{\text{ solvent}}}$

$\chi_{\text{ solvent}}=\frac{\text{n}_{\text{ solvent}}}{\text{n}_{\text{ solute}}+\text{n}_{\text{ solvent}}}$

### Molality (m)

It represents the number of moles of the solute dissolved in 1000g of the solvent.

$=\frac{\text{Number of moles of solute}}{\text{Mass of solvent (in g)}} \times 1000$

$=\frac{\text{mass of solute (in g)}\times 1000}{\text{Molecular weight of solute}\times \text{mass of solvent (in g)}} \times 1000$

### Normality (N)

It is the number of gram equivalents of the substance dissolved per liter of the solution.

w = mass of solute, E = Equivalent weight, V = Volume

$=\frac{\text{Number of g-equivalent of solute}}{\text{Volume of solution (in L)}}$

$=\frac{\text{Mass of solute (in g)}}{\text{equivalent weight}\times \text{Volume (in L)}} \times 1000$

$=\frac{\text{mass of solute (w)(in g)}}{\text{equivalent weight (E)}\times \text{Volume (in mL)}} \times 1000$

OR

$N=\frac{\text{w}\times \text{1000}}{\text{E} \times \text{V}}$

### Molarity (M)

It is the number of moles of solute present in one liter of the solution.

$=\frac{\text{Number of moles of solute}}{\text{Volume of solution (in L)}}$

$=\frac{\text{mass of solute of (in g)}}{\text{Molecular weight of solute} \times \text{Volume of solution (in L)}}$

$=\frac{\text{number of millimoles of solute}}{\text{Volume of solution (in mL)}}$

$=\frac{\text{percent of solute} \times 10}{\text{molecular weight of solute}}$

$=\frac{\text{weight % of solute} \times \text{desnsity of the solution} \times 10}{\text{molecular weight of the solute}}$

### Formality (F)

It represents the number of the gram-formula weight of the substance dissolved per liter of the solution.It is used for ionic compounds which dissolve in a polar solvent to give a pair of ions.

$=\frac{\text{moles of substance added to solution}}{\text{Volume of solution (in L)}}$

## Inter-relation between different concentration of solution

### Relation between molarity and weight percentage

$=\frac{1000 \times\text{weight of % of solute}(x)}{(100-x) \times \text{molecular weight of solute}}$

### Relation between molality and mole fraction

m = Molality, χ = Mole fraction, M1 = Molecular mass of the solvent

$m=\frac{1000 \chi}{(1 \text{ – } \chi) M_{1}}$

### Relation between molality and molarity

m = Molality, M = Molarity, M= Molar mass of the solute, d = Density of the solution

$m=\frac{M}{d – MM_{2}}$

* If d in kg L-1 and M2 in kg mol-1

$m=\frac{M}{1000d – MM_{2}}\times 1000$

* If d in g mL-1 and M2 in g mol-1

$M=\frac{m \times d}{1 + \frac{m \times M_{2}}{1000}}$

### Relation between normality and strength in g/L

N = Normality, $x$ = % of the solute by mass, d = density of solution in g mL-1, E = equivalent mass of solute.

$N=\frac{x \times d \times 10}{E}$

### Relation between Molarity and strength in g/L

M = molarity

$M=\frac{\text{weight of solute (in g)}}{\text{molecular weight of solute}\times \text{volume of solution (in L)}}$

$M=\frac{\text {strength in g }L^{-1} }{\text {molecular weight of solute}}$

### Relation between normality and molarity

$\frac{\text {Normality }}{\text {Molarity}}=\frac{\text{Molecular weight}}{\text{Equivalent weight}}$

For an acid,

$\text{Basicity}=\frac{\text {Molecular weight }}{\text {Equivalent weight}}$

* Normality of a base = molarity $\times$ basicity

For a base,

$\text{Acidity}=\frac{\text {Molecular weight }}{\text {Equivalent weight}}$

* Normality of a base = molarity $\times$ acidity

### Relation between molarity and mole fraction of the solute

d = density of the solution g mL-1, M1 = molar mass of solvent, M2 = molar mass of solute, M = molarity of solution

$\chi=\frac{MM_{1}}{M(M_{1}-M_{2}+1000d)}$