Introduction, Types and Concentration of Solutions

Solution: Homogeneous mixture of 2 (or) more non reacting components is called solution
Homogeneous mixture means that it’s composition and properties are uniform through out the mixture
Concentration: Amount (mass (or) volume (or) number of moles) of solute present in unit amount of solvent (or) solution is called concentration.
The component that is present in largest quantity (more number of moles) is known as solvent
Solvents determines the physical state in which solution exists.
Solute: One (or) more components other than solvent in the solution are called solutes
SOLUTION = SOLVENT + SOLUTE (S)
Standard solution: Solution with known concentration is called standard solution.
Methods of determination of concentration of solutions:
Mass percentage (w/W %) Mass of solute in grams present in 100gm of solution is called mass percentage
\tt \frac{w}{W}\% = \frac{Mass \ of \ solute \ in \ grams}{Mass \ of \ solution \ in \ grams} \times 100
= \tt \frac{Mass \ of \ solute \ (g)}{Mass \ of \ solute \ (g) \ + \ mass \ of \ solvent} \times 100
= \tt \frac{Mass \ of \ solute \ (g)}{Density \ of \ solution \ (g/ml) \times Volume \ of \ solution}
It has no units & it is Independent on temperature
w/W% is commercially used in industrial chemical application
Volume percentage (v\V %): Volume of solute in “ml” present in 100 ml of solution is called volume percentage
\tt \frac{v}{V}\% = \frac{Volume \ of \ solute \ (ml)}{Volume \ of \ solution \ in \ ml \ (or) \ (volume \ of \ solute \ + \ Volume \ of \ solvent)} \times 100
It has number of units ; It depends upon temperature.
Solution containing liquids are commonly expressed in volume percentage
Mass/volume percentage (w/v %): Mass of solute in grams present in 100 ml of solution is called w/v %
\tt \frac{w}{v} \% = \frac{Mass \ of \ solute \ (g)}{Volume \ of \ solution \ (ml)} \times 100
Units:- gm ml⁻¹ : Effect of T : T dependent
Commonly used in medicine & pharmacy this concentration method is used
Strength of the solution:- Mass of solute in grams present in 1L of solution is called strength of the solution.
Strength of the solution = \tt \frac{Mass \ of \ solute \ (gm)}{Volume \ of \ solution \ (L)}
= \tt \frac{Mass \ of \ solute \ (gm)}{Volume \ of \ solution \ (ml)} \times 1000
= \tt \left(\frac{w}{v}\times 100\right) \times 10 = \left(\frac{w}{v}\% \right) \times 10
Units : gm L⁻¹ (or) g dm⁻³
Effect of T: If is T dependent.
Parts per millions (ppm): Number of parts of solute present in 10⁶ parts of solution is called ppm
\tt ppm = \frac{number \ of \ parts \ of \ a \ component}{Total \ number \ of \ parts \ of \ all \ the \ components \ in \ solution} \times 10^{6}
Units: No units
Molarity (M):- Number of moles of solute present in 1L of solution is called molarity.
\tt Molarity (M) = \frac{Number \ of \ moles \ of \ solute(n)}{Volume \ of \ solution (v)L}
(a) \tt M = \frac{n}{v(ml)} \times 1000
(b) \tt M = \frac{G(g)}{GMW} \times \frac{1000}{v(mL)} , G = mass of solute in grams, GMW = Gram molecular mass
(c) \tt M = \left[\frac{G(g)}{V(ml)} \times 1000\right] \times \frac{1}{GMW}
\tt \Rightarrow M = \frac{Strength \ of \ solution}{GMW}
(d) \tt M = \frac{(w/v \%) \times 10}{GMW} \Rightarrow M = \left[\frac{G}{Mass \ of \ solution} \times 100\right] \times \frac{10 \times d(g ml^{-1})}{GMW}
\tt \Rightarrow M = \frac{\left(\frac{w}{w} \%\right)\times 10 \times d}{GMW}
Units:- Mole Liter⁻¹ (or) mole dm⁻³ (or) M
Effect of T: T dependent
If “M” is molarity, “V” is volume of solution & “n” number of moles of solute then
(a) Number of moles of solute n = MV(L)
(b) Number of millimoles of solute = MV(mL)
(c) Number of millimoles of solute = number of moles of solute × 1000 = n × 1000
Dilution principle:-
(i) Addition of solvent to solution is called dilution
(ii) In dilution process number of moles (or millimoles) of solute remains constant
i.e MV = constant
∴ M₁V₁ = M₂V₂
(iii) Volume of solvent added = V₂ − V₁
(iv) When 2 (or) more similar nature of solution are mixed with each other then
\tt M_{resultant} = \frac{M_{1}V_{1} + M_{2}V_{2} + .......}{V_{1} + V_{2} + ......}
Along with these solutions if some amount of solvent is added
\tt M_{resultant} = \frac{(M_{1}V_{1} + M_{2}V_{2} + .......)}{(V_{1} + V_{2} + ......) \ + \ x}
where X = volume of solvent added.
Normality(N):- Number of gram equivalents of solution is present in 1 litre of solution is called normality.
\tt Normality \ N = \frac{Number \ of \ gram \ equivalents \ of \ solute \ (GEN)}{volume \ of \ solution (v) L}
(a) \tt N = \frac{GEW}{V(L)}
(b) \tt N = \frac{GEW}{V(ml)} \times 1000
(c) \tt N = \frac{w(gm)}{GEW} \times \frac{1000}{v(ml)}
(d) \tt N = \frac{strength \ of \ the \ solution}{GEW}
(e) \tt N = \frac{\frac{w}{v}\% \times 10}{GEW}
(f) \tt N = \frac{wt\% \times 10 \times d}{GEW}
Unit of normality:- Gram equivalents L⁻¹ (or) gram equivalents dm⁻³ (or) N
It is also effected by temperature
(a) Number of GEN of solutes = NV (L)
(b) Number of milli equivalents of solute = NV(mL)
(c) Number of milli equivalents of solute = number of gram equivalents of solute × 1000
Dilution Principle:-
In dilution process number of gram equivalents (or milli equivalents) of solute remains constant
(a) NV = constant ⇒ N₁V₁ = N₂V₂
∴ volume of solvent added = V₂ − V1
(b) when 2 (or) more similar nature of solution are mixed with each other
\tt N_{resultant} = \frac{N_{1}V_{1} + N_{2}V_{2} + .......}{V_{1} + V_{2} + .......}
(c) Along with these solutions if some amount of solvent is added
\tt N_{resultant} = \frac{N_{1}V_{1} + N_{2}V_{2} + .......}{\left(V_{1} + V_{2} + .......\right) \ + \ x}
Where x is volume of solvent added.
Relation between M & N
\tt N = \frac{G}{GEW} \times \frac{1}{V(L)}
\tt = \frac{G}{\left(\frac{GMW}{n-factor}\right)} \times \frac{1}{V(L)}
\tt = n-factor \ \left\{\frac{G}{GMW} \times \frac{1}{V(L)}\right\}
N = n-factor × M
Molality (m):- Number of moles of solute present in 1 kg of solvent is known as molality
(a) molality (m) = \tt \frac{number \ of \ moles \ of \ solute \ (n)}{mass \ of \ solvent \ in \ kg}
(b) \tt m = \frac{n}{W_{(A)}}(g) \times 100
(c) \tt m = \frac{w_{(B)}}{GMW_{(B)}} \times \frac{1000}{W_{(A)}}(g)
w_(B) = mass of solute in gms
GMW_(B) = Gram molar mass of solute
w_(A) = Mass of solvent
Units:- mole kg⁻¹ (or) m
'T' effect: no effect of T
Relation between molality & wt %:
\tt m = \frac{x}{GMW} \times \frac{1000}{100-x}
x = Mass of solute
100-x = mass of solvent
Relation between molality & molarity
\tt m = \frac{M \times 1000}{d \times 1000 - M \times GMW}
M = molarity
d = density of the solution
GMW = Gram molecular mass of solute
Mole fraction :- Ratio between number of moles of one component to the total number of moles of all the components present in the solution.
\tt X_{A} = \frac{n_{A}}{n_{A} + n_{B}};X_{B} = \frac{n_{B}}{n_{A} + n_{B}}
X_A = mole fraction of solvent
X_B = mole fraction of solvent
Mole fraction has no units & it is independent an temperature.
Sum of mole fraction of all the component in solution is "1"
Relationship between molality & mole fraction.
\tt m = \frac{X_{B}}{1 - X_{B}}\times \frac{1000}{M_{A}}(or) \ m = \frac{1 - X_{A}}{X_{A}} \times \frac{1000}{M_{A}}
X_B = mole fraction of solute
X_A = mole fraction of solvent
M_A = Molecular mass of solvent
Trick to find out of which is more concentrate.
Case(i): For a given density of solution
If w/W% increases m > M
If w/W% decreases m < M
Case(ii): If d = 1 of solvent m > M
Case(iii): Only Molarity and molality of solutions are mentioned then M > m

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1. Mass percentage: The mass fraction of B in solution is
\tt w_{B}=\frac{Mass\ of\ B\ in\ solution}{Total\ mass\ of\ solution}
Mass percentage of B = w_B × 100

2. Volume percentage: The volume fraction of B in a solution is
\tt \phi_{B}=\frac{Volume\ of\ liquid\ B}{Volume\ of\ solution}
Volume percentage of B = Φ_B × 100

3. Mass by Volume Percentage of B = \tt \frac{Mass\ of\ constituent\ B}{Volume\ of\ solution}\times100

4. Parts per million of B = \tt \frac{Number\ of\ parts\ of\ B}{Total\ number\ of\ parts\ of\ all\ constituents}\times10^{6}

5. Amount (or Mole) Fraction \tt x_{B}=\frac{Amount\ of\ constituent\ B\ in\ the\ solution}{Total\ amount\ of\ constituents\ in\ the\ solution}=\frac{n_{B}}{\Sigma_{B}n_{B}}=\frac{n_{B}}{n_{total}}

6. Molarity M = \tt \frac{Amount\ of\ solute\ in\ solution}{Volume\ of\ solution\ expressed\ in\ dm^{3}\left(i.e.L\right)}=\frac{n_{2}}{V}

7. Molality m = \tt \frac{Amount\ of\ solute}{Mass\ of\ solvent\ expressed\ in\ kg}=\frac{n_{2}}{m_{1}}

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