# 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}}
XA = mole fraction of solvent
XB = 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}}
XB = mole fraction of solute
XA = mole fraction of solvent
MA = 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

### Part5: View the Topic in this Video from 0:08 to 16:55

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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 = wB × 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 = ΦB × 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}}