## Solutions

# 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 10^{6}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

∴ M_{1}V_{1}= M_{2}V_{2}

(iii) Volume of solvent added = V_{2}− V_{1}

(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 ⇒ N_{1}V_{1}= N_{2}V_{2}

∴ volume of solvent added = V_{2}− 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}}