## Chemistry Solutions | Basic Chemical concepts

This blog contains Chemistry Solutions | Basic Chemical concepts - Solutions related all Details like Solute solvent Normality Molarity %w/w %w/v ppm Calculations

### Blog Content - Chemistry Solutions | Basic Chemical concepts

**Solvent Solute Solution**

Types of Solutions -

Types of Solutions -

1. Homogenous solution

2. Heterogeneous solution

3. Aqueous solution

4. Ammonical solution

5. Non aqueous solution

6. Gaseous, Liquid, Solid solution

7. Saturated solution

8. Unsaturated solution

9. Supersaturated solution.

**Mole Concept**

Concentration terms -

Concentration terms -

Normality,

Molarity,

Molality,

Formality,

Mole fraction,

%w/w, % w/v,

PPM

Molarity,

Molality,

Formality,

Mole fraction,

%w/w, % w/v,

PPM

**Calculations**

### Interview QnA - Chemistry Solutions | Basic Chemical concepts

- What is Solute, Solvent, & Solution?
- What are saturated Unsaturated and Supersaturated solutions?
- What is equivalent weight?
- What is difference between 1 Molarity solution & 1 Molality Solution?
- Why Molarity is reduced with temperature increases?

### 1. Definitions - Solvent, Solute, Solution

#### Solvent

The component which present in large quantity#### Solute

The component present in smaller quantity#### Solution

It is mixture of solvent + solute = solution#### Example -

we are making sugar dissolve in waterSolvent - Water (Larger in quality)

Solute - Sugar ( Less in quantity)

Solution = solvent (water) + solute (sugar)

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### 2. Types of solutions

1. Homogenous solution2. Heterogeneous solution

3. Aqueous solution

4. Ammonical solution

5. Non aqueous solution

6. Gaseous, Liquid, Solid solution

7. Saturated solution

8. Unsaturated solution

9. Supersaturated solution.

#### 1. Homogeneous solution

Solution in which two substitutes mixed uniform composition, component can not be identified separately.Example -

Sugar solution in water

#### 2. Heterogeneous solution

Solution in which two or more substances mixed non uniform composition and solution can be identified separately.

*Example -*Oil mixed in water

We can identify both as separate.

#### 3. Aqueous solution

Solution in which Water is solvent#### 4. Ammonical solution

Solution in which ammonia is solvent#### 5. Non aqueous solution

Solution in which Water is absent#### 6. Gaseous, Liquid, Solid solution

*Gaseous solution*1. Gas in Gas solution - H2 and O2 solution

2. Liquid in Gas - Water in Air

3. Solid in Gas - Camphor in Air

*Liquid solution*1. Gas in liquid - Carbon dioxide in water (soda water)

2. Liquid in liquid - Alcohol in Water

3. Solid in Liquid - Sugar solution

*Solid solutions*1. Gas in Solid - H2 palladium

2. Liquid in Solid - Hg in Zn

3. Solid in Solid - Alloys ( Zn in Cu)

#### 7. Saturated solution

- A solution which containing maximum amount of solute
- Solvent can not able dissolve more solute now.
- More addition of solute will be settled down in solution

#### 8. Unsaturated solution

- Solution is able to dissolve more solute.
- Unsaturated solution can still dissolve solute.

#### 9. Supersaturated solution.

- A solution which contains more solute than required at a given temperature.
- Addition of more solute or reduction of temperature will result in form of Crystal.

### 3. Mole Concept

*Definition*
1 mole is 6.023 × 10^23 Particles of substance.

Or

Amount of substance containing which are present in 0.012 gram Carbon12

Mole relation to weight of substance and Molecular weight.

Mole = Weight / Molecular weightMolecular weight of Hydrogen is 2 means

Hydrogen

2 Gram = 1 Mole = 6.023 × 10^23 Particles

*Calculations*Example - 1

How much moles of H2SO4 in 196 Gram

Molecular weight of H2SO4 = 98

Mole = Weight / Molecular weight.

Mole = 196 / 98 =2

Example - 2

What is weight of 0.2 mole H2SO4?

Mole = 0.2

Molecular weight = 98

Mole = Weight / Molecular weight

Weight = Mole × Molecular Wight

= 0.2 × 98 = 19.6 Gram

### 4. Normality, Molarity, Molality,Formality, Mole fraction, %w/w, % w/v, PPM

#### 1. Normality

The number of gram equivalent of the solute present in 1 liter of solution at given temperature.

It is indicated by N

*Example**1 Normal solution - 40 Gram NaOH (100%) dissolved in 1 liter water will become 1 Normality NaOH solution.*
How it is made?

Gram equivalent weight = 40 Gram

Solute NaOH weight = 40 Gram

Solution quantity = 1 Liter = 1000 ml

As per definition

Normality = (Gram / Liter) / Equivalent weight

Here valent is 1 of OH- in NaOH

So Normality =

(1000 × Solute Weight) / (Equivalent weight × Solution Volume)

= (1000 × 40) / (40 × 1000)

= (1000 × 40) / (40 × 1000)

= 1 Normality NaOH solution.

Normality relation for dilution

N1V1 = N2V2

N = Concentration

V = Volume

#### Equivalent weight of substance calculation

Definition - The equivalent weight weight of substance is the weight of it that can react with 1 gram of hydrogen.

It is the ratio of Molecular weight to Basicity of acid, or Acidity of base.

E(Acid or Base)

= Molecular weight / Basicity of acid or Acidity of Base

Here you can see examples of some acids/base with calculations of equivalent weight.

In above equations acidity and basicity can be replaced with

- Total valency of cations, anions or element.
- Numbers of electron gained in Oxidising agent
- Numbers of electron lost in Reducing agent

#### 2. Molarity

When 1 mole of solute dissolved in 1 liter solution at given temperature called 1 Molarity solution or 1 molar solution or 1M solution.

It is indicated by M.

Molarity

= Number of Moles / 1 liter of solutions

= (Weight of substance / Gram molecular weight) / 1 liter solution

= (weight of Substance × 1000) / ( Gram Molecular Weight × Solution quantity)

Relation for dilution

M1V1 = M2V2

M = Molarity of solutions

V = Volume of solutions

Volumatric calculaiton

(M1V1 / n1) = (M2V2 / n2)

n = number of moles in solution

Molarity unit is 'moles/liter' Molarity depends on temperature

Because Charles's law states Temperature is proportional to Volume. So When temperature rises volume also increases and as per molarity equation molarity reduces.

#### 3. Molality

When 1 mole solute weight dissolved in 1 kg solution it is called 1 molality or 1 molal or 1m solution.

Mollity

= (solute mole/ 1 kg solvent) / Molecular weight

= ( 1000 * Solute weight) / (Solute molecular weight * Solution weight)

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= (solute mole/ 1 kg solvent) / Molecular weight

= ( 1000 * Solute weight) / (Solute molecular weight * Solution weight)

#### 4. Formality

- Formality of a solution is defined as the number of gram formula masses of the ionic solute dissolved per liter of the solution.
- Solid Ionic solutions are in ionic form so we will call formula weight instead of molecular weight.
- It is indicated by F
- Simply formality showing concentration of ionic solids which does not exists as molecules but exists as network of ions.
- A solution which is containing gram weight according equal to its formula weight in 1000 ml [1 liter] solution is called 1 formality solution or 1 formal solution or 1 F solution.
- Alum solutions like K
_{2}SO_{4}.Al_{2}(SO_{4})_{3}.24H_{2}O Its formula is KAl(SO_{4})_{2}.12H_{2}O

- If we dissolve 474 Gram alam in 1000 ml [1 liter] water then it will become 1 formality solution of Alum.
- Normality, Molarity, & Formality changes with change in temperature as explained above [Because Charles's law states Temperature is proportional to Volume. So When temperature rises volume also increases and as per molarity equation molarity reduces.

#### 5. Mole fraction

Mole fraction of any substance is defined as the ratio of substance's mole to total moles of solution.

Mole fraction of X-substance = X-substance's mole / Total mole present in solution

Mole fraction of X-substance = X-substance's mole / Total mole present in solution

Total Mole fraction must equal to 1.

*Example -**Calculate mole fraction of 80 gram NaOH dissolved in 324 gram water at 25°C*Mole of Water = Weight / Molecular weight of water

= 324 / 18

__Mole of Water = 18 mole__

Mole of NaOH = Weight / Molecular weight of water

= 80 / 40

= 80 / 40

__Mole of NaOH = 2 mole__*Total mole = Mole of Water + Mole of NaOH*
= 18 + 2

__Total mole = 20 mole__
Mole fraction of Water = Mole of water / Total mole of solution

= 18/20

= 0.9

Mole fraction of NaOH = Mole of NaOH / Total mole of solution

= 2/20

= 0.1

Total of Mole fractions = Mole fraction of Water + Mole fraction of NaOH.

= 0.9 + 0.1

=1

Weight of solute dissolved in 100 gram of solution is called %W/W.

This solutions are made with importance of solute weight.

%W/W = ( 100 * Solute weight ) / ( Weight of Solute + Weight of Solvent )

%W/W = ( 100 * Solute weight ) / ( Weight of Solution )

%W/W = ( 100 * Solute weight ) / ( Weight of Solution )

Solute weight = ( 10 *500 ) / 100

= 18/20

= 0.9

Mole fraction of NaOH = Mole of NaOH / Total mole of solution

= 2/20

= 0.1

Total of Mole fractions = Mole fraction of Water + Mole fraction of NaOH.

= 0.9 + 0.1

=1

#### 6. Weight fraction

Weight fraction is classified in- %W/W - Weight to Weight Percentage
- %W/V - Weight to Volume Percentage.
- PPM - Parts Per Million.

**%W/W - Weight to Weight Percentage**Weight of solute dissolved in 100 gram of solution is called %W/W.

This solutions are made with importance of solute weight.

%W/W = ( 100 * Solute weight ) / ( Weight of Solute + Weight of Solvent )

%W/W = ( 100 * Solute weight ) / ( Weight of Solution )

*Example**How much NaOH required to make 500 gram 10 %W/W solution?*%W/W = ( 100 * Solute weight ) / ( Weight of Solution )

%W/W = 10 %W/W NaOH10 = ( 100 * Solute weight ) / ( 500 )

Weight of Solute = ? Gram

Weight of Solution = 500 Gram

Solute weight = ( 10 *500 ) / 100

__Solute weight = 50 Gram.__So we have to take 50 Gram NaOH(100% Concentration) and then add water till total weight reach 500 Gram.

**%W/V - Weight to Weight Percentage**

Weight of solute dissolved in 100 millilitre of solution is called %W/V.

This solutions are made with importance of solute weight in Solution volume.

%W/V = ( 100 * Solute weight ) / ( Volume of Solution )

*Example***°C**

*How much NaOH required to make 2000 ml 5 %W/V solution at 25*

*?*%W/V = ( 100 * Solute weight ) / ( Volume of Solution )

%W/V = 5 %W/V NaOH5 = ( 100 * Solute weight ) / ( 2000)

Weight of Solute = ? Gram

Weight of Solution = 2000 ml

Solute weight = ( 5 *2000 ) / 100

__Solute weight = 100 Gram.__

So we have to take 100 Gram NaOH(100% Concentration) and then add water till total volume reach 2000 ml.

**PPM - Parts Per Million**

It is milligram weight per liter .

Parts Per Million scale is used when solute is too much less quantity.

Oxygen in sea water

Pollution in air.

### 5. Calculations

#### Normality calculations

**Normality Example-1**

**If We dissolve 2 gram NaOH in 5 liter solution then how much normal solution prepared?**

Molecular weight of NaOH = 40 Gram / mole

Solute weight = 2 Gram NaOH

Equivalent weight of NaOH = 40/1 [ Because Acidity of NaOH is 1 due to its valency of OH- is 1]

Solution Volume = 5 liter = 5000 Millilitre

Normality =

(1000 × Solute Weight) / (Equivalent weight × Solution Volume)= 1000 * 2 / 40*5000

__Normality = 0.01 N__

**Normality**

**Example-2**

**Calculate Normality if 9.8 Gram of H2SO4 dissolved in 2 liter of water.**

**Molecular weight of NaOH = 98 Gram / mole**

Solute weight = 9.8 Gram NaOH

Equivalent weight of H2SO4 = 98/2 = 49[ Because Basicity of NaOH is 2 due to its valency of H+ is 2]

Solution Volume = 2 liter = 2000 Millilitre

Normality =

(1000 × Solute Weight) / (Equivalent weight × Solution Volume)= 1000 * 9.8 / 49*2000

__Normality = 0.1 N__

**Normality**

**Example-3**

**Calculate the weight of oxalic acid (H2C2O4.2H2O) required to prepare 0.05 normal solution in 2 liter.**

**Molecular weight of H2C2O4.2H2O = 126 Gram / mole**

Solute weight = ? Gram H2C2O4.2H2O

Equivalent weight of H2SO4 = 126/2 = 63[ Because Basicity of NaOH is 2 due to its valency of H+ is 2]

Solution Volume = 2 liter = 2000 Millilitre

Normality = 0.05

Normality =

(1000 × Solute Weight) / (Equivalent weight × Solution Volume)0.05= 1000 * Solute weight / 63*2000

Solute weight = 6.3

**Normality**

**Example-4**

**How much Water will added to 500 ml of 0.05N Na2CO3to get 0.01N solution**

**Normality equation for dilution**

N1V1 = N2V2

N1 = 0.05

V1= 500ml

N2= 0.01 N

V2= ?

V2 = N1V1 / N2

V2 = 0.05 * 500 / 0.01

__V2 = 2500 ml__

**Normality**

**Example-5**

**How much normality solution of 20ml NaOH required to neutralise 50 ml 0.02N H2SO4?**

**Normality equation for dilution**

N1V1 = N2V2

N1 = 0.02

V1= 50 ml

N2= ? N

V2= 20

N1V1 = N2V2

N2 = N1V1 / V2 = 0.02 * 50 / 20

__N2 = 0.05 N__

#### Molarity Calculations

**Molarity**

**Example-1**

**How much HCl required to make 2.0 M HCl 500ml solution?**

**Molarity = = (weight of Substance × 1000) / ( Gram Molecular Weight × Solution quantity)**

Molarity = 2 M

weight of Substance = ?

Molecular Weight of HCl = 36.5 Gram/mole

Solution quantity = 500 ml

So

Weight of Substance ( HCl ) = Molarity * Gram Molecular Weight × Solution quantity / 1000

= 2* 36.5*500 / 1000

__Weight of Substance ( HCl ) = 36.5 Gram HCl__

**Molarity**

**Example-2**

**2 moles of a solute is dissolved in 5 liter of solution What is the Molarity of solution?**

Molarity of solution(M) = Number of moles of solute (n) / Volume of solution (V)

Number of moles of solute (n) = 2

Volume of solution (V) = 5 liters of solution.

__Molarity of solution(M) = 2 / 5 = 0.4 M__

**Molarity**

**Example-3**

**Find the molarity of solution containing 171 gram sugar (sucrose) in 2 liter water?**

Molarity = (weight of Substance × 1000) / ( Gram Molecular Weight × Solution quantity)

Molarity = __ M

weight of Substance = 171

Molecular Weight of Sucrose (Molecular formula = C12H22O11) = 342 Gram/mole

Solution quantity = 2000 ml

__Molarity = 171*1 / 342*2 = 0.25 M__

**Molarity**

**Example-4**

**Find the volume of water required to Prepare 0.1 M H2SO4 from 200 ml 0.5 M solution?**

Molarity formula for dilution

M1V1= M2V2

M1= 0.5 M

V1= 200 ml

M2= 0.1 M

V2= ?

V2 = M1V1 / M2

= 0.5*200 / 0.1

V2 = 1000 total volume should be done

Remaining quantity = V2 - V1 = 1000 - 200 = 800 ml water is required.

#### Molality Calculations

**Molality**

**Example-1**

**If we dissolve 0.8 gram NaOH in 2 Kilogram Solvent then what will be molality of solution?**

Molality = (weight of Substance × 1000) / ( Gram Molecular Weight × Solution quantity)

Molality = __ m

weight of Substance = 2000

Molecular Weight of Caustic = 40 Gram/mole

Solution quantity = 2000 mg

= 1000 * 0.8/40*2000

__Molality = 0.01 m__

#### Mole fraction Calculations

**Mole fraction**

**Example-1**

**Calculate mole fractions of solution where 80 gram NaOH dissolved in 324 Gram water?**

Mole fractions of Substance = Substance mole / Total mole of solutionsTotal mole = Water mole + NaOH mole

Moles of Water = Weight of water / Molecular weight = 324 / 18 = 18

Moles of NaOH = Weight of NaOH / Molecular weight = 80 / 40 = 2

Total mole = 18 + 2 = 20

Mole Fraction of Water

= Water mole / Total moles of solution

= 18/20

=0.9

Mole Fraction of NaOH

= NaOH mole / Total moles of solution

= 2/20

=0.1

Total of all substance mole fraction must be 1.

Total mole fractions

= Water mole + NaOH mole.

= 0.9 + 0.1

= 1

#### Weight fraction Calculations

**Weight fraction**

**Example-1**

**Calculate NaOH required to make 10% W/W**

**concentrated**

**500 Gram**

**NaOH?**

%W/W = ( 100 * Solute weight ) / ( Weight of Solution )

%W/W = 10 %W/W NaOH10 = ( 100 * Solute weight ) / ( 500 )

Weight of Solute = ? Gram

Weight of Solution = 500 Gram

Solute weight = ( 10 *500 ) / 100

__Solute weight = 50 Gram.__

*Example***°C**

*How much NaOH required to make 2000 ml 5 %W/V solution at 25*

*?*%W/V = ( 100 * Solute weight ) / ( Volume of Solution )

%W/V = 5 %W/V NaOH5 = ( 100 * Solute weight ) / ( 2000)

Weight of Solute = ? Gram

Weight of Solution = 2000 ml

Solute weight = ( 5 *2000 ) / 100

__Solute weight = 100 Gram.__

So we have to take 100 Gram NaOH(100% Concentration) and then add water till total volume reach 2000 ml.

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