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**CONCENTRATION OF SOLUTION**

We can calculate the concentration of solutions by various methods. Let’s study each method and determine the formulas for this method.

- Percent by weight
- Percent by volume
- Percent by weight/volume
- Molarity
- Molality
- Normality
- Mole Fraction
- ppm
- Formality
- Strength (g/L)

**1. Mass/Weight Percentage or Percentage by Mass/Weight**

It is the amount of solute in grams present in 100 grams of the solution. Therefore, the formula will be:

Mass Percentage = Mass of Solute/Mass of Solution × 100

= mass of Solute/Mass of Solute + Mass of Solvent × 100

= Mass of Solute/Volume of Solution + Density of Solution × 100

The ratio mass of solute to the mass of the solvent is the mass fraction. Thus, the mass percentage of solute = Mass fraction × 100. 10% solution of sugar by mass means that 10 grams of sugar is present in 100 grams of the solution, i.e., we have dissolved 10 grams of sugar in 90 grams of water.**2. Volume Percentage**

It is the volume of solute in mL present in 100 mL solution. The formula will be:**Volume Percentage = Volume of solute/Volume of Solution × 100**

10% solution of HCl by volume means that 10 mL of liquid HCl is present in 100 mL of the solution.**3. Mass by Volume Percentage **

It is the mass of solute present in 100 mL of solution. We can calculate the mass of the solute using the volume percentage.__The formula would be:__**Mass by Volume Percentage = Mass of Solute/Volume of Solution × 100**

It changes on changing temperature.**Example:** A 10% (w/v) urea solution. = 10 gm of urea is present in 100 mL of solution.

Density of solution is required to calculate weight of solute and weight of solvent.**4. Molarity**

The molarity of a solution gives the number of gram molecules of the solute present in one litre of the solution.**Molarity(M) = Number of moles of solute/Volume of Solution in L**

w = Mass of solute in grams

M = molecular weight of solute in gm/mol.

V = volume of solution in ml.**Example. **1 mol L^{-1} solution of KCl means that 1 mol of KCl is dissolved in 1 L of water.__Unit of molarity: mol L__^{-1}

- Molarity is most common way of representing the concentration of solution.
- Molarity is depend on temperature as M ∝ 1/T
- When a solution is diluted (x times), its molarity also decreases (by x times)

**Example.** **Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.****Solution.**

Moles of NaOH = 5g/40 g mol^{-1} = 0.125 mol

Volume of the solution in litres = 450 mL / 1000 mL L^{-1 }

Using equation (2.8),

= 0.278 mol L^{–1}

= 0.278 mol dm^{–3}**5. Molality**

Molality of a solution is the number of moles of solute dissolved in 1 Kg of the solvent.**Molality (m) = Number of moles of solute/Mass of Solvent in Kg**

w = mass of solute in grams

M = molecular wt of solute

W = mass of solvent in gram.

Thus, if one gram molecule of a solute is present in 1 kg of the solvent, the concentration of solutions is said to be one molal.__The unit of molarity is mol kg__^{-1}.

- Molality is the most convenient method to express the concentration of solutions because it involves the mass of liquids rather than their volumes. It is also independent of the variation in temperature.

**Example.** Calculate molality of 2.5 g of ethanoic acid (CH_{3}COOH) in 75 g of benzene.**Solution.**

Molar mass of C_{2}H_{4}O_{2} : 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol^{–1}

Moles of C_{2}H_{4}O_{2} = 2.5/60 g mol^{-1} = 0.0417 mol

Mass of benzene in kg = 75 g/1000 g kg^{–1} = 75 × 10^{–3} kg

Molality of C_{2}H_{4}O_{2}

**6. Normality **

The normality of a solution gives the number of gram equivalents of the solute present in one litre of the solution.**Normality (N) = Number of gram equivalent of solute/Volume of Solution in LEquivalent mass = Molar mass/n - factor**

No. of equivalent = Mass of solute/equivalent mass

= No. of moles of solute x n - factor

w = mass of solute in gram

V = volume of solution in ml

E = equivalent wt of solute

Thus, if one gram equivalent of a solute is present in one litre of the solution, the concentration of solutions is said to be 1 normal.

- 1N = Normal = One gram equivalent of the solute per litre of solution = Normality is 1
- N/2 = Seminormal = 0.5 g equivalent of the solute per litre of solution = Normality is 0.5
- N/10 = Decinormal = 0.1 g equivalent of the solute per litre of solution = Normality is 0.1
- N/100 = Centinormal = 0.01 g equivalent of the solute per litre of solution = Normality is 0.01
- N/1000 = Millinormal = 0.001 g equivalent of the solute per litre of solution = Normality is 0.001

**Relationship between normality and other Concentration Terms**

- Molarity × Molecular mass= Strength of the solution (g/L)
- Normality × Equivalent mass = Strength of the solution (g/L)
- Molarity × Molecular mass= Normality x Equivalent mass
- Normality = n × Molarity

**7. Mole Fraction**

Mole fraction can be defined as the ratio of number of moles of the component in the solution to the total number of moles of all components in the solution.

- It is denoted by the alphabet x and subscript written on the right hand side of x denotes the component of which mole fraction is being calculated.
- Mathematically, Mole fraction of a component =
**Number of moles of the component/Total number of moles of all components** - Let us consider a binary mixture of A and B. let the number of moles of A and B be n
_{A}and n_{B}respectively, then

Mole fraction of A = x_{A}= n_{A}/ n_{A}+ n_{B}

Mole fraction of B = x_{B}= n_{A}/ n_{A}+ n_{B} - For solution where number of components = i

- In a given solution sum of the mole fractions of all the components is
unity.

Mathematically, x_{1}+ x_{2}+ x_{3}…. + x_{i}=1

**Example. **Calculate the mole fraction of ethylene glycol (C_{2}H_{6}O_{2}) in a solution containing 20% of C_{2}H_{6}O_{2} by mass.

**Solution.**

Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water.

Molar mass of C_{2}H_{6}O_{2 }= 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol^{–1}

Mole fraction of water can also be calculated as: 1 – 0.068 = 0.932.**8. Parts per million (ppm)**

When a solute is present in trace quantities, it is convenient to express the concentration of solutions in parts per million (ppm). The formula is as follows:

In case of mass, we may express it as:

In case of volume, we may express it as:

So, we can express the concentration of solutions in parts per million as mass to mass, volume to volume and mass to volume form.

- Atmospheric pollution in cities is also expressed in ppm by volume.
- It refers to the volume of the pollutant in 106 units of volume.
- 10 ppm of SO
_{2}in the air means 10 mL of SO_{2}is present in 106 mL of air.

**9. Formality**

- It is the number of mass in grams present per litre of solution.
- In case, formula mass is equal to molecular mass, formality is equal to molarity.
- Like molarity and normality, the formality is also dependent on temperature. It is used for
**ionic compounds**in which there is no existence of a molecule. - A mole of ionic compounds is called formal and molarity as the formality.

Where,

- w = weight of solute,
- f = formula weight of solute
- V= volume of solution
- nf = no. of gram formula weight

**10. Strength (g/L)**

Weight of solute per litre (1000 mL) of solution.**Example:** 10% w/v sucrose solution, then specify its concentration in g/L

In 100 mL solution, weight of sucrose = 10 g

Therefore, in 1000mL solution= 10/100 = 100 g

Hence strength will be 100g/L**MOLARITY OF MIXING**

Let there be three samples of solution (containing same solvent and solute) with their molarities M_{1} ,M_{2} ,M_{3} and volumes V_{1} ,V_{2} ,V_{3} respectively.

These solutions are mixed; molarity of mixed solution may be given as:

where, MR is the Resultant molarity

V_{1 }+ V_{2} + V_{3} = Resultant volume after mixing**Note:** Molarity is dependent on volume; therefore, it depends on temperature.

1M | molar solution, i.e, molarity is l |

0.5 M or M/2 | Semimolar |

0.1 M or M/2 | Decimolar |

0.01 M or M/100 | Centimolar |

0.001 M or M/100 | Millimolar |

**DILUTION FORMULA**

M_{1}V_{1} = M_{2}V_{2}

(Concentrated solution) (Diluted solution)

**Problem 1. ****Calculate the masses of cane sugar and water required to prepare 250 g of 25% cane sugar solution.****Solution.**

Mass percentage of cane sugar = 25

We know that,

Mass percentage = Mass of solute/Mass of solution × 100

25 = Mass of cane sugar/250g × 100

Mass of cane sugar = 25 ×250/100 = 62.5 g

Mass of water (250 -62.5)g = 187.5 g**Problem 2. ****A tank is charged with a mixture of 1.0 x 10 ^{3} mol of oxygen and 4.5 x 10^{3} mol of helium. Calculate the mole fraction of each gas in the **

The given parameters are

n

Mole fraction can be calculated as

x

=4.5 mol / 5.5 mol = 0.82

xO

= 1.0 x 10

= 0.18

Volume of 100 g of the solution

= 100/d = 100/1.09 mL

= 100/1.09 × 1000 litre

= 1/1.09 × 10 litre

Number of moles of H

Molarity = No. of moles of H

= 1.445 M

We know that,

Normality = Molarity × n

**Problem 4. ****What is the mole fraction of carbon tetrachloride (CCl _{4}) in solution if 3.47 moles of CCl_{4} is dissolved in 8.54 moles of benzene (C_{6}H_{6})?**

Mole Fraction CCI

X

XCCI4 = 0.289

The grams of acetone will need to be converted into moles in order to solve for mole fraction.

Mole Fraction CH

X

X

100.0 g / 18.0 g mol

Add that to the 0.100 mol of NaCl = 5.56 + 0.100 = 5.66

mol total Mole fraction of NaCl = 0.100 mol / 5.66 mol = 0.018

The mole fraction of the H

5.56 mol / 5.66 mol = 0.982

H

C

(2) Determine mole fractions:

H

C

pentane: C

hexane: C

benzene: C

1.62 mol C

1.00 kg = 1000 g of water

1000 g / 18.0152 g/mol = 55.50868 mol

1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three sf)

100.0 g / 342.2948 g/mol = 0.292145835 mol

0.020 = 0.292 / (0.292 + x)

(0.020) (0.292 + x) = 0.292

0.00584 + 0.02x = 0.292

0.02x = 0.28616

x = 14.308 mol of H

Comment: you can also do this:

0.292 is to 0.02 as x is to 0.98

14.308 mol x 18.015 g/mol = 258.0 g

χwater = moles of water / (moles of water + moles of nitrogen)

1.00 x 10

I'm going to carry some guard digits until the end of the calculation.

0.99999x = 7.139440411 x 10

x = 7.139511806 x 10

7.139511806 x 10

1.29 x 10

50.0 g is urea

50.0 g is cinnamic

50.0 g / 60.06 g/mol = 0.8325 mol

cinnamic acid: 50.0 g / 148.16 g/mol = 0.3375 mol

0.3375 mol / 1.1700 mol = 0.2885

Mole of NaCl = 5/58.5 = 0.0854 (Mol. wt. of NaCl = 58.5)

Molality = Moles/Wt. of solvent in gram x 1000

= 0.0854/1000 x 1000 = 0.0854 m

Volume of the solution= Wt. of solution in gram/Density in gram / cc

Again by definition

Molarity = Moles/Volume of solution in litre

= 0.0854/1.008

= 0.085 M

∴ Normality = 0.085 N (for NaCl, eq. wt. = mol. wt)

Further, Mole of H_{2}O = 1000/18 = 55.55

(1000 gram of water = 1000 ml of water, because density = 1 g/cc)

Total mole = Mole of NaCl + Mole of H_{2}O

= .0854 + 55.55 = 55.6354

Mole fraction of NaCl = 7.85/98.66 = 0.0854/55.6409 = 1.535 x 10^{–3}**Problem 14. ****If 20 ml of ethanol (density = 0.7893 gm / ml) is mixed with 40 ml of water (density = 0.9971 gm/ml) at 25°C, the final solution has density of 0.9571 gm / ml. Calculate the percentage change in total volume of mixing. Also calculate the molality of alcohol in the final solution.****Solution.**

Mass of ethanol = v x d

W = 20 × 0.7893 gm = 15.786 gm

Mass of water = v × d

= 40 x 0.9971 gm = 39.884 gm

Total volume = 60 ml

Total mass = 15.786 + 39.884 = 55.67 gm

Let the volume of solution = x ml

X = mass/density = 55.67/0.9571 = 58.165 ml

Change in volume = 60 – 58.165 = 1.835 ml

% change in volume = 1.835/60 x 100

= 3.05%

Molality, m = w/m x 1000/W (in gm) = 15.786/46 x 1000/39.884**Problem 15. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.****Solution.**

Total mass of the solution = 100 g

Mass of benzene = 30 g.

∴ Mass of carbon tetrachloride = (100 - 30)g = 70 g

Molar mass of benzene (C_{6}H_{6}) = (6 × 12 + 6 × 1) g mol^{– 1} = 78 g mol^{-1}

∴ Number of moles of C_{6}H_{6} =30/78 mol = 0.3846 mol

Molar mass of carbon tetrachloride (CCl_{4}) = 1 × 12 + 4 × 355 = 154 g mol^{-1}

∴ Number of moles of CCl_{4} = 70/154 mol = 0.4545 mol

Mole fraction of C_{6}H_{6} =**Problem 16. Determine the mole fraction of methanol CH _{3}OH and water in a solution prepared by dissolving 4.5 g of alcohol in 40 g of H_{2}O. Molar mass of H_{2}O is 18 and Molar mass of CH_{3}OH is 32.**

Moles of CH

mole Moles of H

Therefore, according to the equation

mole fraction of CH

= 0.061

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