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**CONCENTRATION FORMULAE**

There are several methods to express the concentration of solutions. We are going to describe some very important terms used to represent the concentration of solutions.

(a) Percent by weight

(b) Percent by volume

(c) Percent by weight/volume

(d) Mole fraction

(e) Molarity

(f) Molality

(g) Normality

(h) ppm

**(a) Percent by Weight (w / w) : **It is the weight of the solute as a per cent of the total weight of the solution. That is,

% By weight of solute =

For example, if a solution of HCI contains 36 % HCI by weight, it has 36 g of HCI in 100 g of solution.

**(b) Percent by Volume (v / v) :** It is the volume of solute present per 100ml of solution

% By volume =

For example, it we have 35% C_{2}H_{5}OH solution by volume means 35 ml C_{2}H_{5}OH is present per 100 ml of the solution.

**(c) Percent by weight / volume (w / v) : **It is the mass of solute present per 100g of solution, or % =

i.e.30% HCl solution, means 30 g of HCl is present per 100 ml of solution.

**(d) Mole Fraction (X) :** A simple solution is made of two substances; one is the solute and the other is solvent. Mole fraction, X, of solute is defined as the ratio of the number of moles of solute and the total number of moles of solute and solvent. Thus,

If n represents moles of solute and N number of moles of solvent,

Then

X_{solute} = n / n + N

Notice that mole fraction of solvent would be

X_{solvent} = n / n + N

Mole fraction is unitless and (X_{solute} + X_{solvent}) = 1

**(e) Molarity (M) :** In current practice, concentration is most often expressed as molarity. Molarity is defined as the number of moles of solute per litre of solution.

Molarity

w = Mass of solute in grams

M` = molecular weight of solute in gm/mol.

V = volume of solution in ml.

0.01 M NaOH, solution means that 0.01 mole NaOH is present in 1000 ml of its solution.

**(f) Molality (m) :** Molality of a solution is defined as the number of moles of solute per kilogram of solvent:

Molality

or

where

w = mass of solute in grams

M` = molecular wt of solute in gm / mol

W = mass of solvent in grams.

**(g) Normality (N) :** Normality of a solution is defined as number of equivalents of solute per litre of the solution:

Normality

w = mass of solute in gram

V = volume of solution in ml

E = equivalent wt of solute

**(h) ppm (parts per million) : **It is the mass of solute in grams present per 106 grams of solution.

and also

⮚ Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that molality of a solution does not change with temperature since mass remains unaffected with temperature**Example.1. A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.****Solution.** Mass per cent

= 2 / -2 g of A +18 x 100

= 2g / 20g × 100

= 20g

= 10 %**Example.2. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.****Solution. ****Example.3. The density of 3 M solution of NaCl is 1.25 g mL ^{–1}. Calculate molality of the solution.**

M = 3 mol L

Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g

Mass of 1L solution = Mass of solution × density of solution

= 1000 × 1.25 = 1250 g

(since density = 1.25 g mL^{–1})

Mass of water in solution = 1250 –175.5 = 1074.5 g

Molality = mass of solute in number of moles / mass of solvent in kg

= 3 / 1.07468 kg

= 2.7915 Molal.**Example.4. A solution of NaCl 0.5% by wt. If the density of the solution is 0.997 g/ml, calculate (a) The molality (b) Molarity (c) Normality (d) Mole fraction of the solute.**

Number of moles of NaCI = mass of NaCl / Molecular mass of Nacl

= 0.5 / 58.5 = 0.00854

Molality

Volume of the solution = Mass of solution in grams / Density in gm per m

= 100 / 0.997 = 100.3

= 0.0852

**(d) To calculate mole fraction of the solute**

No. of moles of water in 99.5g = 99.5 / 18 = 5.5277

Moles of NaCl = 0.5 / 58.5 = 8.547 x 10^{-3}**Example.5. A sample of tooth-paste contains 10 ^{–4}% F^{–} ions by wt. Find its ppm concentrations.**

Solution.

(∵ Mol.mass of NaOH = 40)

= 0.0125 mole

Thus 500 cm

∴ 1000 cm

= ⇒ 0.025 mole

Hence molarity of the solution = 0.025 M

Volume of the solution = 250 cm

Calculation of molarity

Mol. mass of H

∴ No. of moles of H

= 9.8 / 98 = 0.1

250 cm

= 0.1 mole

∴ 1000 cm

Hence molarity of solution = 0.4 M

Mass of H_{2}SO_{4} dissolved = 15 g

Mass of the solution = 100 g

Density of the solution = 1.02 g/cm3 (given)

Calculation of molality:

Mass of solution = 100 g

Mass of H_{2}SO_{4} = 15g

Mass of water (solvent) = 100 – 15 = 85 g

Mol. mass of H_{2}SO_{4 }= 98

∴ 15 g H_{2}SO_{4} = 15 / 98 = 0.153 moles

Thus 85 g of the solvent contain 0.153 moles 100 g of the solvent contain

Hence the molality of H_{2}SO_{4} solution = 1.8 m

Calculation of molarity:

15 g of H_{2}SO_{4 }= 0.153 moles

Vol. of solution = Wt. of solution / Density of solution = 100 / 1.02 = 98.04 cm^{3}

**Thus 98.04 cm ^{3} of solution contain H_{2}SO_{4} = 0.153 moles.**

**Q.1. The density of 3M solution of Na _{2}S_{2}O_{3} is 1.25g ml^{–1}. Calculate **

(a) The % by weights of Na_{2 }S_{2 }O_{3}

(b) Mole fraction of Na_{2 }S_{2 }O_{3}

(c) The molalities of Na^{+} and ions.

Ans.

(a) Mass of 1000 mL of Na

Mass of Na

= 3 × Mol mass of Na

= 3 × 158 = 474g

Mass percentage of Na

(b) (ii) No. of moles of Na

Mass of water =(1250−474)=776g

No. of moles of water = 776 / 18 = 43.1

Mole fraction of Na

(c) No. of moles of Na^{+} ions

= 2 × No.of moles of Na_{2} S_{2} O_{3} = 2 × 3 = 6

Molality of Na^{+} ions = No. of moles of Na^{+} ions / Mass of water in kg

= 7.73m

No. of moles of S_{2} O_{3} ^{2−} ions = No. of mole of Na_{2} S_{2} O_{3} = 3

Molality of S_{2} O_{3} ^{2−} -ions =

**Q.2. A sample of H _{2}SO_{4} (density 1.787g ml^{–1}) is labelled as 86% by weight. What is molarity of acid ? What volume of acid has to be used to make 1 litre of 0.2M H_{2}SO_{4}?**

Ans.

Let V

H

15.68 x V

From the given data, we find AgNO

∴ 170.0 g of AgN0

∴ 5.77 g 0f AgN0

Moles of NaCl = 3.78 / 58.5 = 0.0646

1 NaCl contains one Cl

∴ moles of Cl

∴ (Cl

∴ concentration of Cl^{⁻} = [Cl^{⁻}] = 0.646

Again, BaCl_{₂} ⇄ Ba^{⁺} + 2Cl^{⁻}

∴ [BaCl] = [Ba^{⁺}] = [Cl^{⁻}] / 2

From above concentration of Cl^{⁻} = 0.646 M

∴ [BaCl_{₂}] = [Cl^{⁻}] / 2 = 0.646 / 2 = 0.323 M

Now, molarity of BaCl_{₂} = weight of BaCl_{₂} / molar weight of BaCl_{₂} × volume in L 0.323 = weight of BaCl_{₂} / (208.3 × 0.25)

[ ∵ molar weight of BaCl_{2} = 208.3g / mol and volume of solution = 250ml = 0.25L

∴0.323 × 208.3 × 0.25 = weight of BaCl_{2}

weight of BaCl_{2} = 16.8g

Hence, weight of BaCl_{₂} = 16.8g.

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