Expressing Concentration of Solutions | Chemistry Class 12 - NEET PDF Download

What is Expression of Concentration of Solution?

We always discuss a solution being diluted or concentrated; this is a qualitative way of expressing the concentration of the solution. A dilute solution means the quantity of solute is relatively very small, and a concentrated solution implies that the solution has a large amount of solute. But these are relative terms and do not give us the quantitative concentration of the solution.

So, to quantitatively describe the concentrations of various solutions around us, we commonly express levels in the following way, let’s study each method and determine the formulas for this method:

1. Percent by weight
2. Percent by volume
3. Percent by weight/volume
4. Molarity
5. Molality
6. Normality
7. Mole Fraction
8. Parts per Million
9. Formality
10. Strength (g/L)
11. Molarity of mixing
    
    

(i) 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:
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.

Example: If 11 g of oxalic acid are dissolved in 500 mL of solution (density = 1.1 g mL^-¹), what is the mass % of oxalic acid in solution?
Solution:11 g of oxalic acid are present in 500 mL of solution.

Density of solution = 1.1 g mL^-¹

Mass of solution = (500 mL) x (1.1. g mL^-¹) = 550 g

Mass of oxalic acid = 11 g

Mass% of Oxalic Acid = 11÷550 ✖ 100 % = 2%

(ii) 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.

Example: Calculate the percent volume of solute if 5 ml of it is dissolved in a solvent of 90 ml.

Solution:Volume of Solute (v) = 5 ml

Volume of Solvent = 90 ml

Volume of Solution (V) = (5 + 90) ml = 95 ml

Using the formula,

P = (v / V) × 100

P = (5/95) × 100

P = 5.26 %

(iii) 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: Calculate the concentration in terms of mass by volume percentage of the solution containing, 3g of potassium chloride in 75ml of potassium chloride (KCl) solution.

Solution:Mass of solute =3 g
Volume of solution =75 ml
Mass by volume % = =3/75×100=4 %

(iv) 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.
• 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 the most common way of representing the concentration of the solution.
• Molarity is dependent 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 

^{(V) 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 the molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.
Solution: Molar mass of C₂H₄O₂ : 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol^–1
Moles of C₂H₄O₂ = 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 C₂H₄O₂

(vi) 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 L
Equivalent 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

Example: Calculate the normality of a 310 mL NaOH solution made by dissolving 0.4 gram of NaOH.

Solution: The formula to calculate the normality is,

Normality (N) = Number of gram equivalents / Volume of the solution in liters

Number of gram equivalents = weight of solute / Equivalent weight of solute

Equivalent weight of solute = 23+16+1 = 40

Since,

Normality (N) = weight of solute / Volume of the solution in liters × Equivalent weight of solute

N = (4/40) × (1000/310)

= 0.1 × 3.2258

= 0.3225 N

(vii) Mole Fraction

Mole fraction can be defined as the ratio of a 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 a solution where a number of components = i

• In a given solution sum of the mole fractions of all the components is 

  unity.
  Mathematically, x₁ + x₂ + x₃ …. + x_i =1

Example :  Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ 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). The solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C₂H₆O₂ = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol^–1

Mole fraction of water can also be calculated as: 1 – 0.068 = 0.932.

(viii) 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 SO₂ in the air means 10 mL of SO₂ is present in 106 mL of air.

(ix) 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

(x) Strength (g/L) 

It is defined as the 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

(xi) Molarity of Mixing

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

where, MR is the Resultant molarity

V₁ + V₂ + V₃ = Resultant volume after mixing

Note: Molarity is dependent on volume; therefore, it depends on temperature.

Dilution Formula

The dilution formula is giving by the following expression:

Practice Questions

Problem 4. What is the mole fraction of carbon tetrachloride (CCl₄) in solution if 3.47 moles of CCl₄ is dissolved in 8.54 moles of benzene (C₆H₆)?
Solution.
Mole Fraction CCI₄ (X_CCI4) = moles of CCI₄/total moles
X_CCI4 = 3.47 moles CCI₄/3.47 moles CCI₄ + 8.54 moles C₆H₆ 
XCCI4 = 0.289

Problem 5. What is the mole fraction of formaldehyde (CH₂O) in solution if 25.7 grams of CH₂O is dissolved in 3.25 moles of carbon tetrachloride (CCl₄)?
Solution. 
The grams of acetone will need to be converted into moles in order to solve for mole fraction.

Mole Fraction CH₂O(X_CH2O) = moles of CH₂O/total moles
X_{CH2O = 0.856 moles CH2O/0.856 moles CH2O + 3.25 moles CCI4}
X_{CH2O = 0.208}

Problem 6. 0.100 mole of NaCl is dissolved into 100.0 grams of pure H₂O. What is the mole fraction of NaCl?
Solution. 
100.0 g / 18.0 g mol^-1 = 5.56 mol of H₂O
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 H₂O:
5.56 mol / 5.66 mol = 0.982

Problem 7. A solution is prepared by mixing 25.0 g of water, H₂O, and 25.0 g of ethanol, C₂H₅OH. Determine the mole fractions of each substance.
Solution. 
(1) Determine the moles of each substance:
H₂O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol
C₂H₅OH ⇒ 25.0 g / 46.07 g/mol = 0.543 mol
(2) Determine mole fractions:
H₂O ⇒ 1.34 mol / (1.34 mol + 0.543 mol) = 0.71
C₂H₅OH ⇒ 0.543 mol / (1.34 mol + 0.543 mol) = 0.29

Problem 8. A solution contains 10.0 g pentane, 10.0 g hexane and 10.0 g benzene. What is the mole fraction of hexane?
Solution. 
Try it yourself Using hints given below !
(1) You need to determine the moles of pentane, hexane and benzene: to do this, you need the molecular weights. Here are the formulas:
pentane: C₅H₁₂
hexane: C₆H₁₄
benzene: C₆H₆
(2) When you have the moles of each, add them together.
(3) Then, divide the moles of hexane by the total.

Problem 9. The molality of an aqueous solution of sugar (C₁₂H₂₂O₁₁) is 1.62m. Calculate the mole fractions of sugar and water.
Solution. 
(1) Molality is moles solute / kg of solvent. Therefore we know our solution is:
1.62 mol C₁₂H₂₂O₁₁
1.00 kg = 1000 g of water
(2) Calculate the moles of water present:
1000 g / 18.0152 g/mol = 55.50868 mol
(3) Determine the mole fraction of the sugar:
1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three sf)
(4) You can calculate the mole fraction of the water by subtraction.

Problem 10. How many grams of water must be used to dissolve 100.0 grams of sucrose (C₁₂H₂₂O₁₁) to prepare a 0.020 mole fraction of sucrose in the solution?
Solution. 
(1) Determine moles of sucrose:
100.0 g / 342.2948 g/mol = 0.292145835 mol
(2) Determine moles of water required to make the solution 0.020 mole fraction of sucrose:
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 H₂O
Comment: you can also do this:
0.292 is to 0.02 as x is to 0.98
(3) Determine grams of water:
14.308 mol x 18.015 g/mol = 258.0 g

Problem 11. Surprisingly, water (in the form of ice) is slightly soluble in liquid nitrogen. At -196 °C, (the boiling point of liquid nitrogen) the mole fraction of water in a saturated solution is 1.00 x 10^-5. Compute the mass of water that can dissolve in 1.00 kg of boiling liquid nitrogen.
Solution.
(1) Use the definition of mole fraction to set up the following:
χwater = moles of water / (moles of water + moles of nitrogen)
1.00 x 10^-5 = x / (x + 71.3944041)
I'm going to carry some guard digits until the end of the calculation.
(2) Some algebra: (1.00 x 10^-5) (x) + 7.139440411 x 10^-4 = x
0.99999x = 7.139440411 x 10^-4
x = 7.139511806 x 10^-4 mol of H₂O
(3) Calculate grams of water from moles of water:
7.139511806 x 10^-4 mol x 18.0152 g/mol = 1.2862 x 10^-2 g
1.29 x 10^-2 g (to three sf)

Problem 12. What is the mole fraction of cinnamic acid in a mixture that is 50.0% weight urea in cinnamic acid (urea = 60.06 g/mol; cinnamic acid = 148.16 g/mol)
Solution.
(1) Let us assume 100.0 g of this mixture are present. Therefore:
50.0 g is urea
50.0 g is cinnamic
(2) Convert grams to moles: urea:
50.0 g / 60.06 g/mol = 0.8325 mol
cinnamic acid: 50.0 g / 148.16 g/mol = 0.3375 mol
(3) Determine mole fraction of cinnamic acid:
0.3375 mol / 1.1700 mol = 0.2885

Problem 13. 5 g of NaCl is dissolved in 1000 g of water. If the density of the resulting solution is 0.997 g per cc, calculate molality, molarity, normality and mole fraction of the solute.
Solution.
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 H₂O = 1000/18 = 55.55
(1000 gram of water = 1000 ml of water, because density = 1 g/cc)
Total mole = Mole of NaCl + Mole of H₂O
= .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 (C₆H₆) = (6 × 12 + 6 × 1) g mol^{– 1} = 78 g mol^-1
∴ Number of moles of C₆H₆ =30/78 mol = 0.3846 mol
Molar mass of carbon tetrachloride (CCl₄) = 1 × 12 + 4 × 355 = 154 g mol^-1
∴ Number of moles of CCl₄ = 70/154 mol = 0.4545 mol
Mole fraction of C₆H₆ =