Mole Concept | Concentration Terms#
- Concentration terms are used to signify the amount of solute dissolved in a solvent.
- Concentration does not change on taking out a part of the solution. For example, if the concentration of 10 L of solution is 'c', then the concentration of same 100 mL solution will also be 'c'.
- Solute + Solvent = Solution. In a solution, solute is in lesser quantity and solvent is in higher quantity.
1. Molarity#
- Molarity is defined as the number of moles of solute dissolved per litre of the solution.
- It is denoted by 'M' and its unit is moles/L.
- If the molarity of a solution is 3 moles/L, then we say that the solution is 3 molar (3 M).
\[Molarity = {Moles\ of\ solute \over Volume\ of\ solution\ in\ L}\]
- Since volume changes with the change in temperature, therefore, molarity is temperature dependent.
2. Grams Per Litre#
- It is defined as the grams of solute dissolved per litre of the solution.
- It is denoted by 'g/L' and its unit is g/L.
\[g/L = {Mass\ of\ solute\ in\ grams \over Volume\ of\ solution\ in\ L}\]
- Since volume changes with the change in temperature, therefore, g/L is temperature dependent.
3. Molality#
- Molality is defined as the number of moles of solute dissolved per kilogram of the solvent.
- It is denoted by 'm' and its unit is moles/Kg.
- If the molality of a solution is 3 moles/Kg, then we say that the solution is 3 molal (3 m).
\[Molality = {Moles\ of\ solute \over Mass\ of\ solvent\ in\ Kg}\]
- Since moles and mass both are independent of temperature, therefore, molality is temperature independent.
4.Mole Fraction#
- Let us consider a solution/mixture consisting of two components: A and B. Mole fractions of A and B are defined as:
\[Χ_A = {n_A \over n_A + n_B}\]
\[Χ_B = {n_B \over n_A + n_B}\]
Here, nA = moles of A, nB = moles of B
- It is worth noting that:
\[Χ_A + Χ_B = 1\]
- Similarly, let us consider a mixture consisting of 'n' components: 1, 2, 3, ...i, ... n. Mole fraction of ith component can be defined as:
\[Χ_i = {n_i \over n_1 + n_2 + n_3 + .... + n_n}\]
\[Χ_1 + Χ_2 + Χ_3 + ... +\ Χ_i\ + ... + Χ_n = 1\]
- Mole fraction is a unitless quantity.
- Mole fraction is temperature independent because mole does not depend on temperature.
5. Parts per million and Parts per billion#
- Parts per million (ppm) is defined as:
\[ppm = {Mass\ of\ solute \times 10^6 \over Mass\ of\ solution}\]
- Parts per billion (ppb) is defined as:
\[ppb = {Mass\ of\ solute \times 10^9 \over Mass\ of\ solution}\]
- Both ppm and ppb are unitless quantities.
- Both ppm and ppb are temperature independent because mass does not depend on temperature.
Questions#
A 1M H2SO4 solution is taken whose density is 1.1 g/mL. Calculate its molality.
\[Molarity\ of\ H_2SO_4\ solution = 1\ M\]
\[Let\ the\ volume\ of\ solution\ be\ 1\ L.\]
\[Then,\ the\ moles\ of\ solute(H_2SO_4) = 1\ mole\]
\[∴ Mass\ of\ solute(H_2SO_4) = 1\times 98 = 98 grams\]
\[Density\ of\ solution = 1.1\ g/mL.\]
\[Mass\ of\ solution = 1.1 \times 1000 = 1100\ g\]
\[Mass\ of\ solvent = Mass\ of\ solution - Mass\ of\ solute\]
\[Mass\ of\ solvent = 1100 - 98 = 1002\ g = 1.002\ Kg\]
\[Molality = {Moles\ of\ solute \over Mass\ of\ solvent\ in\ Kg}\]
\[Molality = {1 \over 1.002} = {500 \over 501}\]
Two solutions of 100 mL and 200 mL are taken. 100 mL solution containing NaCl has the molarity of 0.2 M while 200 mL solution containing CaCl2 has molarity of 0.5 M. The two solutions are mixed in a beaker and an additional 100 mL water is added to it. Calculate molarity of Cl- ions.
- For solution 1:
\[Moles\ of\ NaCl = M \times V\]
\[Moles\ of\ NaCl = 0.2 \times 100 \times 10^{-3} = 0.02\]
\[Moles\ of\ Cl^- = 0.02\]
- For solution 2:
\[Moles\ of\ CaCl_2 = M \times V\]
\[Moles\ of\ CaCl_2 = 0.5 \times 200 \times 10^{-3} = 0.1\]
\[Moles\ of\ Cl^- = 2 \times 0.1 = 0.2\]
- Total moles of Cl- ions:
\[Total\ moles\ of\ Cl^- = 0.02 + 0.2 = 0.22\]
- Total volume of solution:
\[Total\ volume\ of\ solution = 100 + 200 + 100 = 400\ mL = 0.4\ L\]
- Molarity of Cl- ions:
\[Molarity\ of\ Cl^- = {0.22 \over 0.4} = 0.54\ M\]
5 M HCl solution is given from which x mL is taken out in a container. Water is added to it such that volume of solution becomes 5 L and molarity becomes 1.5 mL. Find x.
\[Initial\ moles\ of\ HCl\ taken\ out = Final\ moles\ of HCl\ when\ water\ is\ added\]
\[5 \times x \times 10^{-3} = 1.5 \times 5\]
\[x = 1500 mL\]