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Methods of Expressing the concentration of solution
The concentration of a solution refers to the amount of solute present in the given quantity of solution or solvent. The concentration of a solution may be expressed quantitatively in any of the following ways.
The Strength of a solution is defined as the amount of the solute in grams, present in one litre of the solution. It is expressed in g L-1
Strength = Mass of solute in grams / Volume of solution in litres
If X gram of solute is present in V cm3 of a given solution then
Strength = ( X x 1000) / V
Molarity of a solution is defined as the number of gram-moles of solute dissolved in 1 litre of a solution
Molarity = No. of moles of solute / Volume of Solution in litres
If `X' grams of the solute is present in V cm3 of a given solution, then,
Molarity = (X / Mol. mass) / x( 1000 / V )
Molarity is represented by the symbol M. Molarity can also be calculated from the strength as follows
Molarity = Strength in grams per litre / Molecular mass of the solute
A 0.1M solution of Sugar, C12H22O11 (mol.mass = 342), means that g of sugar is present in one litre (1000 cm3) of the solution.
Normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre of the given solution.
Normality = Number of gram-equivalents of solute / Volume of Solution in litre
If X grams of the solute is present in V cm3 of a given solution, then,
Normality = ( X / Eq.mass ) x ( 1000 mL / V )
Normality is represented by the symbol N. Normality can also be calculated from strength as follows
Normality = Strength in grams per litre / Eq.mass of the solute
A 0.1N (or decinormal) solution of H2SO4 (Eq.mass = 49), means that 4.9 g of H2SO4 is present in one litre (1000 cm3) of the solution.
4. Molality (m)
Molality of a solution is defined as the number of gram-moles of solute dissolved in 1000 grams (or 1 kg) of a Solvent. Mathematically,
Molality = Number of moles of solute / Mass of solvent in kilograms
"If X grams of the solute is dissolved in b grams of the solvent", then
Molality = ( X / Mol. mass ) x ( 1000g/bg )
Molality is represented by the symbol 'm'.
A 0.1m Solution of glucose C6H12O6 (Mol.mass = 180), means that 18g of glucose is present in 1000g (or one kilogram) of water.
5. Mole Fraction
Mole fraction is the ratio of number of moles of one component (Solute or Solvent) to the total number of moles of all the components (Solute and Solvent) present in the Solution. It is denoted by X. Let us suppose that a solution contains 2 components A&B and suppose that nA moles of A and nB moles of B are present in the solution. Then,
Mole fraction of A, XA= nA / (nA + nB) .....(1)
Mole fraction of B, XB = nB / (nA + nB) ......(2)
Adding 1 & 2 we get,
XA + XB = ( nA / nA+nB ) + ( nB / nA+nB ) = nA+nB / nA+Nb
Thus, sum of the two mole fractions is one. Therefore, if mole fraction of one component in a binary solution is known, that of the other can be calculated.
Problems for practice
1. Calculate the volume of 14.3m NH3, solution needed to prepare 1L of 0.1M solution.
2. How would you make up 425 mL of 0.150M HNO3 from 68.0% HNO3? The density of 68.0% HNO3 is1.41g/mL.
Ans: 4.25 mL
3. Calculate the molarity of a solution obtained by mixing 100 mL of M H2SO4 and 200 mL of 1.5M H2SO4
4. 4. Calculate the molality of a solution by dissolving 0.850g of ammonia (NH3) in 100g of water.
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