Crystalization

7.4 CRYSTALLIZATION

Crystals are homogeneous solids, having regular and definite geometrical shape with faces and sharp edges. Pure crystals of compounds have sharp melting points.

When a super saturated solution of a solid is prepared at high temperature and allowed to cool down, then at lower temperature it cannot hold more solute in dissolved state. Some of these dissolved solute particles come out of solution, in solid form having regular and definite geometric shapes. They are called crystals. The process in which dissolved solute comes out of solution and forms crystals is called crystallization.

Following are two practical examples of super saturated solution preparation and crystallization.

Preparation of Crystals of Copper Sulphate (Blue Vitriol CuSO4.5H2O)

Prepare a saturated solution of copper sulphate, in water at room temperature using a beaker. Heat the saturated solution and try to dissolve some more quantity of copper sulphate while stirring the solution with glass rod. Allow the super saturated solution of copper sulphate to cool down at room temperature. Upon cooling and standing, crystals of CuSO4.5H2O, i.e., blue vitriol will form. Filter out the crystals and observe the shape of crystals under a light microscope.

Preparation of Crystals of Potassium Nitrate (KNO3)

Take 100 ml water in a beaker. Prepare a saturated solution of KN03 by dissolving 37g of solute at room temperature by means of stirring with glass rod. Now heat this saturated solution to 50C and dissolve 20g of additional KNO3 while stirring the solution. Filter the hot super saturated solution quickly and collect the filterate in another beaker. Cool the filterate to room temperature. Upon cooling crystals of KNO3 are formed.

Filter out the crystals and observe their shape under light microscope.

Purification of Solids by Crystallization:

An impure solid generally contains two types of impurities. An insoluble impurity and a soluble impurity. Insoluble impurity is totally insoluble in the solvent used for crystallization even at boiling temperature. While the soluble impurity remains in soluble form at room temperature. Therefore, a compound containing such two types of impurities could be easily removed by means of crystallization technique.

For example, a 42g impure sample of KNO3 contains a small quantity of sand and NaCl. To obtain pure crystalline KNO3, we perform the crystallization technique as follows:

Take 50ml of water in a beaker and add the impure sample (40g) of KNO3 to it while stirring with glass rod. Supply heat gently till the temperature of the solution is above 50C. Stirr the solution at this temperature till most of the solid is dissolved. Filter the hot solution and collect the filtrate in a beaker. Sand being in soluble in water will be removed and collected on the filter paper. Upon cooling of the filtrate, crystals of KNO3 will start appearing. When no further crystals are formed, filter it again and collect the filtrate in a beaker. Purified crystals of KNO3 are obtained on the filter paper. The filtrate will contain some quantity of the dissolved KNO3 along with the NaCI, being a soluble impurity.

7.5 STRENGTHS OF A SOLUTION

The strength (concentration) of a solution means the mass or volume of the solute present in known amount of solvent or solution.

The following are the common methods of expressing the strength (concentration) of a solution.

  1. Molarity (M)
  2. Molality (m)  
  3. Mole fraction (x)
  4. Percentage (%)
  5. Normality (N).

It is not used in these days.

1. Molality (M):

It is defined as the number of moles of solute dissolved in 1 liter (1dm3) of a solution; it is denoted by (M).

Thus, 1 mole of NaOH (i.e. its gram formula mass) 40g dissolved in 1 liter (1dm3) of solution is said to be 1 molar (1M) solution. If only half of the mole i.e. 20g of NaOH is dissolved in one liter (1dm3) of solution, the solution is said to be one-half molar (i.e. 0.5M)

The molarity of any solution is found by dividing the number of moles of solute by the number of liters of solution.

Molarity (M) = Moles of Solute / Liters of Solution or dm3 of solution

We know that

 

(i)   Number of moles = Given mass of solute / Gram formula mass of solute

(ii)  Liters of solution= mls of solution / 1000

Thus

     Molarity = (M) moles of solute / Gram formula mass of solute x 100cm3 (ml) / ml(cm3) of solution

For example:

Calculate the molarity of a solution containing 4g of NaOH in 500 cm3 (ml) of solution?

Solution        

          M= Mass of solute x 1000 cm3 / Gram formula mass x cm3 of solution

Data

(i)   Molarity (M)                                =                 ?

(ii)  Mass of solute NaOH                              =                 4g

(iii) Gram formula mass of solute         =                 40g

(iv) cm3 of solution                            =                 500cm3

 

M=4g.1000cm3 / 40g x 500cm3 = 1/5 =0.2

2. Molality (m):

It is defined as the number of moles of solute dissolved per 1000g (1kg) of solvent; it is denoted by (m).

Thus mole of Na2CO3g (i.e. its gram formula mass) 106g dissolved in 1000g of solvent is said to be 1 molal (lm) solution. If only half of the mole i.e. 53g of Na2C3 is dissolved in 1000g of solvent, the solution is said to be one-half molal (i.e. 0.5m)

The molality of a solution is found by the following formula.

Mass of solute * 1000c        No. of moles of solute

Molality (m) =Mass of solute x 1000g / Gram formula mass x grams of solvent or

No. of moles of solute / Kg of solvent

For example:

Calculate the molality of solution containing 5.3g of Na2CO3 in 500g of water.

Solution:

Molality (m) = Mass of solute x1000g / Gram formula mass of solute x grams of solvent

Data:

1.    Molality                                                =       (m) =?

2.    Mass of solute i.e. Na2CO3                       =       5.3g

3.    Gram formula mass of solute Na2CO3        =       106g

4.    Grams of solvent (water)                        =       500g

 

Molarity =m= Mass of solute x 1000g / Gram formula mass of solute x gram of solvent

M = 5.3g x 1000g / 106 x 500g = 1/10 = 0.1m

3. Mole fraction (X):

Mole fraction (X) of any component in a solution is the number of moles of the component divided by total number of moles making up a solution. It is denoted by X.

Mole fraction (X) =        Moles of component

                           Total number of moles making up solution

For example, a solution is prepared by dissolving 1 mole of ethyl alcohol C2H5-OH in 3 moles of water (H2O), where NA and NB the represent the number of moles of ethyl alcohol and water respectively.

 

 Mole fraction of ethyl alcohol = XA = NA/NA+NB =1/1+3 = 1/4 = 0.25

 Mole fraction of water = XB = NB/NA+NB = 3/1+3 = 3/4 = 0.75

 Results: Mole fraction of ethyl alcohol = XA = 0.25

               Mole fraction of water = XB = 0.75

Note, the sum of the mole of fractions is equal to 1.

Mole fraction of ethyl alcohol = 0.25

Mole fraction of water = 0.75

Sum of the mole of fractions = 1.00

Remember

The mole fraction is dimensionless quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present. The sum of mole fractions of all components of a solution must equal 1

4. Percentage (%):

This is based on mass (m) or volume (v) of components of solution. It is of four types.

  1. Percentage in M/M    (mass/mass)
  2. Percentage in M/V    (mass/volume)
  3. Percentage in V/M    (volume/mass)
  4. Percentage in V/V    (volume/volume)

Examples:

  1. 5% (M/M) solution means solute 5g in 95gm solvent.
  2. 10% (M/V) solution means solute 10g in solution 100cm3.
  3. 5% (V/M) solution means solute 5cm3 in solution 100 g.
  4. 15% (V/V) solution means solute 15 cm3 in 85 cm3 solvent.

Problem of %:

6.5g of NaCI are dissolved in 43.5 g water. Calculate the percent by mass of NaCI in this solution.

Solution

% by mass of solute NaCl = Mass of solutex100/Mass of solute+Mass of solvent

= 6.5x100/6.5+43.5 = 6.5x100/50.0

Percent by Mass of NaCl =13.0%

 

7.6 SUSPENSION

If sugar or salt is dissolved in water, the crystals dissolve into molecules or ions. The resulting homogenous mixture is called true solution. If fine sand is stirred in water, the crystals do not dissolve, but even after several days some of the smallest particles remain suspended, such mixture is called a suspension.

Suspension is defined as a heterogeneous mixture consists of visible particles, each of which contains many thousands or even millions of molecules, surrounded by molecules of liquid.

Important to note

Homogenous mixtures can be classified according to size of the constituent particles, as solution, suspension or colloids. In solution the dispersed particles are of molecular size (0.1-lnm). In suspension, the dispersed particles are much larger than molecules (>1000nm).

Between these two extreme type of dispersion, there is another type of homogenous mixture in which dispersed particles are larger than molecules (2.0-1000 nm) but not large enough to settle out. This intermediate type of mixture is called colloid.

 

Difference between Solution and Suspension

6. Components cannot be separated by filtration. Solution

1. The size of particles is between 0.1 to lnm.

2. Particles cannot be seen with low power microscope.

3. It is homogeneous.

4 Particles do not settle down.

5. It is transparent.

6. Components cannot be separated by filtration.

 

Suspension

1. The size of particles is larger than 1000 nm.

2. Particles can be seen by low power microscope.

3. It is heterogeneous.

4. Particles settle down.

5. It is not transparent.

6. Components can be separated by filtration.

Examples of Suspensions in Daily Life:

1) Smoke: A suspension of the particles of carbon in a gas or air.

2) Mud (slime): A suspension of fine particles of solid in small quantity of liquid.

3) Foam (froth): A suspension of fine particles of a gas in a liquid.

4) Emulsion: A suspension of droplets of one liquid into another in which it is not soluble.

 

 
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