Solutions Chapter 2 Class 12 Chemistry NCERT Textbook PDF

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NCERT Class 12 Chemistry Textbook Chapter 2 With Answer PDF Free Download

Solutions

Chapter 2: Solutions

In normal life, we rarely come across pure substances. Most of these are mixtures containing two or more pure substances.

Their utility or importance in life depends on their composition. For example, the properties of brass (mixture of copper and zinc) are quite different from those of German silver (mixture of copper, zinc, and nickel) or bronze (mixture of copper and tin);

1 part per million (ppm) of fluoride ions in water prevents tooth decay, while 1.5 ppm causes the tooth to become mottled and high concentrations of fluoride ions can be poisonous (for example, sodium fluoride is used in rat poison);

intravenous injections are always dissolved in water containing salts at particular ionic concentrations that match with blood plasma concentrations and so on.

In this Unit, we will consider mostly liquid solutions and their formation. This will be followed by
studying the properties of the solutions, like vapor pressure and colligative properties.

We will begin with types of solutions and then various alternatives in which concentrations of a solute can be expressed in liquid solution.

Solutions are homogeneous mixtures of two or more two components. By homogenous mixture, we mean that its composition and properties are uniform throughout the mixture.

Generally, the component that is present in the largest quantity is known as a solvent.

Solvent determines the physical state in which the solution exists. One or more components present in the solution other than the solvent are called solutes.

In this Unit, we shall consider only binary solutions consisting of two components). Here each component may be solid, liquid, or in a gaseous state and is summarised.

The solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature.

It depends upon the nature of solute and solvent as well as temperature and pressure. Let us consider the effect of these factors in the solution of a solid or a gas in a liquid.

Every solid does not dissolve in a given liquid. While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not.

On the other hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not. It is observed that polar solutes dissolve in polar solvents and nonpolar solutes in nonpolar solvents.

In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two, or we may say like dissolves like.

When a solid solute is added to the solvent, some solute dissolves, and its concentration increases in the solution.

This process is known as dissolution. Some solute particles in the solution collide with the solid solute particles and get separated out from the solution.

This process is known as crystallization. A stage is reached when the two processes occur at the
same rate.

Under such conditions, the number of solute particles going into the solution will be equal to the solute particles separating out, and a state of dynamic equilibrium is reached.

Solubility of a Solid in a Liquid

Every solid does not dissolve in a given liquid. While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not.

On the other hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not.

It is observed that polar solutes dissolve in polar solvents and nonpolar solutes in nonpolar solvents. In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.

When a solid solute is added to the solvent, some solute dissolves, and its concentration increases in the solution. This process is known as dissolution.

Some solute particles in the solution collide with the solid solute particles and get separated out from the solution. This process is known as crystallization.

A stage is reached when the two processes occur at the same rate. Under such conditions, the number of solute particles going into the solution will be equal to the solute particles separating out, and a state of dynamic equilibrium is reached.

Solute + Solvent  Solution (2.10) At this stage the concentration of solute in solution will remain constant under the given conditions, i.e., temperature and pressure.

A similar process is followed when gases are dissolved in liquid solvents. Such a solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution.

An unsaturated solution is one in which more solute can be dissolved at the same temperature. The solution which is in dynamic equilibrium with undissolved solute is the saturated solution and contains the maximum amount of solute dissolved in a given amount of solvent.

Thus, the concentration of solute in such a solution is its solubility. Earlier we have observed that the solubility of one substance into another depends on the nature of the substances.

In addition to these variables, two other parameters, i.e., temperature and pressure also control this phenomenon.

Effect of temperature The solubility of a solid in a liquid is significantly affected by temperature changes. Consider the equilibrium represented by equation 2.10.

This, being dynamic equilibrium, must follow Le Chateliers Principle. In general, if in a nearly saturated solution, the dissolution process is endothermic (∆sol H > 0), the solubility should increase with rising in temperature, and if it is exothermic (∆sol H < 0) the solubility should decrease.

These trends are also observed experimentally. Effect of pressure Pressure does not have any significant effect on the solubility of solids in liquids.

It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

AuthorNCERT
Language English
No. of Pages30
PDF Size3.4 MB
CategoryChemistry
Source/Creditsncert.nic.in

NCERT Solutions Class 12 Chemistry Chapter 2 Solution

Q 2.1) If 22 g of benzene is dissolved in 122 g of carbon tetrachloride, determine the mass percentage of carbon tetrachloride (CCl4) and benzene (C6H6).

Answer 2.1:

Mass percentage of Benzene (C6H6) = \frac{Mass\; of\; C_{6}H_{6}}{Total \;mass \;of \;the \;solution}  v\times 100TotalmassofthesolutionMassofC6​H6​​ v×100

= \frac{Mass\; of\; C_{6}H_{6}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100MassofC6​H6​+MassofCCl4​MassofC6​H6​​×100

= \frac{22}{22 + 122}\times 10022+12222​×100

= 15.28%

Mass percentage of Carbon Tetrachloride (CCl4) = \frac{Mass\; of\; CCl_{4}}{Total \;mass \;of \;the \;solution} \times 100TotalmassofthesolutionMassofCCl4​​×100

= \frac{Mass\; of\; CCl_{4}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100MassofC6​H6​+MassofCCl4​MassofCCl4​​×100

= \frac{122}{22 + 122}\times 10022+122122​×100

= 84.72%

Q 2.3) Determine the molarity of each of the solutions given below:

(a) 30 g of Co(NO)3. 6H2O in 4.3 L of solution

(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Answer 2.3:

We know that,

Molarity = \frac{Moles\;of\;Solute}{Volume\;of\;solution\;in\;litre}VolumeofsolutioninlitreMolesofSolute

(a) Molar mass of Co(NO)3. 6H2O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g mol^{-1}mol−1

Therefore, Moles of Co(NO)3. 6H2O = \frac{30}{291}29130​ mol

= 0.103 mol

Therefore, molarity = \frac{0.103\; mol}{4.3\; L}4.3L0.103mol

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol

Therefore, Number of moles present in 30 mL of 0.5 M H2SO4 = \frac{0.5\times 30}{1000}\; mol10000.5×30​mol

= 0.015 mol

Therefore, molarity = \frac{0.015}{0.5\; L}\; mol0.5L0.015​mol

= 0.03 M

Q 2.4) To make 2.5 kg of 0.25 molar aqueous solution, determine the mass of urea (NH2CONH2) that is required.

Answer 2.4:

Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g mol^{-1}mol−1

0.25 molar aqueous solution of urea means:

1000 g of water contains 0.25 mol = (0.25 × 60) g of urea = 15 g of urea

That is,

(1000 + 15) g of solution contains 15 g of urea

Therefore, 2.5 kg (2500 g) of solution contains = \frac{15\times 2500}{1000 + 15}\; g1000+1515×2500​g

= 36.95 g

= 37 g of urea (approx.)

Hence, mass of Urea required is 37 g.

Solutions Class 12 NCERT PDF Free Download

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