NCERT Solutions for Class 12 Science Chemistry Chapter 2 - Solutions

Answer:

According to Raoult's law, for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. Therefore, mole fraction is useful in relating concentration of solution with its vapour pressure.

Hence, the correct answer is option (i).

Answer:

The solubility of sugar in water is significantly affected by temperature changes and a dynamic equilibrium is established.
Sugar + Water ⇌ Solution
It follows Le Chatelier's Principle. Since the given dissolution process is endothermic (∆_solH > 0), the solubility should increase with rise in temperature. Powdered sugar dissolves faster in hot water than it does in cold water. When water is heated, the molecules gain energy and thus, move faster. Hence, the fast moving water molecules come in contact with the sugar more often, causing it to dissolve faster. Also , when sugar is in the powdered form, surface area increases and hence, reaction between sugar and water becomes more rapid.

Hence, the correct answer is option (iv).

Answer:

Crystallisation is the process of formation of solid crystals precipitating from a solution. At equilibrium , the rate of dissolution of a solid solute in a volatile liquid becomes equal to the rate of crystallisation.
Hence, the correct answer is option (iii).
 

Answer:

When a small amount of solute is added to the solution and it does not dissolve and rather gets precipitated then it is known as a supersaturated solution. Since the substance 'A' is precipitated on its addition to the solution, the solution is supersaturated.

Hence, the correct answer is option (ii).
 

Answer:

Solids and Liquids are highly incompressible and hence, remain unaffected by the change in pressure. So, the maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend on pressure.
Therefore, the correct answer is option (iii).

Answer:

At high altitudes, the partial pressure of oxygen is less than that at the ground level. This decreased atmospheric pressure causes the release of oxygen from the blood and hence, decreases the oxygen concentration in blood and tissues. Low blood oxygen causes these people to become weak and unable to think clearly, symptoms of a condition known as anoxia.

Hence, the correct answer is option (ii).

Answer:

A mixture of methanol and acetone shows positive deviation from Raoult's law. Pure methanol possess hydrogen bonding and introduction of acetone between molecules of methanol results in breaking of some of the hydrogen bonds. Due to weakening of interactions , the solution shows positive deviation from Raoult's law.
Hence, the correct answer is option (i).

Answer:

Colligative properties are important properties of solutions because they describe how the properties of the solvent will change as solute is added. They depend only on the number of solute particles in the solution and not on the identity or physical properties of the solution.

Hence, the correct answer is option (ii).

Answer:

Greater is the van't Hoff factor, higher is the elevation in boiling point and hence, higher will be the boiling point of the solution.
In 1.0 M Na₂SO₄ solution, the van't Hoff factor is highest among the given options, i.e., close to 3 because Na₂SO₄ is a strong electrolyte and dissociates completely into cation and anion in aqueous solution as follows:
${\mathrm{Na}}_2 {\mathrm{SO}}_4\to 2{\mathrm{Na}}^{+} +\mathrm{S}{{\mathrm{O}}_4}^{2-}$. 

Hence, the correct answer is option (ii) .

Answer:

Ebulioscopic constant, $K_b=\frac{∆T_b}{m}$
∆T_b = elevation in boiling point, its unit is K (kelvin)
m = molality of the solution, its unit is mol kg^–1
Thus, the unit of K_b becomes K kg mol^–1 or K (molality)^–1.

Hence, the correct answer is option (i).
 

Answer:

Depression in freezing point depends on the van't Hoff factor. More is the van't Hoff factor, more is the depression in freezing point. 
∆$T_{f }= i{K}_{f }m$
where, 
∆T_f = Depression in freezing point 
i = van't Hoff factor
K_f = molal freezing point depression constant 
m = concentration of the solution

Glucose is not an ionic compound and does not dissociate or associate in aqueous solution. Therefore, i is 1.
$∆T_{f_{glu\cos e}}=1\times K_f\times m$
MgCl₂ is an ionic compound and dissociates completely into cation and anion in aqueous solution. Therefore, i is close to 3.
$∆T_{f_{MgC{l}_2} }=3\times K_f\times m$
Hence, depression in freezing point of MgCl₂ is about three times than that of glucose.

Hence, the correct answer is option (iii).

Answer:

When a raw mango is placed in a concentrated salt solution, it loses water due to osmosis, i.e., water flows from a region of lower concentration to a region of higher concentration through a semipermeable membrane. Thus, it shrivels into pickle.

Hence, the correct option is option (iv).

Answer:

According to the definition of osmotic pressure , we know that,

$\prod _{}=CRT$ , where,

$\prod _{}$= Osmotic Pressure
C= Concentration of the solution
R= Universal gas constant
T= Temperature

So, higher is the concentration of the solution, higher is the osmotic pressure. Therefore, at a given temperature osmotic pressure of a concentrated solution of a substance is higher than that of a dilute solution.
Hence, correct answer is option (i).

 

Answer:

The depression in freezing point depends on the concentration (molarity or molality) of the solution and also on the solvent taken.
As ∆$T_f= K_f\times m$ , where,
∆$T_f$ = Depression in freezing point
$K_f$= Molal freezing point depression constant , which , depends on the solvent taken 
$m$= Molality or concentration of the solution

So, two different solutions of same molality and different concentrations will have different depression in freezing point. 
Hence, the correct answer is option (i) .

 

Answer:

$\mathrm{KCl}\to {\mathrm{K}}^{+}+{\mathrm{Cl}}^{-}$ 
Since KCl gives two ions, it's van't Hoff factor (i) is 2.

$\mathrm{NaCl}\to {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}$
Since NaCl gives two ions, it's van't Hoff factor (i) is 2.

${\mathrm{K}}_2 {\mathrm{SO}}_4 \to 2{\mathrm{K}}^{+}+\mathrm{S}{{\mathrm{O}}_4}^{2-}$ 
Since K₂SO₄ gives three ions, it's van't Hoff factor (i) is 3.
Thus, the van't Hoff factors for KCl, NaCl and K₂SO₄, respectively, are 2, 2 and 3.

Hence, the correct answer is option (ii).

Answer:

In reverse osmosis , solvent molecules move from a region of higher concentration to a region of lower concentration of solute through a semipermeable membrane.
Hence, the correct answer is option (ii).

Answer:

According to Henry's law, the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution.
p ∝ χ
p = K_H χ
K_H = Henry's constant
Henry's constant depends upon the nature of gas, nature of liquid and temperature (T). The value of Henry’s constant K_H increases with increase in temperature.

Hence, the correct answer is option (i).

Answer:

According to Henry's law,
p = K_H χ
where,
p = partial pressure of gas in vapour phase
K_H = Henry's constant
χ = Mole fraction of the gas in the solution
$K_H\propto \frac{1}{\chi }$
The value of K_H is inversely proportional to χ. Therefore, the value of Henry's constant is greater for gases with lower solubility.

Hence, the correct answer is option (ii).
   

Answer:

Water will move from side B (concentrated sodium chloride solution) to side A (fresh water) if a pressure greater than osmotic pressure is applied on piston B. This phenomena of movement of solvent from a region of high concentration to a region of low concentration is known as reverse osmosis.

Hence, the correct answer is option (ii) .

Answer:

The van't Hoff factor for a strong electrolyte depends on the total number of ions formed after dissociation. It does not depend on the concentration of the solution. 
Since NaCl dissociates into two ions in aqueous solution, the van't Hoff factor for NaCl is 2.
$\mathrm{NaCl}\to {\mathrm{Na}}^{+} + {\mathrm{Cl}}^{-}$
Thus, van't Hoff factor for all the three aqueous solutions of NaCl labelled as 'A' , 'B' and 'C', remains the same, i.e., i_A = i_B = i_C = 2.

Hence, the correct answer is option (iii).

Answer:

When the A-A(solute-solute) and B-B(solvent-solvent) type interactions are nearly same as A-B(solute-solvent) type interactions , then the solution formed is nearly ideal. Hence, it obeys Raoult's law. Therefore, the mixture of bromoethane and chloroethane is nearly ideal and obeys Raoult's law.

Hence, the correct answer is option (ii).

Answer:

Since NaCl is a non-volatile solute, relative lowering in vapour pressure of the solution is observed. Therefore, NaCl solution in container (B) will have lower vapour pressure as compared to vapour pressure of water in container (A) at a given temperature.

Hence, the correct answer is option (i).

Answer:

If two liquids A and B form minimum boiling point azeotrope  at a specific composition, then,  A-B interactions are weaker than that of A-A and B-B interactions. Since the boiling point for such azeotropes is minimum, the vapour pressure is maximum and hence, will show large positive deviation from Raoult's law. They have the same composition in liquid and vapour phase and boil at a constant temperature

Hence, the correct answer is option (iv).

Answer:

$\mathrm{Molarity}=\frac{\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{litres}}$
$0.02=\frac{n}{4}$
n = 0.08 moles
Mass of solvent = Density of solvent × Volume of solvent
​Mass of water = 1 × (4 + 1) = 5 Kg
Now, $\mathrm{Molality}=\frac{\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Mass} \mathrm{of} \mathrm{solvent} \mathrm{in} \mathrm{Kg}}$
$\mathrm{Molality}=\frac{0.08}{5}=0.016 \mathrm{m}$

Hence, the correct answer is option (iv).

Answer:

On adding acetone, its molecules get in between the host molecules of methanol and break some of the hydrogen bonds between them. Due to weakening of interactions, the solution shows positive deviation from Raoult’s law. The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition. Therefore, at a specific composition, methanol-acetone mixture will form minimum boiling point azeotrope. 

Hence, the correct answer is option (i).

Answer:

According to Henry's law,
p = K_H χ
where,
p = partial pressure of gas in vapour phase
K_H = Henry's constant
χ = Mole fraction of the gas in the solution
$K_H\propto \frac{1}{\chi }$
The value of K_H is inversely proportional to χ. 
Therefore, higher is the value of Henry's constant for a gas, lower is their solubility.
Increasing order for values of K_H: $HCHO<C{H}_4<C{O}_2<Ar$
Increasing order of solubility: $Ar<C{O}_2<C{H}_4<HCHO$

Hence, the correct answer is option (iii).

Answer:

At constant temperature, the solubility of a gaseous solute in a fixed amount of solvent depends upon nature of solute and pressure. At constant pressure, the solubility of gaseous solute in a fixed amount of solvent depends upon nature of solute and temperature.

Hence, the correct answers are options (i) and (ii) .

Answer:

Benzene and toulene form nearly ideal solution and follow Raoult’s law over the entire range of concentration because the intermolecular forces between two benzene molecules are nearly of the same strength as those between two toulene molecules.
For an ideal solution, 
∆_mix H = zero
∆_mix V = zero
Also, ideal solutions do not form minimum boiling point azeotropes. Thus, statements (iii) and (iv) are not true.

Hence, the correct answers are options (iii) and (iv).

Answer:

Since, the solute molecules are non volatile, the vapour above the solution consists of only solvent molecules. After the addition of solute, vapour pressure of solution is found to be lower than that of pure liquid at a given temperature. This is known as relative lowering in vapour pressure.
Relative lowering in vapour pressure depends on the concentration of a non electrolyte solute and the number of particles of electrolyte solute in the solution. It does not depend on the nature of solute particles.

Hence, the correct answers are options (i) and (ii).

Answer:

 Van't Hoff factor (i) is the measure of association or dissociation of solute particles. It is measured as follows:
$i= \frac{Normal molar mass}{Abnormal molar mass}$ or $i=\frac{Observed colligative property}{Calculated colligative property}$
Here abnormal molar mass is the experimentally determined molar mass and calculated colligative properties are obtained by assuming that the non-volatile solute is neither associated nor dissociated.

Hence, the correct answers are options (i) and (iii).

Answer:

Isotonic solutions are those solutions which have same osmotic pressure at a given temperature. Osmotic pressure is given by the formula:
Π = CRT    .......(1)
where,
Π = Osmotic pressure
C = Concentration of solution
R = Universal gas constant
T = Temperature
Since the osmotic pressure for isotonic solutions is same at a given temperature, then, according to equation (1), molar concentrations of the solutions should be the same. Osmotic pressure is a colligative property and since colligative properties depend on the molar concentration of solute, isotonic solutions have the same elevation in boiling point and depression in freezing point which are also colligative properties.

Hence, the correct answers are options (iii) and (iv).

Answer:

Some liquids on mixing, form azeotropes which are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. There are two types of azeotropes: minimum boiling point azeotrope (showing large positive deviation from Raoult's law) and maximum boiling point azeotrope (showing large negative deviation from Raoult's law). 
Water-Ethanol mixture containing approximately 95% by volume of ethanol forms a minimum boiling point azeotrope and Water-Nitric acid mixture with approximate composition, 68% nitric acid and 32% water by mass forms a maximum boiling point azeotrope.

Hence, the correct answers are options (ii) and (iii).

Answer:

For an isotonic solution, osmotic pressure has to be same. The solute and solvent may or may not be same.

Hence, the correct answers are options (ii) and (iii).

Answer:

For a binary ideal solution, the variation in total vapour pressure v/s composition of solution is linear. As the mole fraction of more volatile component increases, the total vapour pressure increases. In figure (i), component 2 is more volatile while in figure (ii), component 1 is more volatile. Thus, the curves in the figures (i) and (iv) show a linear variation of total vapour pressure and composition of the solution.

Hence, the correct answers are options (i) and (iv).

Answer:

Those properties of a solution that depend only on the number of solute particles present and not on the identity of the solute particles are known as colligative properties.
Colligative properties are observed when a non-volatile solute is dissolved in a volatile liquid. The non-volatile solute can be a liquid or a solid.

Hence, the correct answers are options (i) and (ii).

Answer:

Since, both the components appeared in the distillate and the composition of  A and B in vapour phase became same as that of liquid phase , this shows the formation of an azeotropic mixture. Azeotropic mixtures are non-ideal mixtures showing constant boiling point and same composition in vapour and liquid phase. Due to the formation of an azeotropic mixture, it becomes difficult to separate A and B by distillation.

Answer:

NaCl is a non-volatile solute, therefore, addition of NaCl to water lowers its vapour pressure. As the vapour pressure of water decreases, the boiling point increases. This happens because vapour pressure and boiling point are inversely related. On the other hand, methyl alcohol is more volatile than water, therefore, its addition increases the total vapour pressure of the solution and hence, decreases the boiling point of the solution.
Therefore, on addition of 1 mol of NaCl to 1 litre of water, boiling point increases whereas on addition of 1 mol of methyl alcohol to one litre of water, boiling point decreases.

Answer:

A substance dissolves in a solvent if the intermolecular interactions are similar in both the components. For example, polar solutes such as ammonia dissolve in a polar solvent such as water. Similarly, non polar solutes such as anthracene dissolve readily in a non polar solvent such as benzene. Thus, we can say that "like dissolves like".

Answer:

Molarity of a solution is defined as the number of moles dissolved per litre of the solution.

$Molarity=\frac{No. of moles of solute}{Volume of solution \left(Litres\right)}$ 

This expression shows that molarity depends upon the volume of the solution. Since volume depends on temperature, as the temperature changes, volume will change due to which molarity also changes. On the other hand, mass does not change with temperature. Therefore, concentration terms such as mass percentage, ppm , mole fraction and molality which do not involve volume are independent of temperature.

Answer:

According to Henry's law,
p = K_H χ
where,
p = partial pressure of gas in vapour phase
K_H = Henry's constant
χ = Mole fraction of the gas in the solution

The Henry's constant signifies the following:

• Higher the value of K_H at a given pressure, lower is the solubility of the gas in liquid. 
• The value of K_H depends on the nature of gas. Different gases have different values of K_H.

Answer:

At a given pressure, the solubility of oxygen in water increases with the decrease in temperature. Therefore, the concentration of oxygen in sea is more in cold water and thus, the presence of more oxygen at lower temperature makes the life of aquatic species more comfortable in cold water in comparison to warm water.

Answer:

(a) 
(i) According to Henry's law , pressure of a gas is directly related to solubility of the gas. Scuba divers when come towards the surface, air pressure gradually decreases. This reduced pressure releases the dissolved gases present in blood and leads to the formation of bubbles of nitrogen in blood. This blocks the capillaries and creates a medical condition known as bends, which is painful and life-threatening.

(ii) At high altitudes ,the partial pressure of oxygen is less than that at the ground level. This results in low concentration of oxygen in the blood and tissues of the people living at high altitudes. The low blood oxygen level causes them to become weak and also causes discomfort in breathing at high altitudes.

(b) To increase the solubility of $C{O}_2$ in soft drinks , the soda water bottles are sealed under high pressure. When the bottle is opened at room temperature under normal atmospheric conditions, the pressure inside the bottle decreases to atmospheric pressure and excess $C{O}_2$ fizzes out.

Answer:

In pure liquid water, the entire surface of liquid is occupied by the molecules of water. The escape of water molecules can take place easily and hence, the vapour pressure increases. Whereas, when a non-volatile solute, such as glucose is dissolved in water, some of the surface is covered by non-volatile glucose molecules. As a result, number of solvent molecules (water molecules) escaping from the surface is reduced and consequently the vapour pressure of an aqueous solution of glucose also decreases.

Answer:

When salt is sprinkled over snow covered roads, it lowers the freezing point of water to such an extent that water does not freeze to form ice. As a result, the snow starts melting from the surface and therefore, it helps in clearing the snow covered roads in hilly areas. This phenomena is known as depression in freezing point. 

Answer:

A continuous sheet or film (natural or synthetic) which contain a network of submicroscopic holes or pores that allows the flow of smaller solvent molecules and not the bigger solute molecules to pass through it, is known as a semi-permeable membrane. This membrane is used during osmosis and reverse osmosis where only solvent molecules can move across this membrane.

Answer:

An example of a material used for making semipermeable membrane for carrying out reverse osmosis is cellulose acetate.

Answer:

The correct match is listed as:

$$\begin{gathered} \left(i\right)\to \left(d\right) \\ \left(ii\right)\to \left(c\right) \\ \left(iii\right)\to \left(a\right) \\ \left(iv\right)\to \left(b\right) \\ \left(v\right)\to \left(f\right) \\ \left(vi\right)\to \left(e\right) \end{gathered}$$

Answer:

The correct match is listed as:

$$\begin{gathered} \left(i\right)\to \left(e\right) \\ \left(ii\right)\to \left(c\right) \\ \left(iii\right)\to \left(d\right) \\ \left(iv\right)\to \left(b\right) \\ \left(v\right)\to \left(a\right) \end{gathered}$$

Answer:

The correct match between column I and column II is listed as:

$$\begin{gathered} \left(i\right) \to \left(c\right) \\ \left(ii\right)\to \left(e\right) \\ \left(iii\right)\to \left(d\right) \\ \left(iv\right)\to \left(a\right) \\ \left(v\right)\to \left(b\right) \end{gathered}$$

Answer:

The correct match between column I and column II is listed as:

$$\begin{gathered} \left(i\right)\to \left(d\right) \\ \left(ii\right)\to \left(c\right) \\ \left(iii\right)\to \left(b\right) \\ \left(iv\right)\to \left(e\right) \\ \left(v\right)\to \left(a\right) \end{gathered}$$

Answer:

Volume is a function of temperature, i.e., as the temperature changes, volume of the solution also changes. 

Now, $Molarity=\frac{No. of moles of solute}{Volume of solution in litres}$

Since molarity is dependent on volume, molarity of a solution also changes with the temperature.
Thus, assertion and reason are both correct statements and reason is the correct explanation for assertion.

Hence, the correct answer is option (i).

Answer:

Methyl alcohol has a volatile nature. Thus, when it is added to water, boiling point of water decreases because vapour pressure increases when a volatile solute is added to a volatile solvent.
Since the boiling point decreases, no elevation in boiling point is observed. Both assertion and reason are incorrect statements.

Hence, the correct answer is option (iv).

Answer:

When a non-volatile solute (NaCl) is added to water, depression in freezing point is observed due to lowering of vapour pressure. This happens because, the amount of water that can vaporize in the solution has been affected by the addition of non volatile solute (NaCl).
Therefore, both assertion and reason are correct statements and reason is the correct explanation for assertion.

Hence, the correct answer is option (i).

Answer:

Solvent molecules pass through the semipermeable membrane from a region of low concentration of solution to a region of high concentration of solution. This process is called osmosis. Thus, assertion is a correct statement but reason is wrong statement as osmosis takes place through a semipermeable membrane from a region of low concentration to a region of high concentration.

Hence, the correct answer is option (iii).

Answer:

$$\begin{gathered} \left(i\right) w/w (Mass percentage) =\frac{Mass of component in the solution(in grams)}{Total mass of the solution}\times 100 \\ \left(ii\right) V/V(Volume percentage) = \frac{Volume of the component(solute or solvent)}{Total volume of the solution}\times 100 \\ \left(iii\right) w/V(Mass by volume percentage)=\frac{Mass of solute(in grams)}{Volume of solution}\times 100 \\ \left(iv\right) ppm(Parts per million)= \frac{Number of parts of component(solute or solvent)}{Total number of parts of all components of the solution}\times {10}^6 \\ \left(v\right) x(Mole fraction) =\frac{Number of moles of component(solute or solvent)}{Total number of moles of all components} \\ \left(vi\right) M\left(Molarity\right) = \frac{Number of moles of solute}{Volume of solution in Litres} \\ \left(vii\right) m \left(Molality\right) = \frac{Mass of solute}{Mass of solvent in Kg} \end{gathered}$$

Mass percentage , molality, parts per million and mole fraction depends on mass or moles only. Mass is independent of change in temperature and hence, these mass dependent properties do not change with the change in temperature. Only those properties that depend on volume change with the change in temperature.
 

Answer:

(i) $CHC{l}_{3 }\left(l\right) and C{H}_{2}C{l}_2\left(l\right)$

Both  $CHC{l}_{3 }\left(l\right) and C{H}_{2}C{l}_2\left(l\right)$ are volatile liquids.
For a binary mixture containing two volatile liquids, the total pressure is given by:
$p={p^0}_A{\chi }_A+{p^0}_B{\chi }_B$ , where,

$p$= Total vapour pressure
${p^0}_A$= Vapour pressure of pure component A
${p^0}_B$= Vapour pressure of pure component B
${\chi }_{A }\mathrm{and} {\chi }_B$= Mole fractions of component A and B respectively

(ii) $\mathrm{NaCl}\left(\mathrm{s}\right) \mathrm{and} {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Here, NaCl(s) is a non-volatile solute whereas H₂O(l) is a solvent.
When a non-volatile solute is dissolved in water then vapour pressure is lowered (relative lowering in vapour pressure). Vapour pressure of the solution is calculated as:
$$\begin{gathered} p_A={\chi }_A {p^0}_{A }.....\left(1\right) \\ {p}_A=Vapour pressure of solvent\left(A\right) \\ {p^0}_A= Vapour pressure of pure solvent\left(A\right) \\ {\chi }_A= Mole fraction of solvent\left(A\right) \\ The lowering in pressure of component A is given by: \\ ∆p_A={p^0}_A-p_{A } \\ ∆p_A={p^0}_A-{p^0}_A{\chi }_A.....( us\mathrm{in}g equation 1) \\ ∆p_A={p^0}_A(1-{\chi }_A) ........\left(2\right) \\ As it is a binary solution , {\chi }_B=1-{\chi }_A \\ So, equation \left(2\right) can be written as: \\ ∆p_A={p^0}_A{\chi }_B \\ where {\chi }_B= mole fraction of component B\left(solute\right) \\ Now, {\chi }_B=\frac{∆p_A}{{p^0}_A} ......\left(3\right) \\ Equation \left(3\right) gives the relative lowering in vapour pressure \left(\frac{∆p_A}{{p^0}_A}\right) which is equal to the mole fraction of solute (component B) i.e. {\chi }_B \end{gathered}$$



 

Answer:

Ideal Solutions: Those solutions that obey Raoult's law over the entire range of concentrations are known as ideal solutions.
For an ideal solution,
${∆}_{mix}H=0 and {∆}_{mix}V=0$ , which means the change in enthalpy of mixing and change in volume of mixing is zero or unchanged for an ideal solution.
Also, for ideal solutions , A-B interactions are nearly equal to the A-A and B-B type interactions.
A-B = Solute-solvent interactions
A-A= Solute-solute interactions
B-B= Solvent-solvent interactions

Non-Ideal Solutions: Those solution that do not obey Raoult's law over the entire range of concentrations are known as non-ideal solutions.
These solutions show positive or negative deviations from Raoult's law behaviour.

Positive Deviations: Those solutions having vapour pressure higher than the predicted value. In positive deviation:
${∆}_{mix}H=+ve and {∆}_{mix}V=+ve$
Also, A-B interactions < A-A and B-B type interactions

Negative Deviations: Those solutions having vapour pressure lower than the predicted value. In negative deviation
${∆}_{mix}H=-ve and {∆}_{mix}V=-ve$
Also , A-B interactions > A-A and B-B type interactions

 

Answer:

Pure ethanol cannot be separated using fractional distillation because the ethanol-water mixture forms an azeotrope. The solutions or mixtures having same composition in liquid as well vapour phase and boil at a constant temperature are known as azeotrope. Due to constant composition, it cannot be separated out by fractional distillation.

General name given to binary mixtures which show deviation from Raoult's law is azeotrope. There are two types of azeotropes:

(i) Minimum boiling azeotrope:  The solutions which show large positive deviation from Raoult's law form minimum boiling azeotrope at a specific composition, e.g., composition of 95% ethanol and 5% water by mass.

(ii)Maximum boiling azeotrope: The solutions which show large negative deviation from Raoult's law form maximum boiling azeotrope at a specific composition, e.g., composition of 68% nitric acid and 32% water by mass.

Answer:

Raisins swell in size on keeping in water. This happens due to the phenomenon of osmosis. The outer skin of raisin acts as a semipermeable membrane. Water moves from a place of lower concentration to a place of higher concentration through the semipermeable membrane. Thus, water enters inside the raisins and make them swell. The following diagram shows the phenomena of osmosis:

                                                                                                           
                                                       

Applications of osmosis

• Movement of water from soil into plant roots and subsequently into upper portion of plant is partly due to osmosis.
• Preservation of meat against bacterial action by adding salt.
• Preservation of food against bacterial action by adding sugar. Bacterium in canned fruits loses water through the process of osmosis, shrivels and dies.

Answer:

• Plant roots absorb water from soil due to osmosis. Concentration of cell inside root hair cells is higher than that of water present in the soil. Water enters the root cells due to endosmosis.
• Water absorbed by plant roots is circulated in the entire plant body and reaches to the top of tall tree due to osmosis.
• Use of salt and sugar in pickles and jams act as preservatives. It prevents the growth of bacteria and fungi by osmosis.
• Red blood cells burst when they are placed in water containing less than 0.9% (mass/ volume) salt due to endosmosis.

Answer:

This can be achieved as follows:

• Place the egg in mineral acid solution for about 2 hours. The outer shell of  egg dissolves.
• Place the egg in a saturated solution (hypertonic) of sodium chloride for about 3 hours. Size of the egg is reduced as the egg shrivels due to osmosis.
• As the size of the egg shrivels , it can now be easily inserted into a bottle with narrow neck.
• Insert the egg in a bottle with narrow neck. Add water to the bottle. Water will act as hypotonic solution. Egg regains shape due to osmosis.

Answer:

The compounds which associate or dissociate in the solvent show abnormal molecular masses.

a) Association: Compounds like benzoic acid and acetic acid dimerise in benzene due to hydrogen bonding as a result of which the number of particles in the solution decreases. Since, colligative properties depend on the number of particles, such solutes going association show lower colligative property.

b)Dissociation: Electrolytes such as NaCl, KCl, etc. dissociate into ions which results in increase in the number of particles. Hence, such solutions show higher colligative property.
To account for association or dissociation, Van't Hoff factor i.e. i was introduced. It can be defined as:

$$\begin{gathered} i=\frac{Expected molar mass}{Abnormal molar mass} \\ =\frac{Observed colligative property}{Calculated colligative property} \\ =\frac{Total number of moles after association/dissociation}{Total number of moles before association/dissociation} \\ When i>1, dissociation takes place as the number of particles increases \\ When i<1, association takes place as the number of particles decreases \\ When i=1, neither association nor dissociation takes place \end{gathered}$$