Chemistry Test - 2

Concept:

• Liquids vapourise at a given temperature and under equilibrium conditions, the pressure exerted by the vapours of the liquid over the liquid is called vapour pressure.
• If a non-volatile solute is added to a pure liquid, the vapour pressure of the solution at a given temperature is found to be lower than the pure solvent.
• Colligative properties of solutions connected with this decrease of vapour pressure are - 
  • Relative lowering of the vapour pressure of the solvent.
  • Depression of the freezing point of the solvent.
  • Elevation of the boiling point of the solvent, ex: Boiling point of water will increase above 373K.

Explanation:

From the above explanation, we can see that depression in freezing point of a solvent is considered as colligative property. 

Whereas freezing point, boiling point and melting point are the point at which substance feezes, boiled and melted at certain specific temperature and pressure 

Explanation:-

Raoult's Law: It states that for any solution the partial vapour pressure of the volatile component in the solution is directly proportional to its mole fraction.

 p1 = x1. p10 

Important PointsDifferent statements and conditions of Raoult's law:

• The Law state's that for the solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in the solution.

p1 ∝ x1 and p1  =  0p1. x1 and p2 = 0p2. x2

• Raoult's law is said to be a special case of Hendry's law.
• It states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or the solution.

​p = KH . x

Concept:

Molality:

• Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent. 
• It is independent of temperature.
• It is expressed as:

Molality (m) =Moles of solute/Mass of solvent in kg

Elevation in Boiling point:

• For dilute solutions, the elevation of boiling point (∆T_b) is directly proportional to the molal concentration of the solute in a solution.

\(\Delta {T_b} \propto \,m\)

\(\Delta {T_b} = {K_b}\,m\)

• Here m is the molality; K_b is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).

Depression in Freezing point:

• According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of a solid solvent at a lower temperature.
• Thus, the freezing point of the solvent decreases.
• Let  \(T_f^0\) be the freezing point of pure solvent and \({T_f}\) be its freezing point when non-volatile the solute is dissolved in it.
• The decrease in freezing point \(\Delta {T_f} = T_f^0 - {T_f}\) is known as depression in freezing point.
• The depression of freezing point (\(\Delta {T_f}\)) for a dilute solution (ideal solution) is directly proportional to molality, m of the solution. 

\(\Delta {T_f} \propto m\)

\(\Delta {T_f} = {K_f}\,m\)

Where the proportionality is constant, K_f is known as Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant.

Explanation:

We know that,\(\Delta {T_b} = {K_b}\,m\)  and \(\Delta {T_f} = {K_f}\,m\)

Given:  Two solutions are equimolar and the solvents are the same.

• In an equimolar solution, the no. of moles of solute is the same.
• So the molality of both solutions will be the same.
• So, the boiling point ΔT and freezing point ΔTb will also be the same for different non-electrolytes. 

​Hence, Equimolar solutions of different non-volatile and non-ionic solutes in the same solvent have the same boiling point and same freezing point.

The correct answer is option 1 i.e.: A-IV, B - I, C - III, D - II

Solution:

a) Parts per million (ppm): When a solute is present in trace quantities, it is convenient to express the concentration in ppm. 

• PPM = mass of solute × 10⁶ / mass of solution.
• Since we are multiplying the mass of solute by 10⁶ it makes the solution more dilute.

b) Molarity (M): Molarity of a solution is the number of moles of the solute per liter of solution (or a number of millimoles per mL. of solution)

• Molarity = no. of moles of solute in gram/volume of solution in L
• Hence unit becomes mol L^-1.

c) Molality (m): It is the number of moles of gram molecules of the solute per 1000 g of the solvent. 

• Molality = no. of moles of solute in gram/mass of solvent in kg.
• Molality (m) does not depend on temperature since it involves the measurement of the weight of liquids.

d) Mole fraction (X): Mole fraction may be defined as the ratio of the number of moles of one component to the number of moles of all the components (solvent & solute) present in the solution.

• Mole fraction = Amount of constituent in moles/total amount of all constituent in a mixture in moles
• Thereby it is a unitless quantity.

Explanation:

Degree of dissociation:

• The phenomenon to generate current-carrying free ions or free radicals when a given concentration of solution components is taken infraction with the solute.
• The degree of dissociation is calculated by -

Degree of dissociation = \(\frac{Concentration\ of \ dissociated\ substance}{Total\ given\ substance}\)

Factors Affecting the Degree of dissociation:

• Dilution or change in concentration.
• Temperature.
• Nature of electrolyte.
• Nature of electrolyte.
• Nature of the solvent.

Calculation:

2HI \(\rightleftharpoons\) H2 + I₂

At t = 0, concentration is 1 mole

At t = t eq concentration

2HI \(\rightleftharpoons\) H2 + I2

HI is 1-2a,

the concentration of H₂ is "a" mole and the concentration of I₂ is "a" mole then,

The degree of dissociation of HI is given by K_c

Where, K_c = \(\frac{[ H_2 ] [ I_2 ] }{[ HI ]^ 2}\)

Kc = a × a / (1 - 2a)² = a² / (1-2a)²

Here, there will be no impact on the degree of dissociation constant due to the addition of Ar.

Concept:

Vapour Pressure- 

• The pressure at which liquid and vapour can co-exist at a given temperature is called the vapour pressure of the liquid.
• When a liquid is kept in a closed vessel with some free space, it starts to vaporise.
• The vapourisation continues until a state of equilibrium is reached between vapourisation and condensation.
• At equilibrium, the state gets saturated and the pressure exerted by the vapour molecules is called vapour pressure.

• The vapourisation occurs from the surface of the liquid.
• When we add non-volatile solute molecules to the solvent, the solute molecules occupy space on the surface.
• This gives lesser space for solvent molecules on the surface and thus lower number of molecules go to the vapour phase.
• Thus, the vapour pressure also becomes less on the addition of non-volatile solute.
• The lowering of vapour pressure was given by Raoult's law which states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
• The lowering of vapour pressure is derived from Raoult's law which is calculated by the formula:

​\(Δ p = p_0 × x_2\), where x₂ = mole fraction of solute and p₀ = vapour pressure of the pure solution.

Calculation:

Given:

Mole fraction of a solute = x₂ = 0.20

Vapour pressure of the solvent =  p0 = 0.75 atm

Lowering of vapour pressure ( Δp )

Δp = p₀ × x₂ 

Δp = 0.75 × 0.20 = 0.15 atm

Hence, the lowering in the vapour pressure of the solution is 0.15 atm.

Concept: 

Osmosis:

• The phenomenon in which the flow of solvent through a semi-permeable membrane from the pure solvent to the solution takes place is osmosis. 
• It involves the movement of solvent molecules only.
• Osmosis is limited to solutions only.
• Osmosis takes place through a semi-permeable membrane.
• The solvent molecules move from the solution of low solute concentration to higher solute concentration.

Osmotic pressure:

• The osmotic pressure of a solution is called the excess pressure need to be applied to a solution to prevent osmosis. i.e., to stop the passage of solvent molecules through a semi-permeable membrane into the solution.
• It is denoted by π.
• Osmotic pressure is one of the colligative properties as it depends on the number of solute molecules and not on their identity.
• According to Van't Hoff, For a dilute solution, Osmotic pressure is given by:

π = cRT

\({\mathbf{π }}{\text{ }} = \;\frac{{{{\text{n}}_2}}}{{\text{V}}}{\mathbf{RT}}\;\) (→ c = n₂/V)

\({\mathbf{π }}{\text{V = }}\frac{{{{\text{w}}_{\text{2}}}{\text{RT}}}}{{{{\text{M}}_{\text{2}}}}}\;\) (\(\because {n_2}{\text{ = }}\frac{{{w_2}}}{{{M_2}}}\))

Where c = molar concentration/ molarity of the solution; T = Temperature; w₂ = mass of solute; M₂ = molecular weight of solute present in solution, R = gas constant; V = volume of solution in litres; n₂ = no. of moles of solute 

For a solution, at a given temperature, R and T are constant, then Osmotic pressure is:

\(π \propto c\)

Explanation:

Factors affecting Osmotic pressure is given below:

Hence, The Osmotic pressure of the solution increases, if the number of solute molecules is increased.

Concept:

The pressure exerted by vapours over the liquid surface at equilibrium is called vapour pressure of the liquid.

Raoult's law is a chemical law that relates the vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution.

Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present:

i.e.,

P_solution = x_solvent P_solvent°

Calculation:

Given,

The vapour pressure of pure liquid ‘M’ \( = \;P_M^0 = 450\;mmHg,\) 

The vapour pressure of pure liquid ‘N’ \( = P_N^0 = 700\;mmHg\)

According to Raoult’s law,

For liquid ‘M’,

\({P_M} = P_M^0{x_M}\)

Let P_T be the total vapour pressure

The partial pressure of one individual gas within the overall mixtures, P can be expressed as follows:

P = x P_T

\({P_M} = P_M^0{x_M} = {y_M}{P_T}\)

\(P_M^0{x_M} = {y_M}{P_T}\)

\(P_M^0 = \frac{{{y_M}}}{{{x_M}}}\left( {{P_T}} \right) \)

Similarly for liquid ‘N’,

\(P_N^0 = \frac{{{y_N}}}{{{x_N}}}\left( {{P_T}} \right) \)

Given, \(P_M^0 < P_N^0\) (i.e., 450 mmHg < 700 mmHg)

\(\frac{{{y_M}}}{{{x_M}}} < \frac{{{y_N}}}{{{x_N}}}\)

\(\frac{{{y_M}}}{{{y_N}}} < \frac{{{x_M}}}{{{x_N}}}\)

Thus, the correct statement is,

Concept:

When we add a solute to a pure solvent, the solution has a change of vapor pressure.

The change in vapor pressure results in depression of freezing point, ie the freezing point of the solution gets lowered as compared to the pure solvent.

The depression of the freezing point is given by:

T_f^o - T_f = K_f × m × i

where

T_f = Freezing Point of the solution, T_f⁰, m is the molality of the solution and

i is the vant Hoff factor, 

Kf = molal depression constant 

The molality is given by:

\({\rm{\;}}m = {\rm{\;molality\;}} = \frac{{{w_{{\rm{solute\;}}}} × 1000}}{{{M_{{\rm{solute\;}}}} × {w_{{\rm{solvent\;}}\left( {{\rm{in\;g}}} \right)}}}}\)

where M = molar mass of solute,

w_solute = weight of solute added w_solvent = mass of solvent

Calculation:

Here, ethylene glycol is the solvent and water is the solvent.

The mass of 8L of water is 1000 × 8 = 8000 g = wsolvent 

Mass of 2L of ethylene glycol is 1.12 × 2000 = 2240 g = wsolute

M = molar mass of solute = 62g

Kf for water is 1.86 Km-1

\({\rm{\;}}m = {\rm{\;molality\;}} = \frac{{{w_{{\rm{solute\;}}}} × 1000}}{{{M_{{\rm{solute\;}}}} × {w_{{\rm{solvent\;}}\left( {{\rm{in\;g}}} \right)}}}}\)

\(m = \frac{{2240 × 1000}}{{62 × 8000}} = 4.5\)

→  ΔTf = Kf × m = 4.5 × 1.86 = 8.4

Hence, on adding the solute, the solvent will have a depression of freezing by 8.4^o C.

The freezing point of water is 0^oC.

The new freezing point is thus 0 - 8.4 = -8.4⁰C.

Hence, the lowest temperature at which the vehicle can be parked without the danger of getting water to freeze in the radiator is -8.40C.

Concept:

Vapour pressure: 

• Vapour pressure of a liquid at any temperature may be defined as the pressure exerted by the vapour pressure above the liquid in equilibrium with the liquid at that temperature.
• Vapour pressure of the solution at a given temperature is lower than the vapour pressure of the pure solvent at the same temperature.
• According to Raoult's law, the vapour pressure of a solution containing non-volatile solute is given by,

P_A = P_A^o x_A   

Where P_A= vapour pressure of solute (A), P_A^o = vapour pressure in pure state, x_A = mole fraction of solute = \(\frac{{{n_A}}}{{{n_B} + {n_A}}}\)

Factors affecting vapour pressure:

1. Nature of the liquid

• If the intermolecular forces of attraction in the liquid are weak, the molecules can easily leave the liquid and come into the vapour phase and hence the vapour pressure is higher.
• If the intermolecular forces of attraction in the liquid are strong, vapour pressure is weaker.

2. Effect of temperature

• As the temperature of a liquid is increased, the vapour pressure of the liquid increases.

Explanation:

It is given, when a substance is dissolved in a solvent the vapour pressure of the solvent is decreased.

Boiling point is inversely proportional to the vapour pressure of the solvent.

i.e,\({{\Delta }}{{\text{T}}_{\text{b}}}\, \propto \frac{1}{{vapour\,pressure}}\)

So, as the vapour pressure of the solvent is decreased, the boiling point of the solution increases.

Hence, When a substance is dissolved in a solvent the vapour pressure of the solvent is decreased. This results in an increase in the b.p. of the solution

Additional Information

Boiling point:

• The temperature at which vapour pressure of the liquid is equal to the external pressure (i.e., the atmospheric pressure) is called boiling temperature at that pressure. 
• When the external pressure is normal atmospheric pressure (i.e 760 mm), the boiling point is called the normal boiling point. 
• The normal boiling point of water is 100 °C (373 K).
• When the external pressure is equal to 1 bar, the boiling point is called the standard boiling point of the liquid.

• The standard boiling point of water is 99.6 °C (372.6 K).

Some applications of Effect of external (atmospheric) pressure on boiling point:

1. If the external pressure is higher, more heat will be required to make the vapour pressure equal to the external pressure and hence higher will be the boiling point. That is why in hospitals, surgical instruments are sterilized in autoclaves in which boiling point of water is raised by using a weight to cover the vent.

2. Similarly, if the external pressure is decreased, the boiling point is lowered. This is the reason that a liquid boils at a lower temperature on the top of a mountain (where pressure is low) than on the seashore. That is why at hills, use of pressure cooker is essential for cooking food.

Osmotic pressure:

• The osmotic pressure of a solution is called the excess pressure need to be applied to a solution to prevent osmosis. i.e., to stop the passage of solvent molecules through a semi-permeable membrane into the solution.
• It is denoted by π.
• Osmotic pressure is one of the colligative properties as it depends on the number of solute molecules and not on their identity.
• According to Van't Hoff, For a dilute solution, Osmotic pressure is given by:

\({\mathbf{\pi }}{\text{ }} = \;c{\mathbf{RT}}\;\)

\({\mathbf{\pi }}{\text{ }} = \;\frac{{{{\text{n}}_2}}}{{\text{V}}}{\mathbf{RT}}\;\)  ( \(\because c{\text{ = }}\frac{{{{\text{n}}_{\text{2}}}}}{{\text{V}}}\))

\({\mathbf{\pi }}{\text{V = }}\frac{{{{\text{w}}_{\text{2}}}{\text{RT}}}}{{{{\text{M}}_{\text{2}}}}}\;\) (\(\because {n_2}{\text{ = }}\frac{{{w_2}}}{{{M_2}}}\))

Where c= molar concentration/ molarity of the solution, T = Temperature; w2 = mass of solute; M2 = molecular weight of solute presentin solution, R = gas constant; V = volume of solution in litres; n2 = no.of moles of solute