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The CORRECT option is (c) 104.7°C

To elaborate: Given,

Weight of solvent, W1 = 222 g

Weight of solute, w2 = 117 g

Kb = 0.53 K KG mol^-1

Now, addition of a non-volatile solute causes elevation in boiling point, ΔTb

ΔTb = (kb x 1000 x w2)/(M2 x w1)

On substituting, ΔTb = (0.52 x 1000 x 117)/(58.5 x 222) = 4.7°C

New boiling temperature = 100 + 4.7 = 104.7°C.

The correct option is (d) 0.0033

Best explanation: Given,

Mass of solute, msolute = 2.0 grams

Volume of solvent, Vsolvent = 1000 ml

P^0water = 17.535 mm Hg

Molecular mass of Cu(NO3)2 = 188 g/MOLE

We know that ΔP1/P^01 = X2, where X2 is the mole fraction of the solute.

We are REQUIRED to find out the difference in VAPOR pressure i.e. ΔP1.

Therefore, ΔPwater= P^0water x X2

Number of MOLES of copper (II) nitrate = 2.0/188 = 0.0106

Number of moles of water = 1000g/(18g/mole) = 55.56 mole (since density of water is 1g/ml)

Mole fraction of solute, X2 = 0.0106/(0.0106 + 55.56) = 1.9 x 10^-4

ΔPwater = 17.535 x 1.9 x 10^-4 = 0.0033.

The correct CHOICE is (c) Raoult’s law

To explain I would say: Raoult’s law QUANTIFIES the relative lowering in vapor pressure. From the law, it follows that p1 = X1 x p^01. If p^01 was the original pressure before X2 mole fraction of SOLUTE was ADDED to the SOLVENT then reduction in vapor pressure is given as:

Δp1 = p^01 – p1

= p^01– p^01 X1

= p^01(1 – X1)

= p^01X2

Therefore, the relative reduction in vapor pressure (∆p1/p^01) = X2 i.e. mole fraction of solute in the solution.

The correct answer is (B) 0.160

The best I can explain: GIVEN,

P^0water =23.8 mm Hg

Pwater= 2.0 CM Hg = 20.0 mm Hg (after addition of solute)

From the LAW of relative lowering of vapor pressure, ∆p1 = X2 x p^01 (where X2 is the mole fraction of solute)

On rearranging, Δp1/p^01 =X2

Δp1 = 23.8 – 20.0 = 3.8 mm Hg

X2 = 3.8/23.8 = 0.160.

The correct option is (a) 0.8889

Explanation: Given,

4P^0A = P^0A ———- (1)

2XA = XB ———— (2)

Using law of relative lowering of vapor pressure:

PA = P^0A x XA ——- (3)

PB = P^0B x XB ——- (4)

Using values of (1) and (2) in (4)

PA = P^0A x XA

PB = 4P^0Ax 2XA =8(P^0A x XA)

Therefore, PB = 8PA

Mole fraction of component B in vapor phase, YB = PB/(PA + PB)

On simplifying, YB = 8PA/(PA + 8PA) = 8/9 = 0.8889.

Correct CHOICE is (a) Colligative PROPERTIES

Easy EXPLANATION: Colligative properties are set of four properties. This set of properties purely depends on the number of solute particles present in the solution/solvent, independent of the NATURE of the particles with respect to the solution/solvent. The properties are 1.) Relative lowering of vapor pressure, 2.) Elevation in boiling point, 3.) DEPRESSION in freezing point and 4.) Osmotic pressure.

Correct answer is (a) > 1

To explain: Van’t HOFF’s FACTOR (i) is defined as the ratio of the observed COLLIGATIVE property to the calculated colligative property. Since, the abnormal mass for solutes that dissociate is lesser than its NORMAL molar mass, the value of Van’t Hoff’s factor will always be > 1.

The correct choice is (c) 0.5

Easy explanation: Given,

Depression in FREEZING point(ΔT) = -0.2 °C

Molality of the solution(m) = 0.1 m

Freezing point depression constant (KF) = 4.0 °C m^-1

Let the Van’t Hoff factor = i

We know that, ΔT = i x Kf x m

0.2 = i x 4 x 0.1

i = 0.2/0.4 = 0.5

Therefore, the ratio between the NORMAL mass and ABNORMAL mass, which is equal to the Van’t Hoff factor is equal to 0.5.

Correct option is (C) 0.75

Explanation: Given,

Degree of association (α) = 50% = 0.5

Number of molecules ASSOCIATED (n) = 2

Let the Van’t Hoff factor = i

We KNOW that for solutes that associate in SOLUTION, i = 1 + ((1/n) -1) x α

= 1 + ((1/2) – 1) x 0.5

= 1 – (0.5 x 0.5)

= 0.75

Therefore, the Van’t Hoff factor is equal to 0.75.

Right answer is (d) Osmosis

To elaborate: Osmotic PRESSURE method has the greatest advantage over other methods that even for very dilute concentrations it GIVES a large magnitude. This would not be true in case other methods. It is HIGHLY useful for biomolecules since they are unstable at extremely high and low TEMPERATURES thus eradicating the method using boiling and freezing points.

The correct answer is (b) Non-ionic solute in dilute solution

Easiest explanation: In order to determine molar masses correctly, it is ESSENTIAL that the solute is non-volatile, non-ionic and present in dilute form only. If it is ionic ONE has to account for its van’t Hoff factor, i, for association in concentrated solution and dissociation in dilute solution. Molar masses can only be calculated from dilute SOLUTIONS CONTAINING non-dissociable non-ionic solutes.

Correct CHOICE is (d) They form maximum boiling azeotropes

Easiest explanation: The INTERACTIONS between the components of a non-ideal solution showing negative deviation are GREATER than the pure components. The change in VOLUME and ENTHALPY after mixing is negative, i.e., ΔVmixing = -ve, ΔHmixing = -ve.

Right ANSWER is (b) False

Explanation: When we correlate molar mass with osmotic pressure as a colligative property, we sometimes ENCOUNTER a situation where the CALCULATED molar mass of the solute is either higher or lower than the actual molar mass. This is called the abnormal molar mass. Abnormal molar mass occurs as a RESULT of dissociation or association of molecules.

Correct answer is (b) Determining molar mass

The explanation is: Colligative properties serve the PURPOSE for determining molar masses of unknown COMPOUNDS. Colligative properties are not used to determine boiling and melting temperatures as it would RESULT in an incorrect value upon the addition of a solute. Equivalent weight can only be determined if the molar mass is known and the van’t HOFF FACTOR is determined in a similar manner.

The correct answer is (d) 1.00005

The BEST I can explain: pH = 4 means [H^+] = 10^-4 M

HA⇌ H^++A^–

Initial

C MOL L^-1

0

0

After dissociation

C – Cα

Cα

Cα

Total = C (1+α)

THUS, [H^+] = Cα, i.e., 10^-4 = 2 x α or α = 5 x 10^-5 = 0.00005

i = 1 + α = 1 + 0.00005 = 1.00005.

Right option is (a) True

For explanation: A non-ideal solution is the solution in which solute and solvent molecules interact with one another with a different FORCE than the forces of INTERACTION between the molecules of the PURE components. Non-Ideal SOLUTIONS do not OBEY Raoult’s law.

The correct option is (b) 29.915 kPa

The best explanation: Given, P^0water = 31 kPa

Concentration of solution, C = 2 molal = 2 moles of glucose/KG of water

From LAW of relative LOWERING of vapor pressure, ΔP/P^0 = X2, where X2 is the mole fraction of glucose in the solution.

Mass of water = 1 kg = 1000 g

Molecular weight of water = 18 g/mole

Moles of water = 1000/18 = 55.556 moles

X2 = 2/(2 + 55.556) = 0.035

ΔP = 31 kPa x 0.035 = 1.085 kPa

Final pressure = 31 kPa – 1.085 kPa = 29.915 kPa.

Right ANSWER is (c) Mole fraction of solute in liquid phase

To ELABORATE: Relative lowering of vapor pressure is the difference in vapor pressure expressed as a fraction of original vapor pressure. Using RAOULT’s law:

ΔP1 = X2 x P1

Therefore, ΔP1/P1 = X2 which implies the relative lowering is EQUAL to mole fraction of solute in liquid phase. Since the solute is non-volatile, it cannot be present in vapor phase.

The correct option is (b) Azeotropic mixtures cannot be SEPARATED into their constituents by fractional distillation

The EXPLANATION is: Azeotropes have the same concentration in the vapour phase and the liquid phase. In case of minimum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the PURE components. In case of maximum boiling azeotropes, the boiling point of the azeotrope is HIGHER than the boiling point of either of the pure components.

The CORRECT option is (a) 21.3 g/mole

Best explanation: GIVEN,

∆p1/p^01 = 0.005

Mass of solute, mW = 0.5 grams

Mass of solvent, mS = 500 grams

Number of moles of solute, NS = 0.5/MW, where MW is the molecular weight of the solute.

Number of moles of solvent, nSolvent = 500 grams/(106 g/mole) = 4.7 mole

Since mS<
Δp1/p^01 = N2/(n1 + n2) ≈ n2/n1

On substituting values,

0.005 = (0.5/MW)/4.7

On SOLVING, MW = 21.3 g/mole.

Right option is (C) Decrease in FREEZING point

To explain I would say: Decrease in freezing point is the correct colligative property, KNOWN as ‘Depression in freezing point’. This is caused by solute particles present on the surface which lowers the equilibrium solid-vapor pressure. Therefore, a lower freezing temperature is required to match the pressure outside. The other correct colligative properties are 1.) Relative LOWERING of vapor pressure, 2.) Elevation in boiling point, 3.) Depression in freezing point and 4.) Osmotic pressure.

The correct option is (a) -23°C

The best I can explain: Given,

kf = -3.83 k kg mol^-1

MASS of solute, w2 = 228 g

Mass of solvent, W1 = 500 g

Molar mass of solute, M2 = 76 g/mole

Moles of solute = w2/M2 = 228/76 = 3 moles

Molality of the solution, m = Number of moles of solute/mass of solvent (kg)

m = 3 moles/0.5 kg = 6 molal

We know, ΔTf = kf x m

ΔTf = -3.83 x 6 = -23°C.

Correct choice is (b) The SOLUTION shows negative deviation

Explanation: Given,

Observed pressure = 76 torr

According to Raoult’s law,

pA = xA x pA^0 = 5/15 x 64 = 21.33 torr

pB = XB x pB^0 = 10/15 x 76 = 50.67 torr

Therefore, pressure EXPECTED by Raoult’s law = 21.33 + 50.67 = 72 torr.

Thus, observed pressure (70 torr) is LESS than the expected value. Hence, the solution shows negative deviation.

Correct answer is (a) Semi-PERMEABLE membrane

The explanation is: A semi permeable membrane is a must CONDITION since it facilitates the blocking of solute particles from diffusing through and allows only water molecules to PASS through. Obviously, water is the solvent in osmosis hence the concentration of solvent cannot be same if osmosis has to occur. If pressure greater than osmotic pressure is applied, REVERSE osmosis takes place. MEANING, water molecules will flow from region of lower water potential to higher water potential.

Right choice is (a) True

Explanation: An azeotrope or a constant BOILING mixture is a mixture that has the same concentration in the VAPOUR phase and the liquid phase. In azeotropes, the COMPONENT ratio of unvaporized SOLUTION is equal to that of the vaporized solution when boiling. Hence, IDEAL solutions don’t form azeotropes.

The correct answer is (b) < 0

Best explanation: The Gibbs free ENERGY of a system at any moment in TIME is defined as the enthalpy of the system minus the product of the temperature TIMES the entropy of the system. For ideal solutions, the value of the Gibbs Free energy is always negative as mixing of ideal solutions is a SPONTANEOUS process.

Correct answer is (d) Osmosis

To explain I WOULD say: Osmosis is a mass transfer process due to which water MOLECULES MOVE from a REGION of higher water potential to lower water potential down the potential gradient. Dialysis and filtration are processes using the concept of diffusion of impurities. Shriveling is the SHRINKING of a cell. In this case, the cell swells up.

Right answer is (c) 3

Best explanation: VAN’t Hoff Factor can also be WRITTEN as the ratio between the total number of moles of particles after association / DISSOCIATION and the total number of moles of particles before association / dissociation. Since BACL2 COMPLETELY dissociates(into one Ba^2+ and two Cl^– ions), the total number of moles after dissociation is equal to 3. Therefore, the Van’t Hoff Factor for BaCl2 is 3.

Correct option is (c) Ethyl alcohol + WATER

To explain I would say: An ideal solution may be defined as the solution in which no volume CHANGE and no ENTHALPY change take PLACE on mixing the solute and the solvent in any proportion. Ethyl alcohol + Water is a Non-Ideal solution.

The correct option is (d) 398.246 mm Hg

For EXPLANATION: Given,

Mass of solute, m2 = 1.5 grams

Molar mass of solute, M2 = 100

Vapor pressure of pure CS2, p^01 = 400 mm Hg

Volume of solvent, ρ = 200 ml

Density of solvent, ρ = 1.3 g/cc

Number of moles of solute, N2 = m2/MW = 1.5/100 = 0.015 mole

From the law of relative LOWERING of vapor pressure, Δp1 = p^01X2, where X2 is the mole fraction of solute and Δp1 is the difference in pressure.

Mass of solvent, m1 = ρ x V = 1.3 x 200 = 260 grams

Number of moles of solvent, n2 = 260g/[(12 + 32 + 32)g/mole]= 3.421

Since the solution is dilute we can APPROXIMATE X2 = n2/(n1 + n2) ≈ n2/n1 (since n2<< n1)

Using ∆p1 = p^01X2,

∆p1 = (400 mg of Hg) (0.015/3.421) = 1.754 mm of Hg

Using Δp1 = p^01 – p1, we get p1 = p^01–Δp1

 HENCE, resulting lowered vapor pressure, p1 = 400 mm Hg – 1.754 mm Hg = 398.246 mm Hg.

Correct option is (a) SODIUM chloride in water

Explanation: The Van’t Hoff factor is LESS than 1 for solutes that dissociate in solution. In the given list, only sodium chloride dissociates in water whereas the remaining carboxylic ACIDS associate in benzene. HENCE, the value of the Van’t Hoff factor is lesser than 1 for a solution of sodium chloride in water.

Right choice is (b) Solutes that ASSOCIATE in WATER have MOLAR mass higher than the molar mass of the solute calculated theoretically

For explanation: On association of solute MOLECULES, the number of particles in a solution is reduced. SINCE, colligative properties are directly proportional to the number of particles in a solution, the colligative properties reduce. The molecular mass of the solute is inversely proportional to its colligative properties. Therefore, solutes that associate in water have a higher molar mass than its value calculated theoretically.

Correct option is (b) Mole fraction of SOLVENT

Easiest explanation: More the solute, lower is the vapor pressure of the solvent when compared to that when it is in its purest form. In case of greater amount of non-volatile solute, more solvent can be added to RAISE the vapor pressure. THEREFORE, it is ALWAYS DIRECTLY proportional to the mole fraction of the solvent.

Correct choice is (c) A plot of vapor pressures of both components gives a linear plot

The best I can explain: Raoult’s law states that the partial pressure of each component in the solution varies directly with its mole fraction in the solution. It is formulated as ptotal = pA + (pB – pA)XB. From this, it is seen that when a graph is plotted, it gives a linear plot passing through ORIGIN. TOTAL vapor pressure can be expressed as mole fraction of one component and that it varies linearly with the latter. HOWEVER, total vapor solution MAY decrease or increase with increase in mole fraction of a component.

Correct ANSWER is (a) Negative deviation

Explanation: Phenol and aniline both EXHIBIT greater magnitude of hydrogen bonding due to the presence of hydroxyl and amine group, both possessing highly electronegative atoms of oxygen and NITROGEN, RESPECTIVELY. The intermolecular hydrogen bonding is far greater than the DEGREE of intramolecular hydrogen bonding which requires more thermal energy to break. Thus, there is an increase in the boiling point of the mixture.

Right option is (d) 13.3 g/kg

The BEST I can explain: Given,

KH = 0.00060 mole/kgbar

PN2 = mole fraction of N2 x Pair (from Dalton’s law)

Air CONSISTS of 79 mole% N2 and 21 mole% O2.

PN2 = 0.79 x 1 BAR= 0.79 bar

Henry’s law –PN2x KH= solubility of N2

0.79 bar x 0.00060 (mole/kgbar) = solubility of N2

Solubility of N2 = 4.74 x 10^-4 moles of N2/kg water

Converting moles of N2 to kg of N2 :

Solubility of N2 = 4.74 x 10^-4 mole x 28 kg/mole = 0.0133 kg N2/kg water = 13.3 g N2/kg water.

The correct answer is (d) Calcium chloride

The best I can explain: Out of the GIVEN OPTIONS calcium chloride is the only salt which releases heat UPON dissolving. Barium chloride and AMMONIUM chloride undergo endothermic dissolution which decreases the temperature of water. However, lead chloride is an insoluble salt and does not dissolve.

Correct OPTION is (b) Positive slope

Easy explanation: CO2 obeys Henry’s law, which governs the solubility of gases in liquids in relation to its pressure. Henry’s law STATES that mole fraction of gas in the solution varies directly with its partial pressure over the surface i.e. p ∝ x (mole fraction). The graph is of the FORM y = MX, a positively SLOPED straight line passing through the origin.

The correct OPTION is (d) The vapor pressure of the solution increases

For explanation I would say: The vapor pressure of a liquid is independent of the pressure of surroundings. Any solution BOILS when its vapor pressure equals the surrounding atmospheric pressure. The atmospheric pressure decreases with increase in altitude. Due to this, the vapor pressure equals the atmospheric pressure in a SHORTER period of TIME when compared to at the solution being present at sea level. Therefore, the solution will vaporize quickly as the bowling point will REACHED faster.

Right answer is (d) Decreases and increases

For explanation I would say: When a solution is saturated a thermodynamic equilibrium exists between the undissolved solute and dissolved solute. Since the dissolution process here is MENTIONED as endothermic INCREASING and decreasing TEMPERATURES will decrease and increase the concentration, respectively, as given by Le Chatelier’s Principle of equilibrium.