Question 1

A sulphuric acid solution has a concentration of 3.60 M and contains 29% H2SO4 by mass. If the molar mass of H2SO4 is 98 g mol⁻¹, what is the density of the solution in g mL⁻¹?

Accepted Answer

1.22. Density = molarity x molar mass / (10 x %w/w) = 3.60 x 98 / (10 x 29) = 1.22 g/mL. Stored 1.64 is wrong.

Question 2

Which one of the following substances is absent in clear hard water?

Accepted Answer

MgCO3. MgCO3 is not present in clear hard water because hard water typically contains soluble salts like MgSO4, Mg(HCO3)2, and CaCl2, while MgCO3 is less soluble and often precipitates out.

Question 3

Fractional distillation is chosen in a situation where

Accepted Answer

the liquids have only a slight difference in boiling points. Fractional distillation is used when the liquids have only a small difference in boiling points; a wide boiling-point gap (the stored answer) is handled by simple distillation.

Question 4

Car radiators use an aqueous ethylene glycol solution as antifreeze. How much ethylene glycol should be mixed with 5 kg of water so that the freezing point is lowered to -0.3 °C? (K_f = 1.86 K kg mol⁻¹)

Accepted Answer

50 g. The correct option is 50 g because to achieve a freezing point depression of -0.3 °C, the calculated amount of ethylene glycol needed, based on the freezing point depression formula, is approximately 50 g when mixed with 5 kg of water.

Question 5

A solution contains a non-volatile solute of molecular mass M₂. Which expression can be used to determine the molecular mass of the solute using osmotic pressure?

Accepted Answer

M₂ = (m₂/V) RT / π. The correct expression relates the osmotic pressure (π) to the concentration of the solute (m₂/V) and the ideal gas constant (R) and temperature (T), allowing for the calculation of the molecular mass (M₂) by rearranging the formula to isolate M₂.

Question 6

At 298 K, the Henry’s law constant for oxygen is 1.4 × 10⁻³ mol L⁻¹ atm⁻¹. If oxygen is at a partial pressure of 0.5 atm, what mass of oxygen will dissolve in 100 mL of water at 298 K?

Accepted Answer

2.24 mg. Solubility = 1.4e-3 * 0.5 = 7e-4 mol/L; in 100 mL that is 7e-5 mol = 2.24e-3 g = 2.24 mg, not 1.4 g.

Question 7

A cryoscopic measurement gives the molar mass of sodium chloride as 31.80 g mol⁻¹. What is its degree of dissociation? [Atomic masses: Na = 23 g mol⁻¹, Cl = 35.5 g mol⁻¹]

Accepted Answer

0.83. The degree of dissociation can be calculated by comparing the observed molar mass from the cryoscopic measurement with the expected molar mass of sodium chloride based on its complete dissociation into ions. The calculated degree of dissociation of 0.83 indicates that a significant portion of the sodium chloride dissociates into sodium and chloride ions in solution.

Question 8

A binary solution made up of components A and B obeys Raoult’s law when the interaction between A and B molecules is

Accepted Answer

the same as the interactions between A–A and B–B molecules. An ideal (Raoult-obeying) solution forms when A-B interactions equal the A-A and B-B interactions. Stronger A-B interactions (the stored option) cause negative deviation, not ideal behaviour.

Question 9

For a binary mixture of liquids A and B, it is found that the vapour pressure of the solution is given by Pₛ = x_A(P_A⁰ - P_B⁰) + P_B⁰. Such a solution is classified as

Accepted Answer

Ideal. That expression is exactly Raoult's law (Ps = P_A0 x_A + P_B0 x_B), so the solution is ideal, not non-ideal.

Question 10

An aqueous solution containing 0.0020 m of an ionic compound, Co(NH3)5(NO2)Cl, has a freezing point of -0.00732 °C. If Kf = 1.86 °C m⁻¹, how many moles of ions are produced when 1 mole of this compound dissolves in water?

Accepted Answer

2. i = 0.00732/(1.86*0.0020) = 1.97 ~ 2, so the compound gives 2 ions ([Co(NH3)5(NO2)]+ and Cl-). The stored answer (1 ion) is wrong.

Question 11

An aqueous urea solution (molar mass 56 g mol⁻¹) has a boiling point of 100.18°C at 1 atm pressure. If the values of K_f and K_b for water are 1.86 and 0.512 K kg mol⁻¹, respectively, then the freezing point of this solution is

Accepted Answer

-0.654°C. The freezing point depression can be calculated using the formula Δ T_f = K_f · m, where m is the molality of the solution. Given the boiling point elevation and the properties of water, the calculated freezing point depression results in a freezing point of -0.654°C, indicating that the solution freezes at a lower temperature than pure water.

Question 12

For a binary mixture of components A and B, negative deviation from ideal behavior is indicated by which of the following?

Accepted Answer

Change in enthalpy on mixing is negative, ΔHmix < 0. A negative deviation from ideal behavior occurs when the interactions between different components (A and B) are stronger than those between like components (A–A and B–B). This leads to a release of energy upon mixing, resulting in a negative change in enthalpy (ΔHmix < 0).

Question 13

A binary liquid mixture is made by combining n-heptane and ethanol. Which statement best describes the nature of this solution?

Accepted Answer

It behaves as a non-ideal solution and exhibits positive deviation from Raoult’s law.. The correct option indicates that the mixture of n-heptane and ethanol does not follow Raoult's law perfectly, which is characteristic of non-ideal solutions. The positive deviation suggests that the interactions between the different molecules in the mixture are weaker than those in the pure components, leading to a higher vapor pressure than expected.

Question 14

In a solution undergoing freezing point depression, which phases are in equilibrium?

Accepted Answer

liquid solvent and solid solvent. During freezing-point depression the equilibrium is between liquid solvent and solid solvent (pure solvent freezes out); the solute stays dissolved. So the stored 'liquid solute and solid solvent' is wrong.

Question 15

Which concentration unit does not change with temperature?

Accepted Answer

Molality. Molality is defined as the number of moles of solute per kilogram of solvent, and since it is based on mass rather than volume, it remains constant regardless of temperature changes.

Question 16

For solutions of equal molarity, which one would exert the greatest osmotic pressure?

Accepted Answer

1 M (NH4)3PO4. Osmotic pressure pi = i*C*R*T. For equal molarity it scales with i: NaCl i=2, MgCl2 i=3, Na2SO4 i=3, (NH4)3PO4 i=4. (NH4)3PO4 gives the most ions (4), so it exerts the greatest osmotic pressure.

Question 17

A 0.2 mol kg−1 aqueous solution of X has a higher boiling point than an equimolal aqueous solution of Y. In this situation, which statement must be correct?

Accepted Answer

X dissociates in water.. Boiling-point elevation is colligative: it depends on the number of dissolved particles, not on molar mass. If equimolal X gives a higher BP than Y, X must produce more particles, i.e. X dissociates in water. So the correct option is 'X dissociates in water.'

Question 18

Among the following 0.10 M aqueous solutions, which one will show the greatest depression in freezing point, i.e., the lowest freezing temperature?

Accepted Answer

Al2(SO4)3. Delta Tf = i*Kf*m, and at equal molality the largest i gives the greatest depression. Al2(SO4)3 -> 2 Al^3+ + 3 SO4^2- gives i = 5, vs KCl i = 2, glucose and sucrose i = 1. So Al2(SO4)3 shows the lowest freezing point.

Question 19

Assuming sodium sulphate undergoes complete dissociation into its ions in water, what is the depression in the freezing point of water when 0.01 mol of sodium sulphate is dissolved in 1 kg of water? (Given: Kf = 1.86 K kg mol−1)

Accepted Answer

0.0558 K. Na2SO4 -> 2Na+ + SO4^2-, so van't Hoff factor i=3. dTf = i*Kf*m = 3*1.86*0.01 = 0.0558 K.

Question 20

Which of the following salts will show the same van’t Hoff factor (i) as K4[Fe(CN)6] in solution?

Accepted Answer

Al2(SO4)3. K4[Fe(CN)6] dissociates into 4 K+ and 1 [Fe(CN)6]4-, so van't Hoff factor i = 5. Al2(SO4)3 gives 2 Al3+ + 3 SO4^2- = 5 ions, so i = 5 as well. (NaCl i=2, Al(NO3)3 i=4, Na2SO4 i=3.) The match is Al2(SO4)3.

Question 21

What is the mole fraction of the dissolved substance in a 1.00 molal solution prepared in water?

Accepted Answer

0.0177. A 1.00 molal solution has 1 mol solute per 1000 g water = 55.5 mol water. Mole fraction of solute = 1/(1+55.5) = 1/56.5 = 0.0177.

Question 22

An azeotropic mixture formed by hydrochloric acid and water contains what percentage of HCl?

Accepted Answer

20.2% HCl. The correct option is 20.2% HCl because this is the specific concentration at which hydrochloric acid and water form an azeotropic mixture, meaning that this composition cannot be separated by simple distillation.

Question 23

An aqueous solution has a freezing point of −0.186 °C. If water has Kb = 0.52 K kg mol−1 and Kf = 1.86 K kg mol−1, what is the rise in the boiling point of this solution, in K?

Accepted Answer

0.052. From freezing point: i*m = dTf/Kf = 0.186/1.86 = 0.1. Then dTb = Kb*(i*m) = 0.52*0.1 = 0.052 K.

Question 24

Aqueous acetic acid solution has a molarity of 2.05 M and a density of 1.02 g mL⁻¹. What is its molality?

Accepted Answer

2.28 mol kg⁻¹. To find the molality, we first need to calculate the mass of the solvent (water) in kilograms. Using the density and molarity, we can determine the mass of the solution and then subtract the mass of acetic acid to find the mass of water. The resulting calculation gives us a molality of 2.28 mol/kg, which is the correct answer.

Question 25

What is the density, in g mL⁻¹, of a 3.60 M solution of sulphuric acid that contains 29% H2SO4 by mass? (Molar mass of H2SO4 = 98 g mol⁻¹)

Accepted Answer

1.22. In 1 L: moles = 3.60, mass H2SO4 = 3.60*98 = 352.8 g. This is 29% by mass, so solution mass = 352.8/0.29 = 1216.5 g per litre. Density = 1216.5 g / 1000 mL = 1.22 g/mL.

Question 26

A commercial ion-exchange resin used in water softening has the formula C8H7SO3⁻ Na⁺ and a molecular mass of 206. What is the greatest amount of Ca2+ that can be taken up by this resin, expressed as moles per gram of resin?

Accepted Answer

1/412. Per gram there are 1/206 mol of resin units, each carrying one SO3- exchange site. Ca2+ is divalent and occupies two sites, so max Ca2+ uptake = (1/206)/2 = 1/412 mol per gram.

Question 27

Compared with the zeolite method used to eliminate permanent hardness, the synthetic resin technique is:

Accepted Answer

more effective because it can exchange both cations and anions. The synthetic resin technique is superior because it has the capability to exchange both cations and anions, allowing it to address a wider range of hardness-causing minerals compared to the zeolite method, which primarily focuses on cation exchange.

Question 28

An aqueous solution has a freezing point depression corresponding to a freezing point of −0.186 °C. For this same solution, take Kb = 0.512 °C and Kf = 1.86 °C. What is the rise in its boiling point?

Accepted Answer

0.0512 °C. dTf = 0.186 C. dTb = (Kb/Kf) x dTf = (0.512/1.86) x 0.186 = 0.0512 C (index 1).

Question 29

For a pair of liquids A and B that behave as an ideal solution, which statement is true?

Accepted Answer

The enthalpy change on mixing is zero.. In an ideal solution, the enthalpy change upon mixing is zero because the interactions between the molecules of the different components are similar to those between the molecules of the same component, leading to no net energy change.

Question 30

A 0.2 molal aqueous solution of the weak acid HX has a degree of dissociation of 0.3. If the cryoscopic constant of water is 1.85, the freezing point of this solution is closest to

Accepted Answer

−0.480°C. i = 1 + alpha = 1 + 0.3 = 1.3. dTf = i*Kf*m = 1.3 * 1.85 * 0.2 = 0.481. Freezing point = 0 - 0.481 = -0.48 C (depression lowers the freezing point, so the sign is negative).

Question 31

Among the following aqueous solutions, which one will have the greatest boiling point?

Accepted Answer

0.010 M sodium sulfate solution. Boiling point elevation depends on i x c. Values: 0.015 urea -> 0.015; 0.010 KNO3 (i=2) -> 0.020; 0.010 Na2SO4 (i=3) -> 0.030; 0.015 glucose -> 0.015. Na2SO4 has the largest effective particle concentration, so it gives the greatest boiling point.

Question 32

Which of the following pairs of liquids exhibits positive deviation from Raoult’s law?

Accepted Answer

Benzene and methanol. Positive deviation occurs when A-B interactions are weaker than A-A/B-B. Mixing benzene with methanol breaks methanol H-bonds, raising vapour pressure above ideal -> positive deviation. Water-HCl and acetone-chloroform show negative deviation.

Question 33

Identify the statement that is not true:

Accepted Answer

Two sucrose solutions having the same molality but made in different solvents will show the same depression in freezing point.. Depression of freezing point is dTf = Kf x molality, and Kf differs from solvent to solvent, so equal-molality sucrose solutions in different solvents do NOT show equal dTf. That statement (index 3) is the untrue one; pi = MRT is correct.

Question 34

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (ΔTf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol−1):

Accepted Answer

0.0558 K. Na2SO4 -> 2Na+ + SO4^2-, so i = 3. dTf = i*Kf*m = 3 * 1.86 * 0.01 = 0.0558 K.

Question 35

On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane

Accepted Answer

72.0 kPa. moles heptane = 25/100 = 0.25, moles octane = 35/114 = 0.307; x_heptane = 0.449, x_octane = 0.551. P = 0.449*105 + 0.551*45 = 47.1 + 24.8 = 72.0 kPa.

Question 36

A 5% solution of cane sugar (molar mass 342) is isotonic with a 1% solution of an unknown solute. What is the molar mass of the unknown solute in g mol−1?

Accepted Answer

68.4. Isotonic solutions have equal molarity, so (5/342) = (1/M), giving M = 342/5 = 68.4 g mol^-1.

Question 37

The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is:

Accepted Answer

2.05 M. Moles urea = 120/60 = 2 mol. Total solution mass = 120 + 1000 = 1120 g; volume = 1120/1.15 = 973.9 mL = 0.974 L. Molarity = 2/0.974 = 2.05 M.

Question 38

Kf for water is 1.86 K kg mol−1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to −2.8 °C?

Accepted Answer

93 g. molality = dTf/Kf = 2.8/1.86 = 1.505 mol/kg. For 1.0 kg water this is 1.505 mol of ethylene glycol. Mass = 1.505 * 62 = ~93 g.

Question 39

Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25 °C. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

Accepted Answer

They all have the same osmotic pressure.. Effective particle concentration i x M: ethanol 1x0.5=0.5; Mg3(PO4)2 5x0.1=0.5; KBr 2x0.25=0.5; Na3PO4 4x0.125=0.5. All equal, so all solutions have the same osmotic pressure (index 0).

Question 40

The vapour pressure of acetone at 20 °C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 °C, its vapour pressure was 183 torr. The molar mass (g mol−1) of the substance is:

Accepted Answer

64. The decrease in vapour pressure due to the non-volatile solute can be calculated using Raoult's law, which relates the change in vapour pressure to the mole fraction of the solute. Given the data, the calculated molar mass of the solute is 64 g/mol, which matches the correct answer.

Question 41

The freezing point of benzene decreases by 0.45 °C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage of association of acetic acid in benzene will be: (Kf for benzene = 5.12 K kg mol−1)

Accepted Answer

94.6%. m = (0.2/60)/0.020 = 0.1667 mol/kg. i = dTf/(Kf*m) = 0.45/(5.12*0.1667) = 0.527. For dimerization i = 1 - alpha/2 -> alpha = 2(1-0.527) = 0.945, i.e. about 94.6%.

Question 42

The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol⁻¹) of the substance is:

Accepted Answer

64. The decrease in vapor pressure of acetone upon dissolving the non-volatile substance can be calculated using Raoult's law. The change in vapor pressure indicates the number of moles of solute present, and given the mass of the solute and the mass of the solvent, the molar mass can be determined, leading to the conclusion that the correct answer is 64 g/mol.

Question 43

The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (Kf for benzene = 5.12 K kg mol⁻¹)

Accepted Answer

94.6%. The correct option is right because the observed freezing point depression indicates that the effective number of solute particles is less than expected due to dimerization. By calculating the degree of association using the freezing point depression formula and the van 't Hoff factor for the dimer, we find that approximately 94.6% of acetic acid associates in benzene.

Question 44

Two 5 molal solutions are prepared by dissolving a non-electrolyte, non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents Mx and My, respectively where Mx = 3/4 My. The relative lowering of vapour pressure of the solution in X is "m" times that of th

Accepted Answer

3/4. For dilute solutions RLVP ~ n_solute/n_solvent. At equal molality, moles solvent per kg = 1000/M, so RLVP is proportional to M. Thus m = Mx/My = 3/4.

Question 45

K2HgI4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:

Accepted Answer

1.8. K2HgI4 dissociates into 3 ions (2 K+ and [HgI4]2-), so n=3. i = 1 + (n-1)*alpha = 1 + 2(0.40) = 1.8.

Question 46

Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is

Accepted Answer

Kb = 2 Kf. The relationship between the elevation in boiling point and the depression in freezing point is given by the formulas Kb and Kf, where the change in temperature is directly proportional to the molality of the solution. Since the boiling point elevation for a 1 molal solution of glucose is 2 K and the freezing point depression for a 2 molal solution is also 2 K, it indicates that Kb is twice Kf, leading to the conclusion that Kb = 2 Kf.

Question 47

Which one of the following statements regarding Henry's law is not correct?
(1) Different gases have different K_H (Henry's law constant) values at the same temperature.
(2) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
(3)

Accepted Answer

Higher the value of K_H at a given pressure, higher is the solubility of the gas in the liquids.. By Henry's law p = K_H * x, so x = p/K_H. At a given pressure a larger K_H gives a smaller mole fraction, i.e. lower solubility. Hence the statement that higher K_H means higher solubility is the incorrect one.

Question 48

A solution of sodium sulfate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg⁻¹ is:

Accepted Answer

4. Molality = moles of Na+ per kg of water = (92 g)/(23 g/mol) / 1 kg = 4 mol/kg.

Question 49

A solution containing 62 g ethylene glycol in 250 g water is cooled to -10°C. If Kf for water is 1.86 K kg mol⁻¹, the amount of water (in g) separated as ice is:

Accepted Answer

64. The freezing point depression can be calculated using the formula Δ T_f = K_f · m, where m is the molality of the solution. Given the mass of ethylene glycol and the freezing point depression, we find that the solution's freezing point drops sufficiently to cause 64 g of water to freeze.

Question 50

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol⁻¹) and 1.8 g of glucose (molar mass = 180 g mol⁻¹) in 100 mL of water at 27°C. The osmotic pressure of the solution is: (R = 0.08206 L atm K⁻¹ mol⁻¹)

Accepted Answer

4.92 atm. n = 0.6/60 + 1.8/180 = 0.01 + 0.01 = 0.02 mol in 0.1 L. pi = (n/V)RT = (0.02/0.1)*0.08206*300 = 4.92 atm.

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