5: Colligative Properties

For a non-electrolyte, the boiling point elevation equation is ΔT_b = K_bm. A one-molal solution has m = 1 mol kg⁻¹, and its van't Hoff factor is i = 1 because the solute does not dissociate. Substitution gives ΔT_b = K_b × 1 = K_b. K_f applies to freezing point depression, while R and π represent other quantities.

The freezing point depression equation for a non-electrolyte is ΔT_f = K_fm. Since the solution is one molal, m = 1 mol kg⁻¹, and since the solute is a non-electrolyte, i = 1. Therefore ΔT_f = K_f × 1 = K_f. K_b belongs to boiling point elevation, whereas R and T are the gas constant and temperature, not the cryoscopic constant.

The symbol ΔT_b denotes the elevation in boiling point, where ΔT means a temperature difference and the subscript b refers to boiling. It is commonly calculated by ΔT_b = K_bm or iK_bm. ΔT_f is used for freezing point depression, π for osmotic pressure, and x generally denotes a mole fraction, so they do not represent boiling point elevation.

The symbol ΔT_f represents depression in freezing point. Here ΔT denotes the magnitude of the temperature change and the subscript f refers to freezing. For a dilute solution, it is calculated as ΔT_f = K_fm, or iK_fm when the effective particle number is corrected. ΔT_b refers to boiling point elevation, while π and R denote different physical quantities.

In the dilute-solution equation π = CRT, C represents the molar concentration of the solute, usually in moles per litre of solution. R is the gas constant and T is absolute temperature. The relation can also be written π = iCRT for electrolytes. C is therefore not molality, which is moles per kilogram of solvent, and it has no relation to colour or solvent mass alone.

The equation π = CRT is analogous to the ideal-gas equation and uses absolute temperature. Therefore T must be expressed in kelvin, not directly in Celsius or Fahrenheit. To convert Celsius to kelvin, add 273.15. Using Celsius directly would change the proportional relationship and produce an incorrect osmotic pressure.

For elevation of boiling point, the relation is \(\Delta T_b=iK_bm\). A non-electrolyte neither ionises nor dissociates appreciably, so its van’t Hoff factor is \(i=1\). For the same solvent, \(K_b\) is constant. Thus, when molality \(m\) changes to \(2m\), the elevation changes from \(K_bm\) to \(2K_bm\), so it becomes double. It does not become zero or remain unchanged.

For a dilute solution, elevation in boiling point is calculated from \(\Delta T_b=iK_bm\). Since the solute is a non-electrolyte, \(i=1\), so the equation becomes \(\Delta T_b=K_bm\). Substituting the given values gives \(0.52\times2=1.04\). Therefore the correct elevation is 1.04, while 0.26 results from division and the other values do not follow the formula.

The depression in freezing point is given by \(\Delta T_f=iK_fm\). In this question no ionisation is specified, so the usual non-electrolyte value \(i=1\) is used. Hence \(\Delta T_f=1.86\times0.25\). Because 0.25 is one-fourth, this is \(1.86/4=0.465\). The other options arise from leaving out the fraction or using an incorrect operation.

Freezing-point depression means that the solution freezes at a lower temperature than the pure solvent. The solution freezing point is calculated as \(T_f(\text{solution})=T_f^0-\Delta T_f\). Therefore, \(0-1.5=-1.5^{\circ}C\). The positive value 1.5 would represent an increase rather than a depression; the other numerical values come from an incorrect subtraction.

Elevation in boiling point means that adding a non-volatile solute raises the boiling temperature of the solvent. The relation is \(T_b(\text{solution})=T_b^0+\Delta T_b\). Substitution gives \(78+0.8=78.8^{\circ}C\). Thus option C is correct. Subtracting 0.8 would describe a lowering, not an elevation, and 79.8 results from an arithmetic error.

For an electrolyte, the effective number of solute particles is represented by the van’t Hoff factor, so the correct equation is \(\Delta T_f=iK_fm\). Substituting the values gives \(3\times1.8\times0.5\). First, \(1.8\times0.5=0.9\), and then \(3\times0.9=2.7\). Omitting \(i\) gives 0.9, so option C is the correct electrolyte result.

For a solution whose solute produces more effective particles, boiling-point elevation is calculated using \(\Delta T_b=iK_bm\). Substituting the data gives \(2\times0.5\times1.5\). Since \(2\times0.5=1\), the result is \(1\times1.5=1.5^{\circ}C\). The value 0.75 is obtained by omitting \(i\), while 3.0 doubles the correct result.

Freezing-point depression follows \(\Delta T_f=iK_fm\). At the same molality and in the same solvent, \(K_f\) and \(m\) are fixed, so the depression is greatest for the largest effective particle factor \(i\). A non-electrolyte has approximately \(i=1\), a two-ion solute about \(i=2\), and a three-ion solute about \(i=3\) when dissociation is complete. Association lowers the effective particle count.

The equation π = CRT is the dilute-solution form of the osmotic-pressure relation, analogous to the ideal-gas equation. It assumes ideal behaviour and uses molar concentration C, absolute temperature T and the gas constant R. Very concentrated solutions can show non-ideal interactions, while a pure solvent has no solute concentration and therefore no ordinary solute osmotic pressure.

For a dilute non-electrolyte solution, osmotic pressure is given by the van’t Hoff equation \(\pi=CRT\). Insert the stated concentration, gas constant, and absolute temperature: \(\pi=0.1\times0.082\times300\). Since \(0.1\times300=30\), the product is \(30\times0.082=2.46\). Thus option A is correct. The temperature is already supplied as 300 K, so no conversion is needed.

For dilute solutions of the same type, osmotic pressure is related to molar concentration by \(\pi=CRT\). The gas constant \(R\) is universal, and the temperature is the same for both solutions. Therefore \(\pi_1/\pi_2=C_1/C_2\). Given \(\pi_1:\pi_2=2:3\), the concentration ratio must also be \(C_1:C_2=2:3\). The inverse ratio would incorrectly suggest pressure decreases as concentration increases.

Two solutions are isotonic when they have equal osmotic pressure at the same temperature, or equivalently equal effective solute-particle concentration under the relevant conditions. Their colours, solute molar masses and boiling points need not be identical. Equal osmotic pressure is the defining property, so option B is the only unambiguous statement.

A hypertonic external solution has a higher effective solute concentration and osmotic pressure than the cell interior. Across a selectively permeable membrane, water moves from the side with lower effective solute concentration toward the side with higher effective solute concentration. Thus water leaves the cell, which may shrink. Individual molecules can move both ways, but the net flow is outward.

For dilute ideal solutions, π = CRT. At the same temperature, R and T are constant, so osmotic pressure is directly proportional to molar concentration, provided the solute particle factor is comparable. Hence the solution with higher π has higher effective molar concentration. Colour and the absence of solute do not provide this conclusion.

For a non-electrolyte, ΔTf = Kf m because i = 1. Substituting the given values gives ΔTf = 2.0 × 0.4 = 0.8. This is the magnitude of the depression; the actual solution freezing point would be the pure-solvent freezing point minus 0.8°C or K. Values such as 2.4 come from addition rather than the required multiplication.

The magnitude of freezing-point depression is the pure-solvent freezing point minus the solution freezing point: \(\Delta T_f=T_f^0-T_f\). Substituting the values gives \(0-(-0.6)=0.6^{\circ}C\). The negative sign in the solution temperature indicates that the solution freezes below zero; it is not retained as the magnitude of the depression. Hence 0.6°C, option C, is correct.

Boiling-point elevation is the increase from the pure-solvent boiling point to the solution boiling point. Therefore, \(\Delta T_b=T_b-T_b^0\). Using the data, \(100.4-100.0=0.4^{\circ}C\). The elevation is reported as a positive difference. The negative value reverses the subtraction, while 100.4 and 200.4 are not temperature differences between the two stated boiling points.

For a non-electrolyte, \(i=1\), so the boiling-point relation is \(\Delta T_b=K_bm\). Rearrange it to isolate molality: \(m=\Delta T_b/K_b\). Substitution gives \(m=1.04/0.52=2.0\). Thus option C is correct. Multiplying instead of dividing gives an incorrect result, and 0.5 is not consistent with the given elevation and constant.

Assuming a non-electrolyte, the freezing-point relation is \(\Delta T_f=K_fm\). Solving for molality gives \(m=\Delta T_f/K_f\). Hence \(m=0.93/1.86\). Since 1.86 multiplied by 0.5 equals 0.93, the quotient is 0.5 mol kg⁻¹. Therefore option B is correct; 0.25 would give only half the stated depression, and 1.0 would give 1.86°C.