Class IX
Depression In Freezing Point (Cryoscopy)
Teaching Task
Q1) Ans: A
Solution: Mass of 8 L water = 8.2 kg or 8200 g because
the density is 1.
Mass of density alcohol added = Density × volume
= 0.8 × 2000
= 1600 g.
ΔTf = Kf m
= 1.86 × 1600/32 × 1/8200 × 1000
= 11.34
The lowest temperature = -11.34°C
Q2) Ans: A
Solution: K3[Fe(CN)6] → 3K+ + Fe(CN)6^3− Before dissociation: 1 0 0 After dissociation: 0 3 1 Total no. of particle = 4. van’t Hoff factor, i = 4.
ΔTf = (1000 × Kf × w)/(m × w) × i
= (1000 × 1.86 × 0.1 × 4)/(3.9 × 100)
= 2.3 × 10^-2 °C.
ΔTf' = 0 - 2.3 × 10^-2 → Tf' = -2.3 × 10^-2 °C
Q3) Ans: D.
Solution: The molal freezing point depression constant is the depression in freezing point for 1 molal solution.
ΔTf = Kf × m
when m = 1, ΔTf = Kf.
When ΔTf is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality m, then quantity
lim (ΔTf/m) = Molal depression constant = Kf
m→0
Q4) Ans: A
Solution: ΔTf = Kf·m.
Freezing point of sucrose is 271 K.
ΔTf = 273.15 K - 271 K = 2.15 K.
5% solution means 5 grams of solute in 100 gram of solution. So 5 grams solute = 95 grams water.
Moles of sucrose = 5/342 = 0.0146 mol.
Molality of sucrose solution = 0.0146 mol / 0.095 kg = 0.1537 mol/kg
Moles of glucose = 5/180 = 0.0278 mol.
Molality of glucose solution = 0.0278/0.095 kg = 0.2926 mol/kg
ΔTf(glucose) / ΔTf(sucrose) = 0.2926 / 0.1537 = 1.903 → ΔTf(glucose) = 1.903 × 2.15 K = 4.09 K.
Freezing point = 273.15 K - 4.09 K = 269.07 K.
Q5) Ans: B.
Solution: Given component 1 = Benzene.
Component 2 = Solute.
w1 = 50 g, w2 = 1.00 g, ΔTf = 0.40 K, Kf = 5.12 K kg mol^-1
Molar mass of solute M2 = ?
ΔTf = Kf · m
m = molality → m = no. of moles of solute / mass of solvent in kg.
no. of moles of solute = actual mass / molar mass = w2/M2
m = (w2/M2) × 1/w1
w1 = 50 g → w1 = 0.05 kg.
ΔTf = Kf · (w2/M2) × 1/w1
M2 = Kf · w2 / (ΔTf · w1)
= 5.12 × 1 / (0.40 × 0.05)
= 256 g/mol.
Q6) Ans: D.
Solution: ΔTf = i · Kf · m/M.
For substance A, ΔTfA = 0.1°C.
mass of solute = 400 g.
mass of solvent = 100 g.
MA = molar mass of A.
For substance B, ΔTfB = 0.2°C.
mass of solute = 4 g.
mass of solvent = 100 g.
MB = molar mass of B.
Kf & solvent mass are constant, so
ΔTfA/ΔTfB = (mA/MA)/(mB/MB)
0.1/0.2 = (400/MA)/(4/MB)
1/2 = 100/MA × MB/4
MA = 200 MB
Q7) Ans: C
Solution: Given, ΔTf = 0.372°C, Kf = 1.85 kg·K mol^-1.
Molar mass of urea = 60 g.
Given mass of water: 1 kg = 1000 - 200 = 800 gm (ice separated)
ΔTf = Kf · w1/M1 × 1000/w2
0.372 = 1.85 × w1/60 × 1000/800
w1 = 0.372 × 60 × 0.8 / 1.85 = 9.7
Q8) Ans: A.
Solution: Given x + y = 0.1.
KCl → x moles.
BaCl2 → y moles.
KCl → K+ + Cl−, i = 2.
BaCl2 → Ba2+ + 2Cl−, i = 3.
Solvent = 1 kg.
ΔTf = iKf m
Kf = 1.85 K/molal.
ΔTf = (2x + 3y)1.85
x + y = 0.1. If x = 0.01, y = 0.09
ΔTf = (2×0.01 + 3×0.09) × 1.85
= 0.5365
or x = 0.09, y = 0.01
ΔTf = (2×0.09 + 3×0.01) × 1.85
= 0.3885
ΔTf varies btw 0.37 to 0.55
Q9) Ans: A.
Solution: ΔTf = -0.30°C, v.p. of pure water (P°) = 23.5 mmHg.
Temperature = 298 K, Kf = 1.86.
V.P. of solution (P) = ?
ΔTf = Kf·m → m = 0.3/1.86 = 0.1613.
n2/n1 = 0.1613 × 18 / 1000 = 2.9 × 10^-3 → n1/(n1+n2) = x1 = 1/(1+2.9×10^-3) = 0.997.
P = P° × x1 = 23.5 × 0.997 = 23.44 mm.
Q10) Ans: A.
Solution: We have ΔTf = Kf × m.
Given ΔTf = 3.554.
Kf = 1.86.
m = ΔTf/Kf = 3.554/1.86 = 1.91.
Molarity = mass of solute / (molar mass of solute × mass of solvent)
1.911 = x / (342 × 1)
x = 342 × 1.911 = 653.562
Mass of solute = 653.562.
Mass of ice separated = 1000 - 653.84 = 353 gms.
Q11) Ans: A, D.
Solution:
→ The vapour pressure of the solution is less than that of pure solvent.
→ Only solvent molecules solidify at the freezing point.
Q12) Ans: B, C, D.
Solution: Kf = ΔTf/m. molality = 0.849/M / 0.050
1.24 × 34.3 [0.849/0.050]
M = 469.68.
The mercurous chloride is present in the form of Hg2Cl2.
Q13) Ans: A.
Solution:
→ Depression of freezing point is a colligative property.
→ ΔTf = Kf·m.
ΔTf ∝ m.
Q14) Ans: D.
Solution: mole of naphthalene (m) = 0.1.
Mole of benzene (n1) = 0.9.
Vapour pressure of benzene (P°) = 760 mm.
B.P of benzene = 353 K.
V.P of solution = 670 mm.
Freezing point of benzene = 278.5 K.
ΔHfus = 10.67 KJ mol^-1.
(P° - Ps)/Ps = (w2 × Mw1)/(Mw2) × w1.
(760 - 670)/670 = 0.1 × 78 / w1
w1 = 58.06 g.
Weight of benzene = 0.9 × 78 = 70.2 g.
Amount of benzene frozen out = 70.2 - 58.06 = 12.14 g.
Q15) Ans: C
Solution: ΔP/P° = (760 - 670)/670 = 0.13
Learner’s Task
Q1) Ans: A
Solution: Addition of glycol lowers the freezing point of water in the radiator, so
Task
Q1) Ans: A
Solution: Addition of glycol lowers the freezing point of water in the radiator, so that the cold winter temperatures wouldn’t burst the pipes and thus glycol-water mixture is used as antifreeze in radiators of cars.
Q2) Ans: C
Solution: The freezing point depression is directly proportional to the number of particles.
→ Glucose & urea do not dissociate, i → 1
→ NaCl (Na+ and Cl−) → i → 2
→ ZnSO4 (Zn2+ and SO4^2−) → i → 2.
NaCl, ZnSO4 has 2 ions, but ZnSO4 has more charge increasing the freezing point depression.
Q3) Ans: B
Solution: Greater the number of particles, lower is the freezing point.
Lower number of particles higher is the freezing point.
→ NaCl → i = 2
→ Sugar → i = 1
→ BaCl2 → i = 3
→ FeCl3 → i = 4
1. Sugar has fewer particles, so it has higher freezing point.
Q4) Ans: a.
Solution: An aqueous solution of any substance if it is non-volatile freezes below 0°C because the vapor pressure of the solution becomes lower than that of pure solvent.
Q5) Ans: b.
Solution: Kf = 1.86, m = 0.1.
ΔTf = Kf m
= 1.86 × 0.1 = 0.186.
Q6) Ans: a.
Solution: HgI reacts with KI to form K2[HgI4].
Due to this, the number of ions decreases thus decreasing the van't Hoff factor.
ΔTf = iKfms
Since i has decreased, it results in less depression in the freezing point causing the freezing point to rise.
Q7) Ans: B
Solution: ΔTf ∝ m, ΔTf/Kf → ΔTf = Kf m.
Kf is constant.
ΔTf depends on molality. Molal depression constant does not depend on the nature of the solute. It is dependent on the nature of solvent.
Q8) Ans: D.
Solution: m = Kf × 1000 × w / w × ΔTf
= 5.1 × 1000 × 4.8 / 60 × 1.02
= 400
Q9) Ans: A
Solution: During the depression of freezing point in a solution, liquid solvent & solid solvent are in equilibrium. During freezing of a solution only the solvent freezes out and the equilibrium exists between solid and liquid form of solvent.
Q10) Ans: B.
Solution: Freezing point is inversely proportional to mass percentage of the non-volatile electrolyte (Glucose).
1% > 2% > 3% > 10%.
JEE Main level Questions
Q1) Ans: D
Solution: ΔTf = Kf × w1/M1 × 1000/w2
Given, w1 = 2.56 g, w2 = 100 g, ΔTf = 0.68°C.
0.68 = 6.8 × 2.56/M × 1000/100
0.68M = 68 × 2.56
M = 100/68 × 2.56 / 0.68 = 256.
Molecular mass of sulphur = 256.
Formula of sulphur: S8
Q2) Ans: C
Solution: Experiment-1 → ΔTf = 1·Kp·mol glucose.
Experiment-2 → ΔTf = (1 mol glucose + 2 mol NaCl)Kf
Molecular weight of glucose in 2nd case will be lower.
Q3) Ans: d.
Solution: When a non-volatile, non-electrolyte is dissolved in a pure solvent the vapor pressure of the solvent is lowered, so addition of non-volatile solute decreases the freezing point of solution and addition of salt lowers the freezing point of water and thus snow melts.
Q4) Ans: C
Solution: ΔTf = iKfms
ΔTf/urea : ΔTf/glucoze : ΔTf/NaCl = iurea : iglucoze : iNaCl
iurea & glucose = 1, iNaCl = 2
ΔTf urea : ΔTf g : ΔTf NaCl = 1 : 1 : 2
Q5) Ans: A
Solution: MAB1 = 1000 × wB × Kf / ΔTf × w1
= 1000 × 1 × 5.1 / 2.3 × 20 = 110.87 g/mol.
MABY = 1000 × 1 × 5.1 / 1.3 × 20 = 196.15 g/mol.
Molar mass of AB2 & ABY are 110.87, 196.15 respectively.
AB2 = 110.87 → x + 2y = 110.87 → (1)
ABY = 196.15 → x + 4y = 196.15 → (2)
Subtract (2) - (1)
x + 4y = 196.15
x + 2y = 110.87
2y = 85.28
y = 85.28/2 = 42.64.
Substitute y in (1)
x + 2(42.64) = 110.87
x = 110.87 - 2×42.64
= 25.59.
Q6) Ans: D.
Solution: ΔTf = Kf × w1/M1 × 1000/w2
Given ΔTf/Kf = 1/1000, w1 = ?, w2 = 1 litre = 1000 ml.
Molar mass of water = 18 → for 1000 gm → 180 gm.
ΔTf/Kf = w1/M1 × 1000/w2
1/1000 = w1/180 × 1000/1000
w1 = 180/1000 = 0.18 g.
Q7) Ans: A
Solution: Ideal solution containing non-volatile solute
ΔTf = Kf m
where Kf = M1 RTf^2 / ΔvapHm
Q8) Ans: C
Solution: ΔTb = Kb·m,
ΔTf = Kf m
ΔTb + ΔTf = m(Kb + Kf) = Tb - Tf + T°f - T°b
= ws/M × 1000/w0 (Kb + Kf)Tb - Tf + T°f - T°b
= x/342 × 1000/500 [0.22] + 1.86[0.5]
5 = 2x/342 {2.38} ⇒ x = 342 × 5 / 2×2.38 = 359.
Q9) Ans: C
Solution: ΔTf = Kf × w1/M1 × 1000/w2
Given Kf = 1.86, w1 = 100, M = 342
at ΔTf = 0.5
0.5 = 1.86 × 100/342 × 1000/w2
w2 = 1.86 × 100 × 1000 / 0.5 × 342
= 1087.71.
at ΔTf = 0.38
0.38 = 1.86 × 2 × 1000 / 342 × 1087.7
x = 77.49.
Given solution = 100 gm.
The ice frozen out = 100 - 77.49 = 22.5 ≈ 23.
/ 342 × 1087.7
x = 77.49.
Given solution = 100 gm.
The ice froze out = 100 - 77.49 = 22.5 ≈ 23.
Q10) Ans: A
Solution: Effective molality = Given molality × No. of particles per molecule in solution
→ Higher the effective molality, more is depression in freezing point.
0.05 M KNO3 > 0.04 M CaCl₂ > 0.14 M Sugar > 0.075 M?
0.1 M < 0.12 M < 0.14 M < 0.15 M
Effective molality order
Advanced level Questions
Q11) Ans: A, C, D.
Solution: For Hg(NO₃)₂
ΔTf = iKf m.
0.0558 = i(1.86)(3.24/329.4)
i = 3.
i.e., Hg(NO₃)₂ → Hg²⁺ + 2NO₃⁻
i = 3.
So, Hg(NO₃)₂ is completely ionized (100%).
For HgCl₂
ΔTf = iKf m.
0.0744 = i(1.86)(2.68/271)/2
i = 0.0744 × 2 × 2 × 1 / (1.86 × 2.68) = 1
HgCl₂ → Hg²⁺ + 2Cl⁻ → i = 3.
i.e., HgCl₂ is un-ionized.
Q12) Ans: A, B, C, D.
Solution: According to Raoult’s law the addition of non volatile solid solute always lowers the vapor pressure of pure solvent.
→ The solution will be in equilibrium with the solid phase at lower pressure, hence at lower temperature
→ The freezing point of a solvent in the solution containing non-volatile solid solute is always lower than F.P.
Q13) Ans: E
Solution: pαT.
On increasing the pressure, the freezing temperature decreases. It is true that the freezing point of water decreases as the pressure decreases. This is because water expands when it freezes, so compressing ice will tend to convert it to more compact liquid water. Thus, increasing the pressure lowers the freezing point.
The density of water is maximum at 4°C.
Q14) Ans: D.
Solution: The molality of benzene,
x_B / x_A = n_B / (n_A + n_B) / n_A / (n_A + n_B)
→ x_B / x_A = n_B / n_A
n_B / n_A = n_B / w_A × M_A × 1000 / 1000 → m = n_B × 1000 / w_A.
x_A / x_B = m × M_A / 1000.
Molar mass of C₆H₆ = (6×12) + 6 = 78.
Given x_B = 0.2, x_A = 0.8 → 0.2/0.8 = m × 78 / 1000
Molality, m = 3.2.
ΔTf = Kf × m
= 5.5 × 3.2
= 17.6 ≈ 17
Q15) Ans: D.
Solution: The freezing point of a liquid is that temperature at which the liquid and its solid state exist in equilibrium with each other. It may be defined as the temperature at which the liquid & solid state of a substance have the same vapour pressure.
When a non-volatile, non-electrolyte is dissolved in a pure solvent, the vapor pressure of the solvent is lowered, so addition of non-volatile solutes decreases the freezing point of solution.
Q16) Ans: B.
Solution: Kf = RTf²M / 1000 ΔHfus.
For larger value of Tf and smaller value of enthalpy of the solid, the value of Kf would be larger.
Hence, a liquid having high freezing point and small enthalpy of freezing will be most suitable for determining the molecular mass of a compound by cryoscopic measurements.
Q17) Ans: 2.
Solution: Depression in freezing point ΔTf = i × Kf × m.
i = ΔTf/Kf·m.
Given ΔTf = 0 - (0.00732) = 0.00732
m = 0.0020 M.
Kf = 1.86°C/m.
i = 0.00732/(1.86×0.0020) = 1.86 ≈ 2
So, the compound [Co(NH₃)₅(NO₂)]Cl in solution will dissociate into Cl⁻ and [Co(NH₃)₅(NO₂)]⁺.
n = 2.
∴ 2 moles of ions are formed.
Q18) Ans: 6.
Solution: Given: ΔTf = 0.5, Kf = 1.86, M.w urea = 60.
Depression will be caused by water which exists as liquid at -0.5°C.
W = 500 - 128 = 372 g.
ΔTf = Kf × molality = 1000 × Kf × w / m × M.
0.5 = 1000 × 1.86 × w / 60 × 372
w = 0.5 × 60 × 372 / 1000 × 1.86
w = 6 g
Q19) Ans: 3.
Solution: Depression in freezing point ΔTf = i × Kf × m.
For NaCl, i = 2 {NaCl → Na⁺ + Cl⁻}.
Kf for water = 1.86°C/m.
ΔTf = 2°C.
ΔTf = i × Kf × m
2 = 2 × 1.86 × m
m = 0.538 moles/kg.
Equimolar means,
Molality of NaCl = Molality of MgCl₂.
For MgCl₂, i = 3 {MgCl₂ → Mg²⁺ + 2Cl⁻}.
ΔTf = i × Kf × m
= 3 × 1.86 × 0.538 = 3.002
So, freezing point of MgCl₂ solution will be -3.002°C
Q20) Ans: A) P-2, Q-4, R-2, S-3.
Solution:
P) Molal cryoscopic constant → 2) 1.86° kg/mol. constant for water
Q) The factor ΔTf/kf → 4) ΔTf ∝ m → ΔTf = Kf m. molality, m = ΔTf/Kf.
R) Kf for water is → 2) 1.86° kg/mol.
S) Depression in freezing point → 3) Kf ∝ osmotic pressure.