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Which of the following aqueous solution has minimum freezing point ? [1991]
  • a)
    0.01 m NaCl
  • b)
    0.005 m C2H5OH
  • c)
    0.005 m MgI2
  • d)
    0.005 m MgSO4.
Correct answer is option 'A'. Can you explain this answer?

Swara Desai answered
ΔTf = i × Kf × m Van't Hoff factor, i = 2 for NaCl, hence ΔTf = 0.02 Kf which is maximum in the present case.
Hence ΔTf is maximum or freezing point is minimum.

Vapour pressure of benzene at 30°C is 121.8 mm.When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute (Mo. wt. of solvent = 78) [1995]
  • a)
    356.2
  • b)
    456.8
  • c)
    530.1
  • d)
    656.7
Correct answer is option 'A'. Can you explain this answer?

Ruchi Chakraborty answered
Give vapour pressure of  pur e solute (P0) = 121.8 mm;
Weight of solute (w) = 15 g
Weight of solvent (W) = 250 g;
Vapour pressure of pure solvent (P) = 120.2 mm and Molecular weight of solvent (M) = 78 From Raoult’s law

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO3 ? The concentrated acid is 70% HNO3             [NEET 2013]
  • a)
    90.0 g conc. HNO3
  • b)
    70.0 g conc. HNO3
  • c)
    54.0 g conc. HNO3
  • d)
    45.0 g conc. HNO3
Correct answer is option 'D'. Can you explain this answer?

Mamali . answered
The required number of Moles =(250/1000)2=0•5 moles
(no of Moles = molarity×vol.)
so reqd.mass of HNO3 =0•5×63
=31•5

given mass = no of moles × molar mass

Given,
70gms of HNO3 are present in 100 gms of the Sol.

so,1gm will be present in 100/70 gms of sol.

hense, 31•5 gms will be present in

100/70 × 31•5 gms of sol.

so amount of concentrated nitric acid solution used is 45 gms.

A solution of acetone in ethanol [2006]
  • a)
    shows a positive deviation from Raoult’s law
  • b)
    behaves like a non ideal solution
  • c)
    obeys Raoult’s law
  • d)
    shows a negative deviation from Raoult’s law
Correct answer is option 'A'. Can you explain this answer?

Shanaya Rane answered
A solution of acetone in ethanol shows positive deviation from Raoult's law. It is because ethanol molecules are strongly hydrogen bonded. When acetone is added, these molecules break the hydrogen bonds and ethanol becomes more volatile. Therefore its vapour pressure is increased.

The mole fraction of the solute in one molal aqueous solution is: [2 00 5]
  • a)
    0.009
  • b)
    0.018
  • c)
    0.027
  • d)
    0.036
Correct answer is option 'B'. Can you explain this answer?

Aashna Mukherjee answered
One molal solution means one mole solute present in 1 kg (1000 g) solvent i.e., mole of solute = 1
Mole of solvent 
Mole fraction of solute  =

Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25ºC are 200 mm Hg and 415 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be : (Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85 u).             [2012 M]
  • a)
    173.9 mm Hg
  • b)
    615.0 mm Hg
  • c)
    347.9 mm Hg
  • d)
    285.5 mm Hg
Correct answer is option 'C'. Can you explain this answer?

Naina Datta answered
The correct answer is option C
Molar mass of CH2​Cl2​ =12×1+1×2+35.5×2=85 g mol–1
Molar mass of CHCl3 ​  =12×1+1×1+35.5×3=119.5 g mol−1
Moles of CH2​Cl2​= 40 g /85 g mol−1 = 0.47 mol
Moles of CHCl3​ = 25.5 g /119.5 g mol−1 = 0.213 mol
Total number of moles = 0.47+0.213=0.683 mol
Mole fraction of component 2
 =0.47mol/0.683mol= 0.688
Mole fraction of component 1
=1.00–0.688=0.312 
We know that:
PT​=p10​+(p20​−p10​)x2​
=200+(415–200)×0.688
=200+147.9
=347.9 mm Hg

A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]
  • a)
    0.654°C
  • b)
    -0.654°C
  • c)
    6.54°C
  • d)
    -6.54°C
Correct answer is option 'B'. Can you explain this answer?

Shruti Chauhan answered
As ΔTf = Kf. m
ΔTb = Kb. m
 Hence, we have  
= [ΔTb = 100.18 - 100 = 0.18°C]
As the Freezing Point of pure water is 0°C,
ΔTf = 0 –Tf
0.654 = 0 – Tf
∴  Tf = – 0.654 thus the freezing point of solution will be – 0.654°C.

At 25°C, the highest osmotic pressure is exhibited by 0.1 M solution of [1994]
  • a)
    CaCl2
  • b)
    KCl
  • c)
    Glucose
  • d)
    Urea.
Correct answer is option 'A'. Can you explain this answer?

Smruti Sucharita answered
Highest osmotic pressure depends on only i value... Bcoz temp and concentration r same....

For cacl2 i=3
Kcl=2
Glucose =1
Urea =1

So as i is more for cacl2.. It has more osmotic pressure

(i is the vant Hoff s factor..)

The relative lowering of the vapour pressure is equal to the ratio between the number of [1991]
  • a)
    solute molecules to the solvent molecules
  • b)
    solute molecules to the total molecules in the solution
  • c)
    solvent molecules to the total molecules in the solution
  • d)
    solvent molecules to the total number of ions of the solute.
Correct answer is option 'B'. Can you explain this answer?

Sankar Banerjee answered
Relative lowering of vapor pressure is a colligative property, which means it depends on the number of particles in a solution rather than their nature. It is defined as the ratio of the decrease in vapor pressure of a solvent to the vapor pressure of the pure solvent. This property is directly related to the concentration of solute particles in a solution.

Explanation:

Colligative Properties:
- Colligative properties are properties of a solution that depend on the number of solute particles present, rather than the specific type of solute.
- These properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Vapor Pressure:
- Vapor pressure is the pressure exerted by the vapor molecules above the surface of a liquid in a closed container at equilibrium.
- It is a measure of the tendency of molecules to escape from the liquid phase and enter the vapor phase.

Relative Lowering of Vapor Pressure:
- The relative lowering of vapor pressure is defined as the ratio of the decrease in vapor pressure of the solvent in the presence of a non-volatile solute to the vapor pressure of the pure solvent.
- It can be calculated using the formula: Relative Lowering of Vapor Pressure = (P° - P) / P°, where P° is the vapor pressure of the pure solvent and P is the vapor pressure of the solvent in the presence of the solute.

Number of Particles:
- According to Raoult's Law, the vapor pressure of a solvent is directly proportional to the mole fraction of the solvent in the solution.
- When a non-volatile solute is added to the solvent, it lowers the vapor pressure of the solvent.
- The extent of vapor pressure lowering depends on the number of solute particles present in the solution.

Explanation of the Correct Answer:
- The relative lowering of vapor pressure is equal to the ratio between the number of solute molecules to the total molecules in the solution.
- This is because the vapor pressure of the solvent is affected by the presence of solute particles, and the relative lowering of vapor pressure is a measure of this effect.
- The more solute particles there are in the solution, the greater the decrease in vapor pressure of the solvent.
- Therefore, option B is the correct answer as it correctly represents the relationship between the number of solute molecules and the relative lowering of vapor pressure.

In conclusion, the relative lowering of vapor pressure is directly related to the ratio between the number of solute molecules to the total molecules in the solution. This colligative property depends on the concentration of solute particles and is independent of the nature of the solute.

A solution containing components A and B follows Raoult's law when [2002]
  • a)
    A – B attraction force is greater than A – A and B – B
  • b)
    A – B attraction force is less than A – A and B – B
  • c)
    A – B attraction force remains same as A–A and  B –B
  • d)
    Volume of solution is different from sum of volume of solute and solvent
Correct answer is option 'C'. Can you explain this answer?

Raghav Khanna answered
Raoult's law is valid for ideal solution only. The element of non - ideality enters into the picture when the molecules of the solute and solvent affect each others intermolecular forces. A solution containing components of A and B behaves as ideal solution when A - B attraction force remains same as A - A and B - B.

Which condition is not satisfied by an ideal solution? [NEET Kar. 2013]  
  • a)
    Δmix H = 0
  • b)
    Δmix V = 0
  • c)
    Δmix S = 0
  • d)
    Obeyance to Raoult’s Law
Correct answer is option 'C'. Can you explain this answer?

Deepak Joshi answered
An ideal solution is that solution in which each component obeys Raoult’s law under all conditions of temperatures and concentrations. For an ideal solution.

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]
  • a)
    – 0.18°C
  • b)
    – 0.54°C
  • c)
    – 0.36°C
  • d)
    – 0.24°C
Correct answer is option 'D'. Can you explain this answer?

Sahil Menon answered
To find the percent ionization of a weak acid, we need to use the equation:

% ionization = (concentration of ionized acid)/(initial concentration of weak acid) * 100

Given that the solution is 0.1 molal, we can assume that the initial concentration of the weak acid is also 0.1 molal.

Let x represent the concentration of ionized acid. Therefore, the concentration of the weak acid that remains unionized is (0.1 - x) molal.

The equation for the percent ionization can be written as:

30 = (x)/(0.1) * 100

Simplifying the equation, we have:

0.3 = x/0.1

x = 0.03 molal

Therefore, the concentration of the ionized acid is 0.03 molal.

Now, let's calculate the molality of water in the solution using the equation:

Kf = molality of water * boiling point elevation constant

Given that Kf for water is 1.86, we can rearrange the equation to solve for the molality of water:

molality of water = Kf / boiling point elevation constant

molality of water = 1.86 / 0.512 (boiling point elevation constant for water)

molality of water = 3.63 molal

Therefore, the molality of water in the solution is 3.63 molal.

Which of the following 0.10 m aqueous solutions will have the lowest freezing point ? [1997]
  • a)
    Al2(SO4)3
  • b)
    C6H12O6
  • c)
    KCl
  • d)
    C12H22O11
Correct answer is option 'A'. Can you explain this answer?

Gowri Nair answered
Depression in F.P. ∝ No. of particles.
Al2(SO4)3 provides five ions on ionisation
while KCl provides two ions
KCl     K    Cl
C6H12O6 and C12H22O11 are not ionised so they have single particle.
Hence, Al2(SO4)3 have maximum value of depression in F.P or lowest F.P

Formation of a solution from two components can be considered as [2003]
(i) Pure solvent → separated solvent                              molecules, ΔH1
(ii) Pure solute   → separated solute                               molecules, ΔH2
(iii) Separted solvent & solute molecules → Solution,  ΔH3
Solution so formed will be ideal if
(a) ΔHsoln = ΔH1 + ΔH2 - ΔH3
(b) ΔHsoln = ΔH1 + ΔH2 + ΔH3
(c) ΔHsoln = ΔH1 - ΔH2 - ΔH​​​​​​​3
(d) ΔHsoln = ΔH3 - ΔH1 - ΔH​​​​​​​2
  • a)
    a
  • b)
    b
  • c)
    c
  • d)
    d
Correct answer is option 'B'. Can you explain this answer?

Nisha Kulkarni answered
For an ideal solution, ΔHmixing = 0
  (Accroding to Hess's law) i.e., for ideal solutions there is no change in magnitude of the attractive forces in the two components present.

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