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51.

A 0.075 molar solution of monobasic acid has a freezing point of `-0.18^(@)C`. Calculate `K_(a)` for the acid , `(k_(f)=1.86)`

Answer» Correct Answer - `8.884xx10^(-3)`
52.

Calculate the amount of ice that will separate out on cooling containing `50 g `of ethylene glycol in `200 g` of water to `-9.3^(@)C (K_(f)` for water =`1.86 K mol^(-1) kg`)A. 38.71 gB. 38.71 mgC. 42 gD. 42 mg

Answer» Correct Answer - A
`HA hArr H^(+) + A^(-)`
`{:(t = 0," "C,0,0),(t_("eq"),C - C alpha,C alpha,C alpha):}`
`[H^(+)] = C alpha, [H^(+)] = 10^(-PH)`
`:. C alpha = 10^(-2)`
`0.1 alpha = 10^(-2)`
`alpha = 0.1`
`alpha = (I - 1)/(n - 1) , 0.1 = (I - 1)/(2 - 1)`
`i = 1.1`
`:. pi = iCRT`
`= 1.1 xx 0.1 xx RT = 0.11 RT`
53.

Calculate the amount of ice that will separate out on cooling containing `50 g `of ethylene glycol in `200 g` of water to `-9.3^(@)C (K_(f)` for water =`1.86 K mol^(-1) kg`)A. `38.71g`B. `38.71mg`C. `42g`D. `42mg`

Answer» `9.3=(50)/(62xx(200-w))xx1000xx1.86xx1`
`W=38.71g`.
54.

A `10 m` solution of urea is cooled to `-13.02^(@)C`. What amount of urea will separate out if the mass of solution taken is `100 g` ? `[K_(f)("water") =1.86 Kkg mol^(-1)]`

Answer» `DeltaT_(f)=0-(-13.02)=13.02 K`
`m=(DeltaT_(f))/(K_(f))`
`:.m=(13.02)/(1.86)=(7.0 mol kg^(-1))`
`:. m=10=(W_(2)xx1000)/(60xx(100-W_(2)))`
`:. W_(2)=37.5 g`
Mass of `H_(2)O` in this solution =`(100-37.5)=62.5 g`
The mass of water does not change as urea separates out to bring down to molality to `7 mol kg ^(-1)`.
`:. m=(W_(2)xx1000)/(Mw_(2)xxW_(1))`
`:. 7=(W_(2)xx1000)/(60xx62.5) rArr W_(2)=26.25 g`
Mass of urea separated =`(37.5-26.25)=11.25g`
55.

The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water that is cooled to `-10^(@)C`, will be [Given : `K_(f) "for" H_(2)O = 1.86K mol^(-1) kg`]

Answer» Correct Answer - 25 gram
56.

Calculate the molality of `K_(2)CO_(3)` solution which is formed by dissolving `2.5g` of it in one litre of solution. Density of solution is `0.85g ml^(-1)`. (At. wt. of `K=39,C=12,O=16)`

Answer» Given Mass of solute `(K_(2)CO_(3))=2.5g`
Volume of solution `=1000ml`
Density of solution `=0.85g ml^(-1)`
Asked Molality of solution `(m)=?`
Formulae: `m=(W_(B))/(M_(B))xx(1000)/(W_(A))`
Explanation: `m=` molality of solution, `M_(B)=` Molecular weight solute
`W_(A)=` Mass of solvent in grams
Calculation: Mass of solute `=2.5g`
Mass of solution`="volume"xx"density"=1000mlxx0.85g ml^(-1)=850g`
Mass of solvent `=(850-2.5)847.5g`.
Substitution in formulae:
`m=(2.5)/(138)xx(1000)/(847.5)`
`m=0.0213`
57.

Calculate the normal boiling point of a sample of sea water found to contain`3.5%` of `NaCl` and `0.13 %`of `MgCl_(2)` by mass. The normal boiling of point of water is `100^(@)C` and `K_(b)("water")= 0.51K kg mol^(-1)` . Assume that both the salts are completely ionised.A. `100.655^(@)C`B. `99.655^(@)C`C. `101.655^(@)C`D. `102.655^(@)C`

Answer» Correct Answer - A
Mass of `NaCl=3.5g`No. of moles of NACl=`(3.5)/(58.5)`
Numbers of ions furnished by one molecules of `NaCl` is `2`
So, the actual numbers of moles of particals functuion by sodium
chloride =`2xx(3.5)/(58.5)`
Similarly , actual number of mole sof particles furnished by magnesium chloride `=3xx(0.13)/(95)`
total number of moles of particles=`(2xx(3.5)/(58.5)+3xx(0.13)/(95))`
`=0.1238`
Mass of water =`(100-3.5-0.13)=96.37g=(96.3)/(1000)kg`
Molarity =`(0.1238)/(96.37)xx1000=1.2846`
`DeltaT_(b)` = Molarity `xxK_(b)=1.2846xx0.51=0.655K`
58.

What is the molality of a solution made by dissolving 3.42 g of table sugar (sucrose, `C_(12H_(22)O_(11))` in 50.0 mL of water? Strategy: Molality is the number of moles of solute dissolved per kilogram of solvent. Thus, we must find how many moles are present in 3.42 g of sucrose and how many kilograms are contained in 50.0 mL of water.

Answer» `n_("sucrose")=(mass)_("sucrose")/("molar mass")`
`=(3.42 g)/(342 g mol^(-1))=0.01 mol`
Since the density of water is `1.00 g mL^(-1)`, we can find the mass of water as follows:
`d_(H_(2)O)=m_(H_(2)O)/V_(H_(2)O)`
or `m_(H_(2)O)=(d_(H_(2)O))(V_(H_(2)O))`
`=(1.00 g mL^(-1))(50.0 mL)`
`=50.0 g`
Thus, the molality of the solution is
`Molality=n_("sucrose")/g_(H_(2)O)xx(1000 g)/(kg)`
`=(0.01 mol)/(50.0 g)xx(1000 g)/(kg)`
`=0.2 m`
59.

River water is found to contain 11.7% NaCl,9.5%` "MgCl"_(2),` and 8.4%. `"NaHCO"_(3)` , by weight of solution. Calculate its normal boiling point assuming 90% ionization of NaCl, 70% ionization of `"MgCl"_(2)` and 50% ionization of `"NaHCO"_(3)("K"_("b") " for water" =0.52)`

Answer» `"n"_("NaCl")=11.7/58.5=0.2, " n"_("MgCl"_(2))=9.5/9.5=0.1, " n"_("NaHCO"_(3))=8.4/8.4=0.1`
`i_(NaCl)=1+alpha=1+0.9=1.9`, i_(MgCl_(2))=1+2alpha=1+0.7xx2=2.4`,
Weight of solvent `=100-(11.7+9.5+8.4)=70.4g`
`Delta T_(b)=((i_(NaCl)xxn_(NaCl)+i_(MgCl_(2))xxn_(MgCl_(2))+i_(NaHCO_(3))xxn_(NaHCO_(3)))xxK_(b)xx1000)/("Weight of solvent")`
`=((1.9xx0.2+2.4xx0.1+2xx0.1)xx0.52xx1000)/70.4=5.94^(@)C`
therefore Boiling point of solution `=100+5.94=105.95^(@)C`
60.

1Kg of an aqueous solution of Sucrose is cooled and maintained at `-4^(@)"C".` How much ice will be seperated out if the molality of the solution is 0.75? `"K"_("F")("H"_(2)"O")=1.86"Kg mol"^(-1)"K"`

Answer» Since molality of solution is .75,
hence.75 moles of sucrose are present in 1000g of solvent(i.e. water)
`therefore"m"_("solution")="m"_("water")+"m"_("sucrose")=1000+.75xx342=1256.5"gm"`
`therefore"m"_("sucrose")("in kg")=256.5/1256.5xx1000=204.14"gm",."m"_("water")=1000-204.14=795.86"gm"`
`therefore Delta "T"_("F")="K"_("F")xx"m"implies4=1.86xx(204.14//342)/("W"//1000)implies"w"=277.15`
`therefore "amount of ice =795.86-277.15=518.31 gm"`
61.

The vapour pressure of two liquid P and Q are 80 torr and 60 torr respectively. The total vapour pressure obtained by mixing `3` moles of P and 2 mole of Q would beA. 68 torrB. 20 torrC. 140 torrD. 72 torr

Answer» Correct Answer - D
`P_(T)=P_(P)xxP_(Q)=P_(P)^(@)xxP+P_(Q)^(@)xxQ`
`P_(T)=80xx(3)/(5)+60xx(2)/(5)=48+24=72` torr
62.

During osmosis, flow of water through a semipermeable membrane is:A. from both sides of semipermeable membrane with unequal flow ratesB. from solution having lower concentrated onlyC. from solution having higher concentrated onlyD. from both sides of semipermeable membrance with equal flow rates

Answer» Correct Answer - A
Osmosis is a bilateral movement of solvent particles through semipermeable membrance and only net flow (more from dilute solution to concentrated solution)is noticed.
63.

In which of the following compound the solubility decreases with increase of temperature? (a) sodium chloride (b) ammonium nitrate (c) cerie sulphate (d) calcium chloride

Answer»

(c) ceric sulphate

64.

Which of the following is not an ideal solution? (a) Benzene & toluene (b) n – Hexane & n – Heptane (c) Ethyliodide & ethyl bromide (d) Ethanol and water 

Answer»

(d) Ethanol and water

65.

Freezing Point Depression: when `15.0 grams` of ethyl alcohol `(C_(2)H_(5)OH)` is dissolved in `750 grams` of formic acid, the freezing point of the solution is `7.20^(@)C`. The freezing point of pure formic acid is `8.40^(@)C` Evalulate `K_(f)` for formic acid. Strategy: First calculate the molality and the depression of the freezing point. Then solve the equation `DeltaT_(f)=K_(f)m` for `K_(f)` by substituting values for `m` and `DeltaT_(f)`.

Answer» Step 1. Calculate depression of the freezing point
`DeltaT_(f)` is the depression of freezing point. It is defined as
`DeltaT_(f)=T_(f("solvent"))-T_(f("solution"))`
So it is always positive.
Here
`DeltaT_(f)=T_(f) ("formic acid")-T_(f)("solution")`
`=8.40^(@)C-7.20^(@)C`
`=1.20^(@)C` (depression)
Step 2. Calculate moles of solute
`n_(solute)=(mass (solute))/(molar mass (solute))`
`=(15.0 g C_(2)H_(5)OH)/(46.0 g C_(2)H_(5)OH//mol C_(2)H_(5)OH)`
`=0.326 mol C_(2)H_(5)OH`
Step 3. Calculate molality of solute
`Molality (m)=n_(solute)/g_(solvent)xx(1000 g)/(kg)`
`=(0.326 mol)/(750 g)xx(1000 g)/(kg)`
`=0.435 mol Kg^(-1)`
Setp 4. Calculate the molal freezing point depression constant, `K_(f)` According to Equation (2.68), we have
`K_(f)=(DeltaT_(f))/m`
Thus
`K_(f)=(1.20^(@)C)/(0.435 m)`
`=2.76^(@)C//m`
66.

The freezing point depression of a `0.109M` aq. Solution of formic acid is `-0.21^(@)C`. Calculate the equilibrium constant for the reaction, `HCOOH (aq) hArr H^(+)(aq) +HCOO^(Theta)(aq)` `K_(f)` for water `= 1.86 kg mol^(-1)K`

Answer» Correct Answer - `1.44 xx 10^(-4)`
`DeltaT_(f) = (1 - alpha) k_(f) xx m`
`0.21 = (1+alpha) xx 1.86 xx 0.109`
`1 +alpha = 1.0358`
`k_(a) = (Calpha^(2))/(1-alpha) = (1.109(0.0358)^(2))/(1-0.0358) = 1.44 xx 10^(-4)`
67.

Which one of the following has found to have abnormal molar mass? hydrogen bonds and (a) NaCl (b) KCI (c) Acetic acid (d) all the above

Answer»

(d) All the above

68.

Which of the following shows negative deviation from Raoults law? (a) Phenol and aniline (b) Benzene and toluene (c) Acetone and ethanol (d) Bcnzene and acetone

Answer»

(a) Phenol and aniline

69.

Assertion (A) When solution is separted from the pure solved semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side. Reason (R ) Diffusion solvent occurs from a region of concentration solution to a region of low concentration soluton.A. Assertion and reason both are correct statements and reason is correcft explanation for assertion.B. Assertion and reason both are correct statements but reason is not correct explanation for assertion.C. Assertion is correct statement but reason is wrong statement.D. Assertion and reason both are incorrect statements.

Answer» Correct Answer - b
Assertion and reason both are correct statement but reason is not the correct explanation of assertion.
When a solution is separted from the pure solcent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side. Solvent molecules always flow from lower concentration to higher concentration of solution.
70.

In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.(i) Assertion and reason both are correct statements and reason is correct explanation for assertion.(ii) Assertion and reason both are correct statements but reason is not correct explanation for assertion.(iii) Assertion is correct statement but reason is wrong statement.(iv) Assertion and reason both are incorrect statements.(v) Assertion is wrong statement but reason is correct statement.Assertion : When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side.Reason : Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.

Answer»

(ii) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

71.

For carrying reverse osmosis for desalination of water the material used for making semipermeable membrane isA. potassium nitrateB. parchment membraneC. cellulose acetateD. cell membrane

Answer» Correct Answer - C
72.

Statement I: Acetic acid deviates from ideal behaviour. Statement II: Acetic acid exists as a dimer by forming inter molecular hence deviates from Raoults law. (a) Statement I & II are correct and II is the correct explanation of I. (b) Statement I & II are correct but II is not the correct explanation of I. (c) Statement I is true but II is wrong. (d) Statement I is wrong but II is correct

Answer»

(a) Statement I & II are correct but II is the correct explanation of I.

73.

Statement I: A solution of potassium chloride in water deviates from ideal behavior.Statement II: The solute dissociates to give K and Cl ion which form strong ion dipole interaction with water molecules. (a) Statement I & II are correct and II is the correct explanation of I (b) Statement I & II are correct but II is not correct explanation of I (c) Statement I is correct but statement II is wrong. (d) Statement I is wrong but statement II is correct.

Answer»

(a) Statement I & II are correct and II is the correct explanation of I

74.

If Kb for water is 0.52 Km-1, the boiling point of 0.2 m solution of a nonvolatile solute will be :(a) 371.96 K (b) 373.104 K (c) 373.52 K (d) 374.0 K

Answer»

Option : (b) 373.104 K

75.

Explain about the factors that are responsible for deviation from Raoult’s law.

Answer»

1. Solute-solvent interactions: 

For an ideal solution, the interaction between the solvent molecules (A – A), the solute molecules (B – B) and between the solvent and solute molecules (A – B) are expected to be similar. if these interactions are dissimilar, there will be a deviation from ideal behaviour.

2. Dissolution of solute: 

When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the solvent and causes deviation from Raoult’s law. e.g., KCI in water deviates from ideal behaviour due to dissociation as K+ and Cl- ion which form strong ion-dipole interaction with water molecules.

3. Association of solute: 

Association of solute molecules can also cause deviation from ideal behaviour. For example in solution acetic acid exists as a dimer by forming intermolecular hydrogen bonds and hence deviates from Raoult’s law.

4. Temperature: An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which cause decrease in the attractive force between them. As result, the solution deviates from Raoult’s law. 

5. Pressure: 

At high pressure, the molecules tends to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus a solution deviates from Raoult’s law at high pressure. 

6. Concentration:

When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from Raoult’s law.

76.

Choose the correct statement.(a) Raoult’s law is applicable to volatile solid solute in liquid solvent (b) Henry’s law is applicable to solution containing solid solute in liquid solvent (c) For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law. (d) For saturated solution containing volatile solid solute in liquid solvent both laws are obeyed.

Answer»

(c) For very dilute solutions. the solvent obeys Raoult’s law and the solute obeys LIenrys law.

77.

Movement of solvent molecules through a semipermeable membrane from pure solvent to the solution side is called osmosis.What are the following :1. Isotonic solution 2. Hypertonic solution?

Answer»

1. Isotonic solution : 

If the osmotic pressure of the two solutions are equal, they are called isotonic solutions.

2. Hypertonic solution :

A solution having higher osmotic pressure than another solution is called hypertonic solution.

78.

Assertion (A): Acetic acid solution deviates from Raoult’s law. Reason (R): Association of solute molecules exists as a dimer by forming intermolecular. hydrogen bonds and hence deviates from Raoult’s law. (a) Both (A) and (R) arc wrong. (b) Both (A) and (R) are correct and (R) is the correct explanation of (A).(c) n-hexane and n-heptanc – Ideal solution (d) Chioro benzene – Ideal solution

Answer»

(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

79.

ΔTb/m for NaBr solution will have value (Kb = 0.52 K kg mol-1) :(a) 0.52 K mol-1(b) 0.104 K kg mol-1(c) 1.24 kg mol-1(d) 1.04 K kg mol-1

Answer»

Option : (d) 1.04 K kg mol-1

80.

Vapour pressure of a solution is different from that of pure solvent.(i) Name the law which helps us to determine partial vapour pressure of a volatile component in solution.(ii) State the above law.

Answer»

(i) Raoult’s law.

(ii) It states that at a given temperature for a solution of volatile liquids, the partial v.p. of each component in the solution is directly proportional to its mole fraction.

81.

Match the following.List-IList-IIA.\(\frac{P°-P}{P°}\)1Depression in freezing pointBΔTb = iKb . m2.Lowering of vapour pressureC.ΔTf = iKf . m3.Osmotic pressureD.π = CRT4.Elevation in boiling pointCode: (a) A - 1, B - 2, C - 3, D - 4(b) A - 2, B - 4, C - 1, D - 3(c) A - 3, B - 1, C - 4, D - 2(d) A - 4, B - 3, C - 2, D - 1

Answer»

(b) A - 2, B - 4, C - 1, D - 3

82.

Assertion : When NaCI is added to water, a depression in freezing point is obseved. Reson : The lowering of vapour pressure of a solution causes depression in the freezing pointA. Assertion and reson both are correct statements and reson is sorrect explanation for assertion.B. Assertion and reson both are correct statements but resons is not correct explanation for assertion.C. Assertion and reason both are incorrect statements.D. Assertion is wrong statement but reason is correct statement.

Answer» Correct Answer - a
Reason is the correct explanation for assertion.
83.

Assertion (A) When NaCl is added to water a depression in freezing point is observed. Reason (R ) The lowering of vapour pressure of a solution causes depression in the freezing point.A. Assertion and reason both are correct statements and reason is correcft explanation for assertion.B. Assertion and reason both are correct statements but reason is not correct explanation for assertion.C. Assertion is correct statement but reason is wrong statement.D. Assertion and reason both are incorrect statements.

Answer» Correct Answer - a
Asserrtion and reason both are correct and reason is correct explanation of assertion. When NaCl is added to water a depressionin freezing point is observed This is due to lowering of vapour pressur of a solution. Lowering of vapour pressure is observed dut to intermolecuolar interaction of solvent-solute particles.
84.

Assertion (A): Ammonia reacts with water does not obey Henry’s law.Reason (R): The gases reacting with the solvent does not obey Henry’s law. (a) Both (A) and (R) are correct and (R) is the correct explanation of (A). (b) Both (A) and (R) are correct but (R) is not the correct explanation of (A). (c) (A) is correct hut (R) is wrong. (d) (A) is wrong but (R) is correct.

Answer»

(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

85.

Depression in freezing point of NaCI is twice that of in urea. Why?

Answer»

1. The electrolyte NaCI dissociates completely into its constituent ions in their aqueous solution. This causes an increase in the total number olparticles present in the solution.

2. When we dissolve 1 mole of NaCI in water it. dissociates and gives 1 mole of Na+ and 1 mole of Cl- . Hence the solution will have 2 moles of particles.

But when we dissolve 1 mole of urea (non electrolyte) in water it appears as 1 mole only. So the colligative property value would be double in NaCl than in urea.

86.

The rise in boiling point of a solution containing `1.8g` glucose in `100g` of a solvent is `0.1^(@)C`. The molal elevation constant of the liquid is-A. `0.01K//m`B. `0.1K//m`C. `1K//m`D. `10K//m`

Answer» Correct Answer - D
`DeltaT_(b) = K_(b) xx m`
`K_(b) = (DeltaT_(b))/(m) = 0.1 xx (180)/(1.8) = (0.1)/(0.01) = 10`
87.

Assertion (A) : When NaCI is added to water, a depression in freezing point is observed. Reason (R): The lowering of vapour pressure of a solution causes the depression in freezing poi nl. (a) Assertion and Reason are correct and R is the correct explanation of A. (b) Both A and R are correct but R is not the correct explanation of A. (c) A is correct but R is wrong (d) A is wrong but R is correct

Answer»

(a) Assertion and Reason are correct and R is the correct explanation of A.

88.

In the determination of molar mass of AB using a colligative property, what may be the value of van’t Hoff factor if the solute is 50% dissociates? (a) 0.5 (b) 1.5 (c) 2.5 (d) 1

Answer»

(b) 1.5 

\(\alpha=\frac{i-1}{n-1}=0.5\)

\(\frac{i-1}{n-1}=0.5\)

i – 1 = 0.5 

i = 0.5 + 1 = 1.5

89.

Explain the effect of temperature gaseous solute in liquid solvent.

Answer»

1. In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature. 

2. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak inter molecular forces when the temperature increases, the average. kinetic energy of the molecules present in the solution also increases. 

3. The increase in kinetic energy breaks (he weak inter molecular forces between the gaseous solute and liquid solvent with results in the release of the dissolved gas molecules to gaseous state.

4. The dissolution of most of the gases in Liquid solvents is an endothermic process, the increase in temperature decreases the dissolution of gaseous molecules.

90.

The rise in boiling point of a solution containing `1.8g` glucose in `100g` of a solvent is `0.1^(@)C`. The molal elevation constant of the liquid is-

Answer» `DeltaT_(b) = 0.1^(@)C, m = 180, W = 100, w = 1.8`
`K_(b) = (DeltaT_(b) xx m xx W)/(1000 xx w) = (180 xx 0.1 xx 100)/(1000 xx 1.8) = 1.0`
91.

Choose the incorrect pair.(a)π = CRTOsmotic pressure(b)\(ΔT_f=\frac{K_f\times W_B\times 1000}{M_B\times W_A}\)Depression in freezing point(c)\(ΔT_b=\frac{K_f\times W_A\times 1000}{M_A\times W_B}\)Elevation in boiling point(d)\(\frac{Δ_P}{P°_A} =\frac{W_B\times W_A}{M_B\times W_B}\)Relative lowering of vapour pressure

Answer»

(c) \(ΔT_b=\frac{K_f\times W_A\times 1000}{M_A\times W_B}\) : Elevation in boiling point

92.

Concentration of solution may be expressed in different ways.1. Mention any one of the concentration terms. 2. What are colligative properties? 3. Show that elevation in boiling point is a colligative property.

Answer»

1. Molarity – It is the number of moles of the solute present in one litre of the solution.

2. Colligative properties are those properties which depends only on the number of solute particles.

3. ΔTb = Kb m

\(=k_b\,\frac{n_B\times1000}{W_A}\)

i.e., ΔTb α nB i.e., elevation of boiling point depends on number of moles of solute. Hence, it is a colligative property.

93.

What is abnormal colligative property? Explain the reasons.

Answer»

Abnormal colligative property : When the experimentally measured colligative property of a solution is different from that calculated theoretically by the van’t Hoff equation or by the laws of osmosis, then the solution is said to have abnormal colligative property.

Explanation : The colligative property depends on the number of solute particles in the solution but it is independent of their nature. Abnormal values of them arise when the dissolved solute undergoes a molecular change like dissociation or association in the solution.

The observed colligative property (or abnormal colligative property ) may be higher or lower than the theoretical value.

(i) Dissociation of the solute molecules : When a solute like an electrolyte is dissolved in a polar solvent like water, it undergoes dissociation, which results in the increase in the number of particles in the solution.

Hence, 

The observed value of the colligative property becomes higher than the theoretical value, e.g., when one mole of KCl is dissolved in the solution then due to dissociation, KCl → K+ + Cl-, the number of particles increases, hence, the colligative properties like osmotic pressure elevation in the boiling point, etc. increase.

(ii) Association of the solute molecules : When a solute like a non-electrolyte is dissolved in a non-polar solvent like benzene, it undergoes association forming molecules of higher molecular mass. Hence, the number of the particles in the solution decreases. Therefore the colligative properties like osmotic pressure, elevation in the boiling point, etc., are lower than the theoretical value, e.g., nA → An.

2CH3COOH → (CH3COOH)2

2C6H5COOH → (C6H5COOH)2

94.

18 g of glucose is dissolved in 1 kg of water in a beaker. At what temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol-1 )

Answer»

Number of moles of glucose = 18/180 = 0.1 mol.

Mass of solvent = 1 kg.

Morality of glucose , m = \(\frac{n_8}{W_A}\)

\(\frac{0.1\,mol}{1\,kg}\)

= 0.1 mol/Kg

Elevation of boiling point ΔTb = Kb × m

= 0.52 K kg/mol × 0.1 mol/kg = 0.052 K

Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be

373.15 K + 0.052 K = 373.202 K

95.

Colligative properties are exhibited by dilute solutions.1. What do you mean by colligative properties?2. Which are the four colligative properties?

Answer»

1. Colligative properties are those properties of dilute solutions of non-volatile solutes whose value depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.

2. The four colligative properties

  • Relative lowering of vapour pressure 
  • Elevation of boiling point 
  • Depression of freezing point 
  • Osmotic pressure  
96.

Compreshension -I The experimental values of colligative properties of many solutes in solution resembles calculated value of colligative properties. However in same cases, the experimental value of colligative property differ widely than those obtained by calculation. Such experimental values of colligative properties are known as Abnormal values of colligative properties are : (i) Dissociation of solute : It increases the colligative properties. (ii) Association of solute : It decreases the colligative properties e.g. : Dimerisation of acetic acid in benzene One mole `I_(2)` (solid) is added in 1 M, 1 litre Kl solution . ThenA. Osmostic pressure of solution increasesB. Freezing point of solution increasesC. Relative lowering in vapour pressure decreasesD. No change in boiling point of solution

Answer» Correct Answer - 4
97.

Choose the incorrect pair.(a) Benzene and acetone – Ideal solution (b) Ethyl alcohol and cyclohexane – Non-ideal solution (C) n-hexane and n-heptanc – Ideal solution (d) Chioro benzene – Ideal solution

Answer»

(a) Benzene and acetone – Ideal solution

98.

0.75 g of an unknown substance is dissolved in 200 g solvent. If the elevation of boiling point is 0.15 K and molal elevation constant is 7.5K kg more then, calculate the molar mass of unknown substance.

Answer»

∆Tb = Kb m = Kb x W2 x \(\frac{1000}{M_2}\) x W1 

M2 = Kb x W2 x \(\frac{1000}{\Delta T_b}\) x W1 

= 7.5 x 0.75 x \(\frac{1000}{0.15}\) x 200 = 187.5 g mol-1

99.

When maximum amount of solute is dissolved in a solvent at a given temperature, the solution is called ……

Answer»

When maximum amount of solute is dissolved in a solvent at a given temperature, the solution is called Saturated solution.

100.

Menthol is a crytalling substance with peppermint taste. A 6.3% solution of menthol in cyclohexane freezes at `1.9^(@)C`. Determine the molar mass of menthol. The freezing point and molal depression constant of cyclohexane are `6.5^(@)C` and 20.2 K `m^(-1)` respectively.

Answer» A 6.2 p-ercent solution of menthol means that 6.2 g of the colute are dessolved in 100 g of the solution. Therefore mass of solvent is 93.8g.
`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
`W_(B)=6.2g, W_(A)=93.8 g=0.0938 kg,`
`DeltaT_(f)=1.95-(-6.5)=8.45 K, K_(f)=20.2 K kg mol^(-1)`
`M_(B)=((20.2K kg mol^(-1))xx(6.2g))/((8.45K)xx(0.0983 kg))=158.0 g mool^(-1)`