Solutions Test -18

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Solutions Test -18

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Question 1
1 / -0

On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?

A
Sugar crystals in cold water.

B
Sugar crystals in hot water.

C
Powdered sugar in cold water.

D
Powdered sugar in hot water.

Solution

(i) The solubility of a solute / sugar increases with increase in surface area of its particles Powdered sugar has higher surface area , therefore , powdered sugar will dissolve most rapidly.

(ii) Since dissolution of sugar is attended with a cool feel to touch , it is an endothermic process . As per LeChatelier's principle an increase in temperature / using hot water would favour the dissolution of sugar.
Question 2
1 / -0

At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is .........

A
Less than the rate of crystallisation

B
Greater than the rate of crystallisation

C
Equal to the rate of crystallisation

D
zero

Solution

This happens as per conditions attained at equilibrium state ; i.e. rate of forward reaction (dissolution) = rate of backward reaction ( crystallisation).
Question 3
1 / -0

A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is

A
Saturated

B
Supersaturated

C
Unsaturated

D
Concentrated

Solution

A supersaturated solution is a solution that contains lesser than maximum amount of solute per given amount of solvent at a particular temperature. If even a small amount of solute is added to such a solution it precipitates/crystallised rapidly. It should be noted that a supersaturated solution differs from unsaturated solution in the sense that NO precipitation or crystalisation would occur by adding even a small amount of solute to it, rather it goes into solution and remains dissolved at a particular temperature.
Question 4
1 / -0

Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon

A
Temperature

B
Nature of solute

C
Pressure

D
Nature of solvent

Solution

Solubility of a solid in liquid does not depend upon pressure since solid and liquids are almost incompressible.
Question 5
1 / -0

Low concentration of oxygen in the blood and tissues of people living at high altitude is due to

A
Low temperature

B
Low atmospheric pressure

C
High atmospheric pressure

D
Both low temperature and high atmospheric pressure

Solution

At high altitude the atmospheric pressure is decreased &,due to low atmospheric pressure the solubility of oxygen in blood and tissues is reduced.
Question 6
1 / -0

Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?

A
Methanol and acetone.

B
Chloroform and acetone.

C
Nitric acid and water

D
Phenol and aniline.

Solution

(A-A)* interaction is greater than the( A-B )** interaction. Intermolecular hydrogen bonding in methanol is more than methanol and acetone separately. So, methanol and acetone mixtures will show a positive deviation from Raoult’s law. ( A- A )* interaction represents interaction between particles / molecules of acetone among which there is no hydrogen bonding. ( A - B )** interaction is the interaction between the particles / molecules of acetone and methanol .
Question 7
1 / -0

Which of the following factor (s) affect the solubility of a gaseous solute in the fixed volume of liquid solvent?

(a) Nature of solute

(b) Temperature

(c) Pressure

(i) (a) and (c) At constant T

(ii) (a) and (b) At constant P

(iii) (b) and (c) only

(iv) (c) only

A
(i) and (ii)

B
(i) and (iii)

C
(iv) and (iii)

D
(ii) and (iv)

Solution

At constant temperature, the solubility of a gaseous solute in liquid depends on nature of solute and pressure. At constant pressure, solubility is dependent upon nature of solute and temperature.
Question 8
1 / -0

Intermolecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?

(i) Δ_mix^H = zero

(ii) Δ_mix^V = zero

(iii) These will form minimum boiling azeotrope.

(iv) These will not form ideal solution.

A
(iii) and (iv)

B
(ii) and (iv)

C
(i) and (iv)

D
(i) and (ii)

Solution

In a mixture of benzene and toluene intermolecular forces between benzene and toluene molecules would be nearly of the same strength as those of two benzene molecules and two toluene molecules separately. The solution will , therefore , form an ideal solution & obey Raoult’s law .
Question 9
1 / -0

Relative lowering of vapour pressure is a colligative property because ...

(i) It depends on the concentration of a non-electrolyte solute in solution and does not depend on the nature of the solute molecules.

(ii) It depends on number of particles of electrolyte solute in solution and does not depend on the nature of the solute particles.

(iii) It depends on the concentration of a non-electrolyte solute in solution as well as on the nature of the solute molecules.

(iv) It depends on the concentration of an electrolyte or non-electrolyte solute in solution as well as on the nature of solute molecules.

A
(i) and (ii)

B
(iii) and (ii)

C
(iv) and (ii)

D
(iv) and (iii)

Solution

Colligative property depends on (i) the concentration of a nonelectrolyte solute in solution, (ii) the number of particles of electrolyte solute in solution , & (iii) It does not depend on the nature of solute molecules / particles.
Question 10
1 / -0

For a very dilute solution of ${ H }_{ 3 }{ PO }_{ 3 }$ , Van’t Hoff factor is:

A
$i=7$

B
$i=3$

C
$i=4$

D
$i=5$

Solution

Dissociation of $H_{3}PO_{3}$ is as follows:

$H_{3}PO_{3}$ $\rightarrow$ $2H^{+} + HPO_{3}^{-2}$

It produces 3 ions on dissociation. Hence, Van't Hoff factor = 3.
Question 11
1 / -0

$12g$ of urea is present in $1$ litre of solution and $68.4 g$ of sucrose is separately dissolved in $1$ litre of another sample of solution. The lowering of vapour pressure of first solution is :

A
Equal to second

B
Greater than second

C
Less than second

D
Double that of seconds

Solution

Moles of urea $= 12/60 = 0.2$

Moles of sucrose $= 68.4/342 = 0.2$

(Molar mass of sucrose is 342gm )

Since there are equal moles of solute in equal volumes of water, the mole fraction is the same.

As per Raoult's law, the relative lowering of pressure is also the same.
Question 12
1 / -0

The pressure at which liquid and vapour can coexist at equilibrium is called the :

A
limiting vapour pressure

B
real vapour pressure

C
normal vapour pressure

D
saturated vapour pressure

Solution

Ordinary evaporation is a surface phenomenon - some molecules have enough kinetic energy to escape. If the container is closed, then an equilibrium is reached where an equal number of molecules return to the surface.

The pressure at which this equilibrium where the coexistence of liquid and vapour is achieved is called the saturated vapor pressure.
Question 13
1 / -0

In which of the following pairs of solutions will the values of the Van't Hoff factor be the same?

A
$0.05$ M ${ K }_{ 4 }\left[ Fe\left( CN \right) _{ 6 } \right]$ and $0.10$ M ${ FeSO }_{ 4 }$

B
$0.20$ M $NaCl$ and $0.10$ M ${ BaCl }_{ 2 }$

C
$0.05$ M ${ FeSO }_{ 4 }\left( NH_{ 4 } \right) _{ 2 }{ SO }_{ 4 }.6{ H }_{ 2 }O$ and $0.02$ M $KCl.{ MgCl }_{ 2 }.{ 6H }_{ 2 }O$

D
All have same value

Solution

Van't Hoff factor does not depends on molarity for strong electrolytes. Also, water does not add in the Van't Hoff factor.

Following are the dissociation reactions,

$K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$ [i=5]

$FeSO_4 \rightarrow Fe^{2+} + SO_4^{2-}$ [i=2]

$NaCl \rightarrow Na^+ + Cl^-$   [i=2]

$BaCl_2 \rightarrow Ba^{2+} + 2Cl^{-}$    [i=3]

$FeSO_4 (NH_4)_2 SO_4. 6H_2O \rightarrow Fe^{2+} + SO_4^{2-} +  2NH_4^+ + SO_4^{2-} + 6H_2O$    [i=5]

$KCl.{ MgCl }_{ 2 }.{ 6H }_{ 2 }O\rightarrow K^+ + 3Cl^- + Mg^{2+}+ 6H_2O$    [i=5]

Hence option C is correct.
Question 14
1 / -0

The van't hoff factor for a very dilute solution of ${ Fe }_{ 2 }\left( { SO }_{ 4 } \right) _{ 3 }$ is:

A
$9$

B
$5$

C
$24$

D
$16$

Solution

$Fe_{2}(SO_{4})_{3}$ dissociates into $2Fe^{3+}+3 SO_{4}^{2-}$

Now
$i=1+\alpha(n-1)$

$i=1+1\times(5-1)$

So, $i=5$
Question 15
1 / -0

At a certain temperature, the vapour pressure of water is 90 mm. At the same temperature the vapour pressure of a solution containing a non-volatile solute is 81 mm. The relative lowering of vapour pressure is :

A
$9$

B
$0.9$

C
$10$

D
$0.1$

Solution

The relative lowering in vapour pressure $= \displaystyle\frac {p^0-p} {p^0}$

$p^0$ is the vapour pressure of pure solvent and p is the vapour pressure of the solution.

Given that,
$p^0=90 \ mm$
$p=81 \ mm$

The relative lowering in vapour pressure $= \displaystyle\frac {p^0-p} {p^0}= \displaystyle\frac {90-81}{90}=0.1$

Option D is correct.
Question 16
1 / -0

Which one of the following represents the graph between log P on Y-axis and $\cfrac { 1 }{ T }$ on X-axis?

A

B

C

D

Solution

The Clausius Clapeyron equation states that,
$P = Ae^{(-\dfrac{\Delta H}{RT})}$

Taking log both sides, we have-

$log\,P = log\,A - \dfrac{\Delta H}{RT}$

Graph of log $P$ vs $\dfrac 1T$ is thus a straight line with slope $-\dfrac{\Delta H} {R}$ and intercept log $A$ .
Question 17
1 / -0

Lowering of vapour pressures of equimolar solution of glucose, sodium chloride and barium nitrate are in the order:

A
Glucose > $NaCl$ > $Ba(NO_{ 3 }) _{ 2 }$

B
Glucose = $NaCl$ = $Ba(NO_{ 3 }) _{ 2 }$

C
$Ba(NO_{ 3 }) _{ 2 }$ > $NaCl$ > Glucose

D
$NaCl$ > $Ba(NO_{ 3 }) _{ 2 }$ > Glucose

Solution

Van't Hoff factor $=i$

For $Ba(NO_3)_2 , i=3;$ $Ba+2NO_{3}^{-}$

For $NaCl$ , $i=2;$ $$Na^
++ Cl^-$$

For Glucose, $i=1$

So higher the van't Hoff factor, higher is the lowering of vapor pressure

So $Ba(NO_3)_2$ will have the highest lowering of vapor pressure.

Option C is correct.
Question 18
1 / -0

Van't Hoff factor for a dilute solution of sodium argento cyanide is :

A
$2$

B
$0.25$

C
$0.50$

D
$3.0$

Solution

The compound is $Na[Ag(CN)_2]$ .

The dissociation reaction is,

$Na[Ag(CN)_2] \rightarrow Na^+ + [Ag(CN)_2]^-$
1 0 0
$1- \alpha$ $\alpha$ $\alpha$

Van't Hoff factor, $i = 1 + \alpha$

For very dilute solution,

$\because \alpha = 1$

Hence, $i = 2$

Therefore, option A is correct.
Question 19
1 / -0

$0.1 M K_{4} [Fe (CN)_{6}]$ is 60% ionized. What will be its van't Hoff factor?

A
1.4

B
2.4

C
3.4

D
4.4

Solution

It a fraction $\alpha$ of the solute dissociates into n ions then
$\alpha = 0\cdot 6$ $n=5$
$i =1+\alpha (n-1)$
$=1+0\cdot 6(5-1)$
$=1+2\cdot 4$
$=3\cdot 4$

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Solutions Test -18

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Solutions Test -18

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SHARING IS CARING

Question 1

Sugar crystals in cold water.

Sugar crystals in hot water.

Powdered sugar in cold water.

Powdered sugar in hot water.

Solution

Question 2

Less than the rate of crystallisation

Greater than the rate of crystallisation

Equal to the rate of crystallisation

zero

Solution

Question 3

Saturated

Supersaturated

Unsaturated

Concentrated

Solution

Question 4

Temperature

Nature of solute

Pressure

Nature of solvent

Solution

Question 5

Low temperature

Low atmospheric pressure

High atmospheric pressure

Both low temperature and high atmospheric pressure

Solution

Question 6

Methanol and acetone.

Chloroform and acetone.

Nitric acid and water

Phenol and aniline.

Solution

Question 7

(i) and (ii)

(i) and (iii)

(iv) and (iii)

(ii) and (iv)

Solution

Question 8

(iii) and (iv)

(ii) and (iv)

(i) and (iv)

(i) and (ii)

Solution

Question 9

(i) and (ii)

(iii) and (ii)

(iv) and (ii)

(iv) and (iii)

Solution

Question 10

i=7i=7i=7

i=3i=3i=3

i=4i=4i=4

i=5i=5i=5

Solution

Question 11

Equal to second

Greater than second

Less than second

Double that of seconds

Solution

Question 12

limiting vapour pressure

real vapour pressure

normal vapour pressure

saturated vapour pressure

Solution

$i=7$

$i=3$

$i=4$

$i=5$

$0.05$ M ${ K }_{ 4 }\left[ Fe\left( CN \right) _{ 6 } \right]$ and $0.10$ M ${ FeSO }_{ 4 }$

$0.20$ M $NaCl$ and $0.10$ M ${ BaCl }_{ 2 }$

$0.05$ M ${ FeSO }_{ 4 }\left( NH_{ 4 } \right) _{ 2 }{ SO }_{ 4 }.6{ H }_{ 2 }O$ and $0.02$ M $KCl.{ MgCl }_{ 2 }.{ 6H }_{ 2 }O$

$9$

$5$

$24$

$16$

$9$

$0.9$

$10$

$0.1$

Glucose > $NaCl$ > $Ba(NO_{ 3 }) _{ 2 }$

Glucose = $NaCl$ = $Ba(NO_{ 3 }) _{ 2 }$

$Ba(NO_{ 3 }) _{ 2 }$ > $NaCl$ > Glucose

$NaCl$ > $Ba(NO_{ 3 }) _{ 2 }$ > Glucose

$2$

$0.25$

$0.50$

$3.0$