MCQ Questions for Class 12 Engineering Chemistry Solutions Quiz 2

The vapour pressure (V.P.) of a dilute solution of non-volatile solute is P and the V.P. of pure solvent is

$P^0$ , the lowering of the V.P. is :

0%
$+ve$
0%
$-ve$
0%
$P/P^0$
0%
$P^0/P$

Explanation

The vapour pressure (V.P.) of a dilute solution of non-volatile solute is P and the V.P. of pure solvent is

$P_0$ , the lowering of the V.P. is +ve.Lowering of vapor pressure is always positive.

Which substance does NOT dissolve well in water?

0%
${NH}_{3}$
0%
${C}_{4}{H}_{10}$
0%
${C}_{12}{H}_{22}{O}_{11}$
0%
${H}_{2}{SO}_{4}$

Explanation

we all know, "Like dissolve like"

water is a polar solvent and

$C_4H_{10}$ is non-polar in nature

so, it is insoluble in water.

The van't Hoff factor for a very dilute aqueous solution of

$K\left[Ag\left( CN \right) _{ 2 } \right]$ is:

0%
$4$
0%
$3$
0%
$2$
0%
$5$

Explanation

$K[Ag(CN)_{2}]$ dissociate into

$K^{+}$ and

$[Ag(CN)_{2}]^{-}$

$i=1+1=2$ .

Hence, the correct option is

$\text{C}$

In case of an electrolyte which dissociates in the solution, the van't hoff factor

$i$ is :

0%
$>1$
0%
$<1$
0%
$1$
0%
$>1 \ or \ <1$

Explanation

We know that,

$i = \dfrac{ number\: of\: molecules\: after\: association/dissociation }{ number\: of\: molecules\: before\: association/dissociation }$
After dissociation number of molecules increases. Therefore i>1

Which condition is not satisfied by an ideal solution?

0%
${\Delta}_{mix}H=0$
0%
${\Delta}_{mix}V=0$
0%
${\Delta}_{mix}S=0$
0%
Obeyance of Raoult's law

Explanation

For an ideal solution

${\Delta}_{mixing}H=0;{\Delta}_{mixing}V=0$
and it should obey Raoult's law.

Aqueous solution of glucose was prepared by dissolving

$18g$ of glucose in

$90g$ of water. The relative lowering of vapour pressure is equal to

0%
$1.1$
0%
$0.0196$
0%
$20$
0%
$196$

Explanation

By Raoult's law:

Relative lowering of vapour pressure

$=\cfrac { \cfrac { w }{ m } }{ \cfrac { w }{ m } +\cfrac { W }{ M } } =\cfrac { 18/180 }{ \cfrac { 18 }{ 180 } +\cfrac { 90 }{ 18 } }$

$=0.0196$

Limiting value of Van't Hoff factor of

$[K_4Fe(CN)_6]$ is

$11$ .

0%
True
0%
False

Explanation

As we know,
Van't Hoff factor = no of ions in solution
and

$[K_4Fe(CN)_6]$ gives 5 ions in solution so its Van't Hoff factor is 5.

Vant Hoff factor for a dilute solution of glucose is:

0%
$0$
0%
$1$
0%
$1.5$
0%
$2$

Explanation

Solution of glucose is a non-electrolytic solution.

For a non-electrolytic solution, vant hoff factor is 1.

Hence, the correct option is

$\text{B}$

Which substance has a higher solubility in cold water and a lower solubility in hot water?

0%
$LiCl$
0%
$Ca{(OH)}_{2}$
0%
${CO}_{2}$
0%
${C}_{6}{H}_{12}{O}_{6}$

Explanation

For gaseous solute, the solubility of gas-particle decreases with an increase in temperature.

Therefore,

$CO_2$ is more soluble in cold water and less soluble in hot water.

A concentrated solution contains high amount of ________.

0%
solvent
0%
solute
0%
both solute and solvent
0%
none of these

100 mL of liquid A and 25 mL of liquid B are mixed to formal solution of volume 125 mL. Then the solution is___________.

0%
ideal
0%
non-ideal with positive deviation
0%
non-ideal with negative deviation
0%
cannot be predicted

Explanation

As change in volume

$(\Delta V)$ is zero so the solution is ideal.

The main reason for the extremely low solubility of carbon dioxide in benzene

$(C_{6}H_{6})$ at room temperature is due to which of the following?

0%
The increased disorder due to mixing of the solute and solvent.
0%
The relatively low strength of the intermolecular forces between carbon dioxide and benzene.
0%
The strong hydrogen bonding in benzene.
0%
The weak solvation of carbon and oxygen ions by benzene.

Explanation

$CO_2$ is a polar compound.

Benzene is non-polar compound.

Like dissolves like, i.e. Polar dissolves polar because the intermolecular force between polar-polar molecules are strong.

Thus the intermolecular force between non polar and polar molecule is weak.

Therefore,

$CO_2$ has low solubility in benzene.

Which of the following solution in water will have the lowest vapour pressure?

0%
$0.1 M, NaCI$
0%
$0.1 M, Sucrose$
0%
$0.1 M,{ BaCI }_{ 2 }$
0%
$0.1 M { Na }_{ 3 }{ PO }_{ 4 }$

Explanation

$Na_3PO_4$ will give maximum ions in water (

$4$ ions)

So, it has lowest vapour pressure.

Which one of the following salts would have the same value of the van't Hoff factor as that of

$K_3Fe(CN)_6?$

0%
$Al_2(SO_4)_3$
0%
$NaCl$
0%
$Na_2SO_4$
0%
$Al(NO_3)_3$

Explanation

$K_3Fe(CN)_6$ dissociates into

$3K^+ + [Fe(CN)_6]^{3-}$

total of 4 particles, in a similar way,

$Al(NO_3)_3$ dissociates into

$Al^{3+}+3[(NO_3)]^-$

total of 4 particles.

Ideal solution is formed when its components:

0%
have zero heat of mixing only
0%
have zero volume change on mixing only
0%
have zero heat of mixing and zero volume change
0%
can be converted into gases

Explanation

An ideal solution one in which the molecules attract one another with equal force irrespective of their nature.

Thus, a solution composed of two components A and B will be an ideal one if the forces between A and A, B and B should be the same.

$V_{mix}=0$

$V_{solvent}+V_{solute}=V_{solution}$

Heat change on mixing should be zero

$H_{mix} = 0$ (Heat is neither absorbed nor evolved).

Which of the following components form an ideal solution?

0%
Benzene and toluene
0%
Acetone and aniline
0%
Ethyl alcohol and benzene
0%
Water and nitric acid

Explanation

The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions.

For ideal solution, the enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,

${\triangle }_{mix}H=0$ and

${\triangle}_{mix} V=0$

Thus type of solution is a solution of benzene and toluene.

Identify which of the following statements is FALSE.

0%
The vapor pressure of a liquid decreases with increasing atmospheric pressure.
0%
The value of an equilibrium constant is dependent on temperature.
0%
The rate of a spontaneous reaction cannot be determined solely by its Gibbs free energy.
0%
During a phase transition, the temperature of a substance must be constant.

Which of the following would have the lowest vapor pressure in the pure state?

0%
Sodium hydroxide because of the strong interparticle forces of ionic solids
0%
Methanol because of the hydrogen bonding in the functional group along with the non-polar nature of the carbon hydrogen bonds
0%
Water because of the very strong hydrogen bonding in the substances
0%
Ammonia because of the unbonded pair of electrons on the central atom coupled with hydrogen bonding in the substance.

Which of the following are true solutions?

0%
ENO salt in water
0%
Tomato soup
0%
Aam panna
0%
Filtered black tea

Explanation

The definition of true solution is a homogeneous mixture of two or more substances in which substance dissolved (solute) in solvent has the size of the particle less than

$10^{-9} m$ or

$1 nm$ . So the perfect example of true solution is ENO salt in water. ENO is perfectly soluble in water and has minimum size. SO the correct option is A

For an ideal solution with

$P^o_A > P^o_B$ , which of the following is true ?

0%
$^x_{A(I)} = ^ x_{A(v)}$
0%
$^x_{A(I)} > ^ x_{A(v)}$
0%
$^x_{A(I)} < ^ x_{A(v)}$
0%
No relationship in their mole fraction

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

0%
Glucose > $NaCl$ > $Ba(NO_{ 3 }) _{ 2 }$
0%
Glucose = $NaCl$ = $Ba(NO_{ 3 }) _{ 2 }$
0%
$Ba(NO_{ 3 }) _{ 2 }$ > $NaCl$ > Glucose
0%
$NaCl$ > $Ba(NO_{ 3 }) _{ 2 }$ > Glucose

Explanation

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

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

Option C is correct.

Which one of the following represents the graph between log P on Y-axis and

$\cfrac { 1 }{ T }$ on X-axis?

Explanation

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

$A$ .

When a solution is diluted at constant temperature, its vapour pressure:

0%
increases
0%
decreases
0%
remains same
0%
may increase or decrease

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

0%
limiting vapour pressure
0%
real vapour pressure
0%
normal vapour pressure
0%
saturated vapour pressure

The vapour pressure of water at

$300$ K in a closed container is

$0.4$ atm. If the volume of the container is doubled, its vapour pressure at

$300$ K will be:

0%
$0.8$ atm
0%
$0.2$ atm
0%
$0.4$ atm
0%
$0.6$ atm

Explanation

Vapour pressure will be the same because it doesn't depend on the volume it depends only on Temperature.

Hence, the correct option is

$A$

The boiling points of

${ C }_{ 6 }{ H }_{ 6 }, { CH }_{ 3 }{ OH }, { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$ and

${ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }$ are

$80^{ 0 }C, { 65 }^{ 0 }C, { 184 }^{ 0 }C$ and

$212^{ 0 }C$ respectively. Which of these will show highest vapour pressure at room temperature?

0%
${ C }_{ 6 }{ H }_{ 6 }$
0%
${ CH }_{ 3 }OH$
0%
${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
${ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }$

Explanation

The lower the boiling point, the higher the vapor pressure. Out of given options,

$CH_3OH$ has the lowest boiling point. So, it has the highest vapor pressure.

The answer is option

$B$ .

Which characteristics the weak inter molecular forces of attraction in a liquid?

0%
High boiling point
0%
High vapour pressure
0%
High critical temperature
0%
High heat of vapourisation

Which of the following solutions will have the maximum lowering of vapour pressure ?

0%
1M - ${ CaCl }_{ 2 }$
0%
1M- $NaCl$
0%
1N - $Phenol$
0%
1M - $Sucrose$

Explanation

Since the some compounds in the options are ionic, the maximum lowering is decided as per the highest vant hoff factor.
For calcium chloride ,V.H. factor

$=$ 3
For salt,V .H. factor

$=$ 2
For phenol , V.H. factor < 1 (molecules associate)
For sucrose , factor

$=$ 1.

A solution is obtained by dissolving 12g of urea (Mol.wt.=60) in a litre of solution. Another solution is prepared by dissolving 68.4g of cane sugar(Mol.wt.=342) in a litre of solution at the same temperature. The lowering of vapour pressure in the first solution is:

0%
nearly 5 times that of the second solution
0%
same as that of second solution
0%
double that of second solution
0%
nearly one fifth of the second solution

Explanation

For urea

$\Delta P_1=X_{urea}P^0$

$\Delta P_1=\dfrac {\dfrac {12}{60}}{total \ mole}P^0$

$\Delta P_1=\dfrac {0\cdot 2}{total \ mole} P^0$ ----(1)

For sugar

$\Delta P_2=X_{sugar}P^0$

$\Delta P_2=\dfrac {\dfrac {68\cdot 4}{342}}{total \ mole}P^0$

$\Delta P_2=\dfrac {0\cdot 2}{total \ mole} P^0$ ----(2)

Hence

$(1)=(2)$

Hence, the correct option is

$B$

$12g$ 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 :

0%
Equal to second
0%
Greater than second
0%
Less than second
0%
Double that of seconds

Explanation

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.

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 :

0%
$9$
0%
$0.9$
0%
$10$
0%
$0.1$

Explanation

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.

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

0%
$0.05$ M ${ K }_{ 4 }\left[ Fe\left( CN \right) _{ 6 } \right]$ and $0.10$ M ${ FeSO }_{ 4 }$
0%
$0.20$ M $NaCl$ and $0.10$ M ${ BaCl }_{ 2 }$
0%
$0.05$ M ${ FeSO }_{ 4 }\left( NH_{ 4 } \right) _{ 2 }{ SO }_{ 4 }.6{ H }_{ 2 }O$ and $0.02$ M $KCl.{ MgCl }_{ 2 }.{ 6H }_{ 2 }O$
0%
All have same value

Explanation

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.

The van't hoff factor for a very dilute solution of

${ Fe }_{ 2 }\left( { SO }_{ 4 } \right) _{ 3 }$ is:

0%
$9$
0%
$5$
0%
$24$
0%
$16$

Explanation

$Fe_{2}(SO_{4})_{3}$ dissociates into

$2Fe^{3+}+3 SO_{4}^{2-}$

Now

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

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

So,

$i=5$

Correct combination among the following is
i) Water can be converted into vapour at

${ 100 }^{ 0 }C$ only
ii) Vapour pressure can be increased more than the atmospheric pressure at a certain temperature.
iii) Lowering of vapour pressure is directly proportional to mole fraction of solute.

0%
All are correct
0%
ii only wrong
0%
iii only correct
0%
All are wrong

Explanation

Using clausius clapeyron equation,
we get

$log \dfrac{P_2}{P_1} = -\dfrac{\Delta H}{R} \times (\dfrac{1}{T_1} - \dfrac{1}{T_2})$
Therefore, at any particluar temperature, the vapour pressure of solution is unique.
Also, conversion to vapour of a liquid can take place at many temperatures and many pressure values.
The third statement is the outcome of Raoult's law.

In case a solute associates in solution, the van’t Hoff factor is :

0%
$i > 1$
0%
$i = 1$
0%
$i < 1$
0%
None of these

Explanation

if a fraction

$\alpha$ of

$n$ moles of solute associate to form one mole of an n-mer (dimer, trimer, etc.), then

$i=1-\left(1-{\frac {1}{n}}\right)\alpha .$

For the dimerisation of acetic acid in benzene

$2CH_3COOH ⇌ {(CH_3COOH)}_2$

2 moles of acetic acid associate to form 1 mole of dimer, so that

$i=1-\left(1-{\frac {1}{2}}\right)\alpha =1-{\frac {\alpha }{2}}.$

For association in the absence of dissociation, the van 't Hoff factor is:

$i<1$

Hence, the correct option is

$\text{C}$

For a very dilute aqueous solution of

${ K }_{ 4 }\left[ Fe\left( CN \right) _{ 6 } \right]$ , van't Hoff factor is :

0%
$i=11$
0%
$i=5$
0%
$i=\dfrac { 1 }{ 11 }$
0%
$i=10$

Explanation

Considering complete dissociation,

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

$i = 5$

For a substance

$A$ when dissolved in a solvent

$B$ shows the molecular mass corresponding to

${ A }_{ 3 }$ . The van't Hoff factor will be:

0%
$1$
0%
$2$
0%
$3$
0%
$\dfrac{1}{3}$

Explanation

$3A\rightarrow A_{3}$

3 moles of A associate to form 1 mole as A solute undergoes trimerization so van't Hoff factor

$\dfrac{1}{3}$ .

Hence, the correct option is

$D$

For a non-electrolytic solution, vant Hoff factor is equal to :

0%
0
0%
1
0%
between 0 and 1
0%
none of the above

Explanation

The van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van't Hoff factor is essentially

$1$ .

The following are some statements about vapour pressure.

I. Vapour pressure of a solution increases with increase in temperature.

II. Liquids with high vapour pressure even at ordinary temperatures are volatile liquids.

III. The temperature at which vapour pressure of solution is equal to atmospheric pressure is called boiling point.

IV. When a non-volatile solute is dissolved in a volatile solvent, its vapour pressure decreases.

Which of the above is/are correct?

0%
All are correct
0%
II, III and IV
0%
I and II
0%
II and IV

For a very dilute solution of

${ H }_{ 3 }{ PO }_{ 3 }$ , Van’t Hoff factor is:

0%
$i=7$
0%
$i=3$
0%
$i=4$
0%
$i=5$

Explanation

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.

For a given value of the degree of dissociation (

$\alpha$ ), which of the following have correct van’t Hoff factor?

0%
$NaCl ; i = 2+ \alpha$
0%
$Ca ( NO_{3})_{2} ; i =1 + 2\alpha$
0%
$K_{4}[(Fe(CN)_{6}] ; i = 1+ 4\alpha$
0%
$(NH_{4})_{3}PO_{4} ; i = 3+ \alpha$

Explanation

The expression to determine van't Hoff factor of a solution is given by:

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

$n$ = number of ions formed after dissociation

$\alpha$ = Degree of dissociation

where,

$i$ = van't Hoff factor

A. For

$NaCl$ ;

$n=2$ so

$i=1+\alpha$

B. For

$Ca(N{ O }_{ 3 }{ ) }_{ 2 }$ ;

$n=3$ so

$i=1+2\alpha$

C. For

${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ ;

$n=5$ so

$i=1+4\alpha$

D. For

$(N{ H }_{ 4 }{ ) }_{ 3 }P{ O }_{ 4 }$ ;

$n=4$ so

$i=1+3\alpha$

So, option B and C are correct.

Which of the following is/are true?

0%
For the same solution, elevation in boiling point = depression in freezing point
0%
Van’t Hoff factor for a dilute solution of $BaCl_{2}$ is 3
0%
Elevation in boiling point is due to increase in vapour pressure
0%
Depression in freezing point is due to decrease in vapour pressure

Explanation

Elevation in boiling point and depression in freezing point is due to decrease in vapour pressure of the solution.
Van't hoff factor for Barium chloride is 3 (complete dissociation)
Also, elevation in boiling point is different than depression in freezing point due to different

$K_b \: and \: K_f$ .

Which of the following form/s an ideal solution?

0%
Ethyl bromide + ethyl iodide
0%
Ethyl alcohol + water
0%
Chloroform + benzene
0%
Benzene + toluene

Explanation

An ideal solution is a solution in which the enthalpy of solution is zero or close to zero.

$C_2H_5Br$ &

$C_2H_5I$ has same enthalpy even Benzene & toluene both has same enthalpy.

$\Delta H=0$
So [A] & [D] are correct option

If a solute under goes dimerization and trimerization, the minimum values of the vant Hoff factors are:

0%
0.50 and 1.50
0%
1.50 and 1.33
0%
0.50 and 0.33
0%
0.25 and 0.67

Explanation

$2A\rightarrow A_{2}$ [2 mole of A-associate to form 1]
Vant hoff factor

$=\dfrac{1}{2}=0.5$

$3A\rightarrow A_{3}$
Vant hoff factor

$=\dfrac{1}{3}=0.33$

Hence, the correct option is

$C$

Highly pure dilute solution of sodium in liquid ammonia:

0%
on evaporation yield metals.
0%
exhibits electrical conductivity.
0%
produces sodium amide and hydrogen gas instantly.
0%
acts as powerful reducing agent.

Explanation

Correct option: $A, B$ and $D.$

Explanation:

Dilute solution of sodium in liquid ammonia forms solved electrons as shown below;

$Na + (x + y)NH_3 → [Na_{(NH_3)_x}]^+ + [e_{(NH_3)_y}]^-$

A highly pure dilute solution of sodium in liquid ammonia exhibits electrical conductivity due to the presence of ions.
The solution on evaporation yields $Na$ metal.
The solution of $Na$ metal in liquid ammonia is strongly reducing due to the presence of solvated electrons. the deep blue color of the ammonia, when dissolved with sodium, is actually due to solvated electrons.
Here there is formation of solvated electrons and not sodium amide and hydrogen gas.

A solution containing components A and B follows Raoults law

0%
A - B attraction force is greater than A - A and B - B
0%
A - B attraction force is less than A - A and B - B
0%
A - B attraction force remains same as A - A and B - B
0%
Volume of solution is different from sum of volume of solute and solvent

The Van't Hoff factor will be highest for:

0%
Sodium chloride
0%
Magnesium chloride
0%
Sodium phosphate
0%
Urea

Explanation

As the Van't Hoff factor depends on number of ions or particles produced on dissociation.

$i_{NaCl}=2$

$i_{MgCl_2}=3$

$i_{Na_3PO_4}=4$

$i_{urea}=1$

thus, the correct answer is C.

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

0%
$2$
0%
$0.25$
0%
$0.50$
0%
$3.0$

Explanation

The compound is

$Na[Ag(CN)_2]$ .

The dissociation reaction is,

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

1 0 0

$1- \alpha$

$\alpha$

Van't Hoff factor,

$i = 1 + \alpha$

For very dilute solution,

$\because \alpha = 1$

Hence,

$i = 2$

Therefore, option A is correct.

The incorrect statement is :

0%
Vapour pressure of a liquid always increases by increasing temperature.
0%
Vapour pressure only depends on temperature and not on the nature of substance.
0%
Vapour pressure does not depend on the quantity of the liquid taken and the surface area of the liquid.
0%
Vapour pressure is not a colligative property & is independent of the concentration of the liquid.

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

0%
1.4
0%
2.4
0%
3.4
0%
4.4

Explanation

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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