MCQ Questions for Class 12 Engineering Chemistry Solutions Quiz 10

$1.2$ % of solution of

$NaCl$ is isotonic with

$7.2$ % of solution of glucose. Calculate the van't Hoff's factor of

$NaCl$ solution.

0%
$2.36$
0%
$1.50$
0%
$1.95$
0%
$1.00$

Explanation

The molar masses of NaCl and glucose are 58.5 g/mol and 180 g/mol respectively.

The expression for the osmotic pressure is as follows:

$\pi =iCRT$

Since the two solutions are isotonic, they have equal osmotic pressure.

$\pi (NaCl)=\pi (glucose)$

$i_{NaCl}\times \dfrac{1.2}{58.5}=1\times \dfrac{7.2}{180}$

$i_{NaCl}=1.95$

Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of

$Al_2(SO_4)_3$ (if all are 100% ionised)?

0%
$Al(NO_3)_3$
0%
$K_4[Fe(CN)_6]$
0%
$K_2SO_4$
0%
$K_3[Fe(CN)_6]$

Explanation

Van't Hoff factor

$i =\dfrac {\text{number of solute particles present in solution}}{\text{theoretical number of solute particles due to solution of non electrolyte}} =\dfrac {n(observed)}{n(theoretical)}$ .

1 molecule of

$Al_2(SO_4)_3$ ionizes in solution to produce 5 ions.

$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$

Hence,

$i=\dfrac {n(observed)}{n(theoretical)}=\dfrac {5}{1}=5$ .

1 molecule of

$K_4[Fe(CN)_6]$ ionize in solution to produce 5 ions.

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

Hence,

$i=\dfrac {n(observed)}{n(theoretical)}=\dfrac {5}{1}=5$ .

Thus,

$K_4[Fe(CN)_6]$ has the same value of van't Hoff's factor (i) as that of

$Al_2(SO_4)_3$ (assuming 100% ionization).

Note: 1 molecule each of

$Al(NO_3)_3$ ,

$K_2SO_4$ and

$K_3[Fe(CN)_6]$ ionizes in solution to produce 4, 3 and 4 ions respectively.

For an ideal liquid solution, which of the following is unity?

0%
Activity coefficient
0%
Fugacity coefficient
0%
Fugacity
0%
Activity

Two liquids

$X$ and

$Y$ form an ideal solution at

$300\ K$ , the vapour pressure of the solution containing

$1\ mole$ of

$X$ and

$3\ moles$ of

$Y$ is

$550$ mm Hg. At the same temperature, if

$1\ mole$ of

$Y$ is further added to this solution, the vapour pressure of the solution increases by

$10\ mm\ Hg$ . Vapour pressure (in mm Hg) of

$X$ and

$Y$ in their pure states will be respectively :

0%
$200$ and $300$
0%
$300$ and $400$
0%
$400$ and $600$
0%
$500$ and $600$

Explanation

$P_{total}=P^o_A\cdot X_A+P^o_A\cdot X_B$

$550=P^o_A+\times \dfrac{1}{4}+P^o_B\times \dfrac{3}{4}$

$P^o_A+3P^o_B=2200$ ...(i)

When

$1\, mol$ of

$y$ is further added to the solution

$560=P^o_A\times \dfrac{1}{5}+P^o_B\times \dfrac{4}{5}$

$P^o_A+4P^o_B = 2800$ ...(ii)

On subtraction, (ii) - (i)

$P^o_B=2800-2200=600$

On putting the value of

$P^o_B$ in Eq. (i)

$P^o_A+3\times 600 = 2200$

$P^o_A=2200-1800=400$

Hence, the correct option is

$C$ .

Which of the following shows behavior of binary liquid solution?

0%
Plot of ${ 1 }/{ { \rho }_{ total } }$ vs ${ 1 }/{ { Y }_{ A } }$ , mole fraction of $A$ in vapour phase
0%
Plot of ${ 1 }/{ { \rho }_{ total } }$ vs ${ 1 }/{ { Y }_{ B } }$ is linear
0%
Plot of ${ 1 }/{ { \rho }_{ total } }$ vs ${ 1 }/{( { Y }_{ B } {Y}_{A})}$ is linear
0%
Plot of ${ 1 }/{ { \rho }_{ total } }$ vs ${ Y }_{ A }$ is linear

Explanation

For binary liquid solution, plot of

$\displaystyle \dfrac { 1 }{ { P }_{ total } }$ vs

$\displaystyle { Y }_{ B }$ is linear.

$\displaystyle \dfrac { { P }_{ B } }{ {P }_{ total } } = { Y }_{ B }$

$\displaystyle \dfrac { 1 }{ {P }_{ total } } \propto { Y }_{ B }$

The following is a graph plotted between the vapour pressure of two volatile liquids against their respective mole fractions.
Which of the following statements is/are correct ?

0%
When $x_A=1$ and $x_B=0$ , then $p =p^{\circ}_A$
0%
When $x_B=1$ and $x_A=0$ , then $p =p^{\circ}_B$
0%
When $x_A=1$ and $x_B=0$ , then $p =p^{\circ}_B$
0%
When $x_B=1$ and $x_A=0$ , then $p =p^{\circ}_A$

Explanation

In the graph, when

$x_A =1\,\, and \,\,x_B = 0$ , then

$p = p^oA$ since only vapour of A exist and when

$x_B =1\,\, and \,\,x_ A = 0$ then

$P=P^o_B$ since in that case only vapour of B exist.

Which one of the following is the ratio of the lowering of vapour pressure of 0.1 M aqueous solutions of

$BaCl_2 , NaCl$ and

$Al_2(SO_4)_3$ respectively?

0%
3 : 2 : 5
0%
5 : 2 : 3
0%
5 : 3 : 2
0%
2 : 3 : 5

Explanation

Lowering of vapour pressure is a colligative property. It depends on the number of particles of the solute.

$BaCl_2\rightleftharpoons Ba^{2+}+2Cl^- ; \ 3$ ions

$NaCl \rightleftharpoons Na^+ + Cl^- ; \ 2$ ions

$Al_2(SO_4)_3 \rightleftharpoons 2Al^{3+} + 3SO^{3-}_4 ; \ 5$ ions

Hence, the ratio of lowering of vapour pressure is

$3 : 2 : 5$ since the concentration of the solutions are same.

The correct statement among the following is:

0%
smoke is carbon dispersed in air
0%
butter is water dispersed in fat
0%
greater is the valency of ion more will be its coagulating power
0%
more is the gold number of a lyophobic sot, more is protecting power

Explanation

Smoke is carbon (solid particle) dispersed in air. It is the type of solid in gas type of solution.

Hence option

$A$ is correct.

Van't Hoff factor of centimolal solution of

${ K }_{ 3 }\left[ Fe{ \left( CN \right) }_{ 6 } \right]$ is

$3.333$ . Calculate the percent dissociation of

${ K }_{ 3 }\left[ Fe{ \left( CN \right) }_{ 6 } \right]$ .

0%
$33.33$
0%
$0.78$
0%
$78$
0%
$23.33$

The volume vs. temperature graph of 1 mole of an ideal gas is given below.

The pressure of the gas (in atm) at

$X,\ Y$ and

$Z$ , respectively, are:

0%
$0.328, 0.820, 0.820$
0%
$3.28, 8.20, 3.28$
0%
$0.238, 0.280, 0.280$
0%
$32.8, 0.280, 82.0$

Explanation

Using ideal gas equation,

$PV = nRT$

For gas

$X$ ,

$p=\dfrac{nRT}{V}; P=\dfrac{1 \times 0.082 \times 200 }{50}=0.32\ atm$

For gas $$Y$,

$P=\dfrac{1 \times 0.082 \times 500}{50}=0.820atm\,\,\,$

For gas

$Z$ ,

$P=\dfrac{1\times0.082 \times 200}{20}=0.820\ atm$

Hence, option

$A$ is correct.

The lowering in vapour pressure is maximum for:

0%
0.1 M urea
0%
0.1 M NaCl
0%
0.1 M MgCl
0%
0.1 M $K_4[Fe(CN)_6]$

Explanation

The lowering in vapour pressure is maximum for 0.1 M

$\displaystyle K_4[Fe(CN)_6]$ .

All the given solutions have the same molarity. For 0.1 M

$\displaystyle K_4[Fe(CN)_6]$ solution, the dissociation of one molecule of

$\displaystyle K_4[Fe(CN)_6]$ gives the maximum number of ions, (5 ions).

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

The lowering in vapour pressure is a colligative property and depends on the number of solute particles.

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

0%
It depends on the concentration of a non-electrolyte solute in solution and does not depend on the nature of the solute molecules.
0%
It depends on the number of particles of electrolyte solute in solution and does not depend on the nature of the solute particles
0%
It depends on the concentration of a non-electrolyte solute in solution as well as on the nature of the solute molecules.
0%
It depends on the concentration of an electrolyte or non-electrolyte solute in solution as well as on the nature of the solute molecules.

In which of the following solvents, KI has highest solubility?

0%
$C_6H_6(\in = 0)$
0%
$(CH_3)_2CO(\in = 2)$
0%
$CH_3OH(\in = 32)$
0%
$CCl_4(\in = 0)$

Explanation

KI is an ionic compound. Hence, it will be most soluble in a polar solvent, which has a high value of dielectric constant.

Hence,

$CH_3OH$ with the highest value of dielectric constant(32) among the given options will provide high solubility.

When a liquid that is immiscible with water was steam distilled at

$95.2^o$ C at a total pressure of

$99.652$ kPa, the distillate contained

$1.27$ g of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is

$85.140$ kPa at

$95.2^o$ C?

0%
$99.65\ g mol^{-1}$
0%
$18 \ g mol^{-1}$
0%
$134.1 \ g mol^{-1}$
0%
$105.74\ g mol^{-1}$

Explanation

Total pressure,

$P_{total}=99.652$ kPa

$P_{water}=p_B= 85.140$ kPa

$P_{liquid}=p_A=(99.652-85.140)$ kPa

$=14.512$ kPa

and

$\displaystyle\dfrac{m_A}{m_B}=\dfrac{1.27}{1}$

$\displaystyle\dfrac{m_A}{m_B}=\dfrac{p_AM_A}{p_BM_B}$

or,

$M_A=\left(\dfrac{m_A}{m_B}\right)\left(\dfrac{p_BM_B}{p_A}\right)$

$M_A=1.27\times \left(\displaystyle \frac{85.140kPa\times 18\ g \ mol^{-1}}{14.512\ kPa}\right)$

$M_A\simeq 134.1 \ g \ mol^{-1}$

Van't Hoff's factor of aq.

${ K }_{ 2 }S{ O }_{ 4 }$ at infinite dilution has value equal to:

0%
$1$
0%
$2$
0%
$3$
0%
Between $2$ and $3$

Explanation

${ K }_{ 2 }S{ O }_{ 4 }\longrightarrow 2{ K }^{ + }+S{ O }_{ 4 }^{ 2- }$

At infinite dilution, each molecule gives

$3$ ions, hence, the Van't Hoff's factor,

$i=3$ .

If the various terms in the below-given expressions have usual meanings, the van't Hoff factor (

$i$ ) cannot be calculated by which of the following expression?

0%
$\pi V=\sqrt { i } nRT$
0%
$\Delta { T }_{ f }=iK_{ f }.m$
0%
$\Delta { T }_{ b }=i{ K }_{ b }.m\quad$
0%
$\cfrac { { P }_{ solvent }^{ o }-{ P }_{ solution } }{ { P }_{ solvent }^{ o } } =i\left( \cfrac { n }{ N+n } \right)$

Explanation

Van't Hoff factor is simply a ratio. It is a ratio of abnormal colligative property to normal colligative property

$i=\cfrac {Observed\quad colligative\quad property }{ Normal\quad colligative\quad property }$

It can be calculated by the given formula

$i=\cfrac { \pi V }{ nRT } ;\quad i=\cfrac { \Delta { T }_{ f } }{ { K }_{ f }m } ;\quad i=\cfrac { \Delta { T }_{ b } }{ { K }_{ b }m }$ and

$i=\left( \cfrac { { P }^{ o }-{ P }_{ S } }{ { P }^{ o } } \right) \left( \cfrac { n+N }{ n } \right)$

$12$ g of a non-volatile solute dissolved in

$108$ g of water produces the relative lowering of the vapour pressure of

$0.1$ . The molecular mass of the solute is:

0%
$80$
0%
$60$
0%
$20$
0%
$40$

Explanation

$12$ g of a non-volatile solute dissolved in

$108$ g of water produces the relative lowering of vapour pressure of

$0.1$ . The molecular mass of the solute is 20.

$\displaystyle \dfrac { \Delta P }{ P^0 }=0.1$

$\displaystyle \dfrac { \Delta P }{ P^0 }=\dfrac { W_2M_1 }{ W_1M_2 }$

$\displaystyle 0.1=\dfrac { \text { 12 g } \times \text { 18 g/mol } }{ \text { 108 g } \times M_2 }$

$\displaystyle M_2=\text { 20 g/mol }$
Note:

$W_1$ and

$W_2$ are weights of water and solute respectively.

$M_1$ and

$M_2$ are molecular weights of water and solute respectively.

Two liquids

$X$ and

$Y$ from an ideal solution. At

$300K$ , a vapour pressure of the solution containing

$1$ mol of

$X$ and

$3$ mol of

$Y$ is

$550$

$mm \ Hg$ . At the same temperature, if

$1$ mol of

$Y$ is further added to this solution, a vapour pressure of the solution increased by

$10$

$mm \ Hg$ . Vapour pressure ( in mmHg) of

$X$ and

$Y$ in their pure states will be respectively:

0%
$300$ and $400$
0%
$400$ and $600$
0%
$500$ and $600$
0%
$200$ and $300$

Explanation

$\begin{array}{l} { P_{ Total } }={ X_{ x } }{ P_{ x } }+{ X_{ y } }{ P_{ y } }\, \, So,\, \, 550=\frac { 1 }{ 4 } { P_{ x } }+\frac { 3 }{ 4 } { P_{ y } } & \left[ \begin{array}{l} Here\, \, { p_{ y } }\, \, and\, \, { p_{ y } }\, \, and\, \, vapour\, \, pressure \\ of\, \, X\, and\, \, Y\, \, in\, \, pure\, \, states. \end{array} \right] \\ { P_{ x } }+3{ P_{ 4 } }=550\times 4;{ P_{ x } }+3{ P_{ y } }=2200\to \left( i \right) & \\ { 2^{ nd } }\, \, case & \\ { P_{ total } }=560\, \, mm\, \, Hg. & \\ \frac { 1 }{ 5 } { P_{ x } }+\frac { 4 }{ 5 } { P_{ y } }=560\, \, { P_{ x } }+{ 48_{ y } }=2800\to \left( { ii } \right) & \\ \left( { ii } \right) -\left( i \right) & \\ { p_{ y } }=600\, \, mm\, \, Hg\, \, { p_{ x } }=2200-1800\Rightarrow 400\, \, mm\, \, 07ng & \end{array}$

1 mole of liquid A and 9 moles of liquid B are mixed to form a solution. If

$P_B^o=400mm$ of Hg and

$P_B^o=200mm$ of Hg at a temperature 'T' and normal boiling point of liquid B is 300 K then answer the questions that follow.

Given data:

$K_b=2.7\ K\ kg\ mol^{-1},$ Molar mass of

$B=100$

If 'A' is perfectly non-volatile and it dimerises to an extent of 60% then what will be the vapour pressure of the solution.

0%
$360 mm$ of Hg
0%
$\displaystyle \frac{3600}{9.7} mm$ of Hg
0%
$\displaystyle \frac{4000}{9.7} mm$ of Hg
0%
$36 mm$ of Hg

The values of van't Hoff factors for

$KCl, NaCl$ and

$K_2SO_4$ respectively are:

0%
$1, 1, 2$
0%
$1, 1, 1$
0%
$2, 2, 3$
0%
$2, 3, 2$

Explanation

The values of van't Hoff factors for

$KCl, NaCl$ and

$K_2SO_4$ are

$2, 2, 3$ respectively.

$KCl \rightarrow K^+ + Cl^-$

$i = 2$

$NaCl \rightarrow Na^+ + Cl^+$

$i = 2$

$K_2SO_4 \rightarrow 2K^+ + SO^{2-}_4$

$i = 3$

Hence, option

$\text C$ is correct.

$CuSO_4$

$\cdot$

$5H_2O(s)$

$\rightleftharpoons$

$CuSO_4\cdot$ 3

$H_2O(s)$ +

$2H_2O(g)$ ;

$K_p$ = 4

$\times10^{-4}$

$atm^2$ if the vapour pressure of water is 38 torr then percentage of relative humidity is: ( Assume all data at constant temperature)

0%
$4$
0%
$10$
0%
$40$
0%
$none\ of\ these$

Explanation

${K_p} = P_{{H_2}O}^2 = 4 \times {10^{ - 4}}$

${P_{{H_2}O}} = 2 \times {10^{ - 2}}\,\,atm \Rightarrow 0.02\,\,atm$

$P_{{H_2}O}^1 = 38\,\,atm \Rightarrow \frac{{38}}{{760}}\,\,atm \Rightarrow \frac{1}{{20}}\,\,atm \Rightarrow 0.05\,\,atm$

$\% \,\,{\text{Relative}}\,\,humidity = \frac{{{P_{{H_2}O}}}}{{P_{{H_2}O}^1}} \Rightarrow \frac{{0.02}}{{0.05}} = \frac{2}{5} \times 100 \Rightarrow 40\,\,\%$

Option C is correct.

Two closed vessels of an equal volume containing air at pressure

$P_1$ and temperature

$T_1$ are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at

$T_1$ and that in the other at

$T_2$ , what will be the pressure in the vessels?

0%
$\dfrac{2P_1T_1}{T_1+T_2}$
0%
$\dfrac{T_1}{2P_1T_2}$
0%
$\dfrac{2P_1T_2}{T_1+T_2}$
0%
$\dfrac{2P_1}{T_1+T_2}$

Explanation

Initial mole

$=$ final mole

according to the ideal gas equation

$n=\dfrac{PV}{RT}$

$\dfrac{P_1V}{T_1R}+\dfrac{P_1V}{RT_1}=\dfrac{PV}{RT_1}+\dfrac{PV}{RT_2}$

$\dfrac{V}{R}\left(\dfrac{P_1}{T_1}+\dfrac{P_1}{T_1}\right)=\dfrac{V}{R}\left(\dfrac{P}{T_1}+\dfrac{P}{T_2}\right)$

$\dfrac{2P_1}{T_1}=P\left(\dfrac{1}{T_1}+\dfrac{1}{T_2}\right)$

$\dfrac{2P_1}{T_1}=P\left(\dfrac{T_2+T_1}{T_1T_2}\right)$

$\dfrac{2P_1T_1T_2}{T_1}=P(T_2+T_1)$

$P=\dfrac{2P_1T_2}{T_1+T_2}$ .

The vapour pressure of pure benzene at a certain temperature is

$640\ mm\ Hg$ . A non-volatile solute weighing

$2.175\ g$ is added to

$39.0\ g$ of benzene. The vapour pressure of solution is

$600\ mm\ Hg$ . What is the molar mass of the solute?

0%
$72.1\ g\ mol^{-1}$ .
0%
$70.6\ g\ mol^{-1}$ .
0%
$69.4\ g\ mol^{-1}$ .
0%
$56.2\ g\ mol^{-1}$ .

Explanation

$\begin{aligned} P_{A}^{0} &=640 \mathrm{~mm}{\mathrm{Hg}} \\ P_{S} &=600 \mathrm{~mm} {\mathrm{Hg}} \\ & \omega_{A}=39 \mathrm{~g} ; \quad \omega_{B}=2.175 \mathrm{~g} \end{aligned}$

$\begin{array}{l}\text{According to Rault's law (solution }\\ \text { is ideal or dilute), } \frac{P_{A}^{\circ}-P_{S}}{P_{S}}=\frac{n_{B}}{n_{A}}=\frac{\omega_{B}}{M_{B}} \times \frac{M_{n}}{\omega_{A}}\end{array}$

$\begin{array}{l} \Rightarrow \frac{640-600}{600}=\frac{2.175}{M _B} \times \frac{78}{39} \\ \Rightarrow \quad M_{B}=\frac{2.175 \times 78}{39} \times \frac{600}{40}=65.25 \\ \therefore \text { Molar mass of solute }=65.25 \mathrm{~g/mol} \end{array}$

Equal mass of

$H_2$ ,

$He$ and

$CH_4$ are mixed in empty container at

$300$ K, when total pressure is

$2.6$ atm. The partial pressure of

$H_2$ in the mixture is:

0%
$2.1$ atm.
0%
$1.6$ atm.
0%
$0.8$ atm.
0%
$0.2$ atm.

Explanation

According to Dalton's law, the partial pressure of a gas is equals to the product of total pressure and mole fraction.

We have

$H_2,He\:and\:CH_4$ gas with equal mass, m.

Total Pressure = 2.6 atm

$Partial\:Pressure of {H_2}=Total\:Pressure\times\chi_{H_2}$

$=2.6\times\dfrac{n_{H_2}}{n_{H_2}+{n_{He}}+{n_{CH_4}}}$

$=2.6\times\dfrac{\frac{m}{2}}{\frac{m}{2}+\frac{m}{4}+\frac{m}{16}}$

$=2.6\times\dfrac{8}{13}$

$Partial\:Pressure_{H_2}=1.6\:atm$

$18$ g glucose (

$C_6H_{12}O_6$ ) is added to

$178.2$ g of water. The vapour pressure of this aqueous solution at

$100^0C$ in torr is:

0%
$7.60$
0%
$76.00$
0%
$752.40$
0%
$759.00$

Explanation

Molecular mass of water

$=\left( 2\times 1 \right) +\left( 1\times 16 \right) =18g$

For

$178.2$ g water,

${ n }_{ A }=\dfrac{178.2}{18}=9.9\ mol$

Molecular mass of glucose

$=(6)(12)+(12)(1)+6(16)=180$ g.

For

$18$ g glucose,

${ n }_{ B }=\dfrac{18}{180}=0.1\ mol$

${ X }_{ B }=\dfrac{0.1}{\left( 0.1+9.9 \right) }=0.01$

${ X }_{ A }=1-0.01=0.99$

For lowering of vapour pressure,

$P={ P }_{ A }^{ 0 }{ X }^{ A }={ P }_{ A }^{ 0 }\left( 1-{ X }_{ B } \right)$

$P=760\left( 1-0.01 \right)$

$P=760-7.6$

$P=752.40$ torr

Vapour pressure of water is

$752.40$ torr.

Therefore, the correct option is

$C$ .

$CaCl_2.6H_2O(s) \rightleftharpoons CaCl_2(s) +6H_2O(g);$

$K_p=6.4\times 10^{-17}$ . Excess solid

$CaCl_2.6H_2O$ and

$CaCl_2$ are taken in a container containing some water vapours at a pressure of 1.14 torr at a particular temperature. Which of the following is true?

0%
$CaCl_2(s)$ acts as drying agent under given condition.
0%
$CaCl_2(s)$ acts as hygroscopic substance given condition.
0%
$CaCl_2.6H_2O(s)$ acts as effluoroscent substance
0%
Mass of $CaCl_2.6H_2O(s)$ increases due to some reaction.

Explanation

Effluroscence is a phenomenon when precipitation occurs directly on the surface.

$CaCl_2.6H_2O\:\leftrightharpoons\:CaCl_2(s)+6H_2O(g)$

$K_p=6.4\times 10^{-17}$

$K_p=(P_{H_2O})^6$

$P_{H_2O}=(2^6\times10^{-3\times 6})^{\dfrac{1}{6}}$

$P_{(H_2O)}=2\times 10^{-3}$

$P_{(H_2O)}=0.002\:atm\:=1.52\:torr$

The partial pressure of water is greater than the pressure of water vapour present in the vessel.

So, the solid will aggregate to form on the surface.

So,

$CaCl_2.6H_2O(s)$ acts as an effluoroscent substance.

Option C is correct answer

A gaseous mixture contains

$56\ g\ N_{2}. 44\ g\ CO_{2}$ and

$16\ g\ CH_{4}$ . The total pressure of the mixture is

$720\ mm\ Hg$ . The partial pressure of

$CH_4$ in mm Hg is:

0%
$620\ mm$ of $Hg$
0%
$180\ mm$ of $Hg$
0%
$160\ mm$ of $Hg$
0%
$200\ mm$ of $Hg$

Explanation

Given: Mass of $N_{2}$ = 56g

Mass of $CO_{2}$ = 44g

Mass of $CH_{4}$ = 16g

Solution: Moles of $N_{2}$ = $\dfrac{56}{28}$ = 2

Moles of $CO_{2}$ = $\dfrac{44}{44}$ = 1

Moles of $CH_{4}$ = $\dfrac{16}{16}$ = 1

Now, as we know, partial pressure of a gaseous component = mole fraction of the component * total pressure

Thus, partial pressure of $CH_{4}$ = ( $\dfrac{1}{2 + 1 +1}$ * 720) mm Hg

= 180 mm Hg

Thus, the correct option is (B).

${ PCl }_{ 5 }$ is 80% dissociated at

$250$ then its vapour density at room temperature will be:

0%
56.5
0%
104.25
0%
101.2
0%
52.7

Two containers, X and Y at 300K and 350K with water vapour pressures 22 mm and 40 mm respectively a are connected, initially closed with a valve. If the valves opened.

0%
The final pressure in each container is 31 mm
0%
The final pressure in each container is 40 mm
0%
Mass of $H_{2}O$ (l) in X increases
0%
Mass of $H_{2}O$ (l) in Y decreases roar.

Explanation

$\begin{array}{l} \text { - The vapour pressur in the cointainer } y \text { is higher } \\ \text { than the vapour pressure in the container } x \text { . } \\ \text { - When the value is opened, the vapour from } \\ \text { container y diffuse in the container } x \text { } \\ \text { and condense there. } \\ \text { - When the equilibrium is re-established, the } \\ \text { mass of liquid water in the contaier y } \\ \text { decreases and container } x \text { increases. } \\ \text { Hence the answer is, option ( } A \text { ) } \\ \text { i.e. The final pressure in each container is } 31 \mathrm{~mm} \text { . } \end{array}$

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 k Calculate vapour pressure of water at 298K?

0%
$4.53kPa$
0%
$3.53kPa$
0%
$5.53kPa$
0%
$6.53kPa$

The vapour pressure of

$C_6H_6$ and

$C_7H_8$ mixture at

$50^{\circ}C$ is given by

$p=179X_B+92$ where

$X_B$ is the mole fraction of

$C_6H_6$ .

Calculate (in mm) Vapour pressure of liquid mixture obtained by mixing

$936 \, g\ C_6H_6 \,$ and

$736\ g$ toluene is:

0%
300 mm Hg
0%
250 mm Hg
0%
199.4 mm Hg
0%
180.6 mm Hg

Explanation

Given,PM = 179

$X_B$ +92

For pure

$C_{6} h_{6}$ ,

$x_{B},$ i.e. mole fraction of benzene

$=1$

$\Rightarrow \quad P_{B}^{0}=179+92=271\mathrm{~mm}$

$x_{B}=\frac{\frac{936}{78}}{\frac{936}{78}+\frac{736}{92}}=\frac{12}{12+8}=0.60$

$\begin{aligned} x_{T} &=\frac{\frac{736}{92}}{\frac{736}{92}+\frac{936}{78}}=\frac{8}{12+8}=0.40 \\ P_{M}=& 179 x_{B}+92 \\=& 179 \times 0.6+92 \\=& 107.4+92 \\=& 199.4 \mathrm{~mm} \text { of } \mathrm{Hg} \\ \text { Hence, }(\mathrm{C}) \text { is correct option. } \end{aligned}$

The vapour pressure of two pure liquids A and B which form an ideal solution are 1000 and 1600 torr respectively at 400 K. A liquid solution of A and B for which the mole fraction of A is 0.60 is contained in a cylinder by a piston on which the pressure can be varied. The solution use slowly vapourised at 400 K by decreasing the applied pressure. What is the composition of last droplet of liquid remaining in equilibrium with vapour?

0%
$X_A= 0.30, X_B= 0.70$
0%
$X_A= 0.40, X_B= 0.60$
0%
$X_A= 0.70, X_B= 0.30$
0%
$X_A= 0.50, X_B= 0.50$

Explanation

$\begin{array}{l}\text{Here,} Y_A =\dfrac{ P_A^0 X_A }{P_A^0 X_A+ P^0_B(1-X_A)}\implies 0.6 =\dfrac{1000 X_A}{1000X_A+1600(1-X_A)}\\ \quad \therefore X_A=0.70 \quad X_B = 1-X_A =0.30 \end{array}$

2004004_878189_ans_b792462a43ab4499a61d477b0f33aea7.jpeg

Pressure of the gas in column (1) is :

0%
60 cm of Hg
0%
55 cm of Hg
0%
50 cm of Hg
0%
45 cm of Hg

$P_A \, = \, X_AP_A$ and

$P_B \, = \, X_BP_B$

$P_T \, = \, X_AP_A \, + \, X_BP_B$
Vapour pressure of mixtures of Benzene

$(C_6H_6)$ and toluene

$(C_7H_8)$ at

$50^{\circ}C$ are given by

$P_M \, = \, 179X_B \, + \, 92$ where

$X_B$ is mole fraction of

$C_6H_6$ .
What is the vapour pressure of pure liquids?

0%
$P_B \, = \, 92 mm, \, P_T \, = \, 179 mm$
0%
$P_B \, = \, 271 mm, \, P_T \, = \, 92 mm$
0%
$P_B \, = \, 180 mm, \, P_T \, = \, 91 mm$
0%
None of these

Explanation

$x_A+x_B=1$

$PT=x_AP_A+x_BP_B= (1-xB)P_A+x_BP_B$

$P_T=P_A+x_B(P_B-P_A) \longrightarrow (1)$

Given,

$P_M= 92+179x_B\longrightarrow (2)$

on comparing equation

$(1)$ &

$(2)$ we have

$P_A= 92 mm, P_B-P_A= 179mm$

$P_B= 179+P_A=179+92$

$P_A=92 mm, P_B=271mm$

The vapour density of undecomposed

${ N }_{ 2 }{ O }_{ 4 }$ isWhen heated, vapour density decreases to

$24.5$ due to its dissociation to

${ N O}_{ 2 }$ . The percentage dissociation of

${ N }_{ 2 }{ O }_{ 4 }$ at the final temperature is:

0%
$88$
0%
$60$
0%
$40$
0%
$70$

Explanation

The molecular weight of the substance

$=2\times$ vapor density.

Before dissociation,

$MW=2\times 46=92$

After dissociation,

$MW=2\times 24.5=49$

vant half factor

$(i)=\cfrac{\text{MW before dissociation}}{\text{MW after dissociation}}=\cfrac{92}{49}=1.8775$

Also,

$\alpha=\cfrac{i-1}{n-1}=\text{degree of dissociation which is the number of particles after dissociation.}$

$N_2O_4\longrightarrow 2NO_2;\quad n=2;\quad \alpha=\cfrac{1.8775-1}{2-1}=0.8775$

$\therefore\%\alpha=87.75\%\simeq 88\%$

Vapour pressure of

$CC{l}_{4}$ at

$25^{O}C$ is 143mm Hg. 0.5gm of a non-volatile solute (mol. wt. 65) is dissolved in 100ml of

$CC{l}_{4}$ . Find the vapour pressure of the solution.

(Density of

$CC{l}_{4} = 1.58 gm/{cm}^{3}$ )

0%
141.93 mm
0%
94.39 mm
0%
199.34 mm
0%
143.99 mm

Formation of a solution form two component can be considered as:
(1) Pure solvent

$\rightarrow$ separated solvent molecules,

${ \Delta H }_{ 1 }$
(2) Pure solute

$\rightarrow$ separated solute molecules,

${ \Delta H }_{ 2 }$
(3) Separated solvent and solute molecules

$\rightarrow$ solution,

${ \Delta H }_{ 3 }$

The solution so formed will be ideal if:

0%
${ \Delta H }_{ soln }\ =\ { \Delta H }_{ 1 }-{ \Delta H }_{ 2 }-{ \Delta H }_{ 3 }$
0%
${ \Delta H }_{ soln }\ =\ { \Delta H }_{ 3 }-{ \Delta H }_{ 1 }-{ \Delta H }_{ 2 }$
0%
${ \Delta H }_{ soln }\ =\ { \Delta H }_{ 1 }+{ \Delta H }_{ 2 }+{ \Delta H }_{ 3 }$
0%
${ \Delta H }_{ soln }\ =\ { \Delta H }_{ 1 }{ +\Delta H }_{ 2 }-{ \Delta H }_{ 3 }$

The vapour pressure of two pure liquids

$A$ and

$B$ , that form an ideal solution are

$100$ and

$900$ torr respectively at temperature

$T$ . This liquid solution of

$A$ and

$B$ is composed of

$1\ mole$ of

$A$ and

$1\ mole$ of

$B$ . What will be the pressure, when

$1$ mole of mixture has been vapourized?

0%
$800\ torr$
0%
$500\ torr$
0%
$300\ torr$
0%
None of these

Explanation

$P_A^o = 100 \, torr \,\,\,\, P_B^o = 900 \, torr$

$n_A = 1 \, mole \,\,\,\, n_B = 1 \, mol$

$x_A = \dfrac{1}{1 + 1} = 0.5 mole$

$x_B = \dfrac{1}{1 + 1} = 0.5 mole$

Pressure

$= P_A^o x_A + P_B^o x_B$

$= 100 \times 0.5 + 900 \times 0.5 = 50 +450 = 500 \, torr$

Solubility of

$CO_{2}(g)$ in water at

$1\ bar$ and

$25^{\circ}C$ involves the following equilibria:

$CO_{2}(g) \rightleftharpoons CO_{2}(aq); K_{1}^{\circ} = 0.04$

$CO_{2}(aq.) + H_{2}O(l) \rightleftharpoons H^{+}(aq.) + HCO_{3}^{-}(aq); K_{2}^{\circ} = 4\times 10^{-6}$

$HCO_{3}^{-}(aq.) \rightleftharpoons H^{+}(aq.) + CO_{3}^{2-}(aq); K_{3}^{\circ} = 5\times 10^{-11}$

$[H^{+}]$ in saturated solution of

$CO_{2}(g)$ water is ?

0%
$0.04\ M$
0%
$4\times 10^{-3}M$
0%
$4\times 10^{-4}M$
0%
$2\times 10^{-3}M$

The vapour pressure of two pure liquids A and B, that form an ideal solution are

$100$ and

$900$ torr respectively at temperature T. This liquid solution of A and B is composed of

$1$ mole of A and

$1$ mole of B. What will be the pressure, when

$1$ mole of mixture has been vapourized?

0%
$800$ torr
0%
$500$ torr
0%
$300$ torr
0%
None of these

Explanation

Vapour pressure of pure A

$P_{A}^{\circ}=100 torr$

Vapour presuure of pure B

$P_{B}=900 torr$

$n_{A}=1 mole$

$n_{B}= 1 mole$

$x_{A}=\dfrac{1}{1+1}=0.5$

$x_B=\dfrac{1}{1+1}=0.5$

$P=P_{A}^{\circ}x_{A}+P_{B}^{\circ}x_{B}$

$=100\times 0.5+900\times 0.5=500torr$

Two moles of pure liquid 'A'

$(P_{A}^{0} =80mm$ of Hg) and

$3$ moles of pure liquid 'B' (

$P_{B}^{0} = 120mm$ of Hg) are mixed. Assuming ideal behaviour?

0%
Vapour pressure of the mixture is $104$ mm of Hg
0%
Mole fraction of liquid 'A' in Vapour pressure is $0.3077$
0%
Mole fraction of 'B' in Vapour pressure is $0.692$
0%
Mole fraction of 'B' in Vapour pressure is $0.785$

Explanation

$P_{A}^{\circ}=80 mm Hg$

$P_{B}^{\circ}=120 mm Hg$

$n_{A}=2 moles$

$n_{B}=3 moles$

$x_{A}=\dfrac{2}{2+3}=\dfrac{2}{5}=0.4$

$x_{B}=\dfrac{3}{5}=0.6$

$P_{A}=P_{A}^{\circ}x_{A}=80\times 0.4=32 mm Hg$

$P_{B}=P_{B}^{\circ}x_{B}=120\times 0.6=72 mm Hg$

Total vapour pressure

$P=P_{A}+P_{B}$

$=32+72=104 mm Hg$

Vapour pressure of

$CCl_4$ at

$25^0$ C is

$143$ mm Hg.

$0.5$ gm of a non-volatile solute (mol.wt.

$65$ ) is dissolved in

$100$ ml of

$CCl_4$ . Find the vapour pressure of the solution. (Density of CCl_4 =

$1.58 gm/cm^3$ )

0%
$141.93$ mm
0%
$94.39$ mm
0%
$199.34$ mm
0%
$143.99$ mm

Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice:

Assertion
Amalgam of mercury with sodium is an Assertion as: Amalgam of solid solutions.
Reason
Mercury is solvent and sodium is solute in the solution.

0%
If both assertion and reason are true and reason is the correct explanation of assertion.
0%
If both assertion and reason are true but reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both assertion and reason are false.

Two liquids

$X$ and

$Y$ are perfectly immiscible. If

$X$ and

$Y$ have molecular masses in ration

$1 : 2$ , the total vapour pressure of a mixture of

$X$ and

$Y$ prepared in weight ratio

$2 : 3$ should be:

$(P_{X}^{0} = 400\ torr, P_{Y}^{0} = 200\ torr)$ .

0%
$314\ torr$
0%
$466.7\ torr$
0%
$600\ torr$
0%
$700\ torr$

Explanation

Let molecular mass of

$X = M$ and molecular mass of

$Y = 2M$

vapour pressure of

$X \, P_{x_0}^0 = 400$ torr

vapour pressure of

$Y \, P_y = 200$ torr

Let weight of

$X = 2w$

weight of

$Y = 3w$

Mole fraction

$X_x = \dfrac{\dfrac{2w}{m}}{\dfrac{2w}{m} + \dfrac{3w}{2m}} = \dfrac{4}{7}$

Mole fraction =

$Y_x = 1 - \dfrac{4}{7} = \dfrac{3}{7}$

$P = P_x^0 X_x + P_y^o Y_x$

$= 400 \times \dfrac{4}{7} + 200 \times \dfrac{3}{7} = \dfrac{2200}{7}$

$= 314 \, torr$

The vapour pressure of three liquids

$P, Q$ and

$R$ , of nearly equal molecular masses is shown as a function of temperature.

The correct statement is:

0%
The normal boiling points follow the order $P > Q > R$
0%
The variation of pressure with respect to temperature for each liquid is given by $\displaystyle\frac{dP}{dT}=\frac{K}{T^2}$ , where K is a constant
0%
The strength of intermolecular interactions follows the order $P > Q > R$
0%
The normal boiling point of $Q$ is close to $65^o$ C

Explanation

The normal boiling points follow the order

$P < Q < R$ . The normal boiling point can be obtained from the graph. It is temperature at which vapour pressure is close to

$100 kPa$ .
The variation of pressure with respect to temperature for each liquid cannot be given by

$\displaystyle\frac{dP}{dT}=\frac{K}{T^2}$ , where

$K$ is a constant. This is because as temperature increases, the slope (

$\displaystyle\frac{dP}{dT}$ ) increases.
The strength of intermolecular interactions follows the order

$P < Q < R.$ Higher is the normal boiling point, stronger are the intermolecular interactions.
The normal boiling point of

$Q$ is close to

$65^oC$ .

This can be obtained from the graph.

$X,Y$ and

$Z$ in the given graph are?

0%
$X={p}_{1}+{p}_{2},Y=1,Z=0$
0%
$X={p}_{1}+{p}_{2},Y=0,Z=0$
0%
$X={p}_{1}\times {p}_{2},Y=0,Z=0$
0%
$X={p}_{1}-{p}_{2},Y=1,Z=0$

Explanation

Here,

$X =P_T=p_1+p_2=$ Total pressure of the solution.

$p_1$ and

$p_2$ is the partial pressure of component 1 and 2 respectively.

$Y=x_2=$ mole fraction of component 2 in the liquid phase.

$Z=y_2=$ mole fraction of component 2 in the vapour phase.

When mole fraction of component 2 in the liquid phase is 0 then mole fraction of component 2 in the vapour phase is also 0 which means the liquid is pure of component 1 only.

So, the correct option is B.

A chemist has

$m$ gm of salt water that is

$m$ % salty. How many gram of salt must he add to make a solution that is

$2m$ % salty?

0%
$\dfrac {m}{100 + m}$
0%
$\dfrac {2m}{100 - 2m}$
0%
$\dfrac {m^{2}}{100 - 2m}$
0%
$\dfrac {m^{2}}{100 + 2m}$

Two liquids X and Y are perfectly immiscible. If X and Y have molecular masses in ratio 1 : 2, the total vapour pressure of a mixture of X and Y prepared in weight ratio 2 : 3 should be (

$Px^0 = 400 torr, Py^0 = 200 torr$ )

0%
314 torr
0%
466.7 torr
0%
600 torr
0%
700 torr

Explanation

Molecular mass of

$X = M$

Molecular mass of

$Y = 2M$

weight of

$X = 2W$

weight of

$Y = 3w$

no.of mole of

$X = \dfrac{2w}{M}$

no. of mole of

$Y = \dfrac{3w}{2m}$

mole fraction of

$X = Xx = \dfrac{\dfrac{2w}{m}}{\dfrac{2w}{M} + \dfrac{3w}{2M}}$

$X_x = \dfrac{4}{7}$

Mole fraction of

$Y = Xy = \dfrac{3}{7}$

$P_x^o = 400 \, torr$

$P^o_y = 200 \, torr$

$P = \dfrac{4}{7} \times 400 + 200 \times \dfrac{3}{7}$

$= \dfrac{2200}{7} = 314 torr$

option

$A$ may be correct approximately

CuSO

$_4.$ 5

$H_2O$ (s)

$\rightleftharpoons$ CuSO

$_4$ .3H

$_2O$ (s) + 2H

$_2O$ (g);

$k_p$ = 4

$10^{4}$ atm

$^2.$ If the vapour pressure of water is 38 torr then percentage of relative humidity is : (Assume all data at constant temperature)

0%
4
0%
10
0%
40
0%
None of these

Explanation

$\begin{aligned} K_ P &=\left(P_{ H_{2} O}\right)^{2} \\ P_{ H_{2} O }&=\sqrt{K_ P} \\ &=\sqrt{4 \times 10^{-4}} \\ &=2 \times 10^{-2} \end{aligned}$

$\text { % Relative Humidity }=\dfrac{P_{H_2O} \times 760}{\text { v.p of water}}$

$\begin{aligned} =& \dfrac{2 \times 10^{-2} \times 760 \times 100}{38} \\ &=40 \end{aligned}$

The relative lowering in vapour pressure is proportional to the ratio of number of :

0%
solute molecules to solvent molecules
0%
solvent molecules to solute molecules
0%
solute molecules to the total number of molecules to the total number of molecules in solution
0%
solvent molecules to the total number of molecules in solution

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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