MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 14

The figure shows the graphs of pressure versus density for an ideal gas at two temperatures,

$T_1$ and

$T_2$ , according to this which is correct?

0%
$T_1 > T_2$
0%
$T_1 = T_2$
0%
$T_1 < T_2$
0%
none of the above

Henry's law constant K of

$CO_{2}$ in water at

$20^{\circ}C$ is

$3.0\times 10^{-2}molL^{-1}atm^{-1}$ . Calculate the mass of

$CO_{2}$ present in 100 L of soft drink bottled with a partial pressure of

$CO_{2}$ of 4 atm at the same temperature

0%
5.28 g
0%
12.0 g
0%
428 g
0%
528 g

When gases are heated from

${20}^{o}C$ to

${40}^{o}C$ at constant pressure, then the volume

0%
increase by the same magnitude
0%
become double
0%
increase in the ratio of their molecular masses
0%
increase but to different extent

Two flasks of equal volume, connected by a narrow tube of negligible volume, contain $2$ moles of $H_2$ gas at $1$ atm pressure and $300$ K. Now, one of the flasks are heated to $400$ K and the other is maintained at $300 K$ , then find final pressure:

0%
$\cfrac {8}{5}$ atm
0%
$\cfrac {5}{3}$ atm
0%
$\cfrac {7}{5}$ atm
0%
$\cfrac {8}{7}$ atm

Two gases X and Y have densities,

${d_{(x)}} = 3{d_{(y)}}$ and molecular masses,

${M_{(x)}} = 0.5{M_{(y)}}$ . Then, the ratio of their pressures, i.e,

${P_x}:{P_y}$ would be

0%
1/4
0%
1/6
0%
4
0%
6

The vapour pressure of water at T(K) is 20 mm Hg. The following solutions are prepared at T(K) :

0%
6 g of urea (mol. wt. = 60) is dissolved in 178.2 g of water
0%
0.01 mole of glucose is dissolved in 179.82 g of water
0%
5.3 g of $N{a_2}C{O_3}$ (mol. wt. =106) is dissolved in 179.1 g of water.
0%
non of these

${X_A}$ and

${X_B}$ represent mole fraction in liquid phase while

${y_A}$ and

${y_B}$ represent mole fraction in vapour phase in binary solution. Which of the following statements is/are correct?

0%
At point C liquid and vapour phase have same composition
0%
Vapour pressure of B is less than A
0%
Intermolecular forces in the solution are weaker than between like particles
0%
All of these

Two liquids A & B form an ideal solution. What is the vapour pressure of solution containing 2 moles of A and 3 moles of B at 300 K? [Given: At 300 K, Vapour pr. of pure liquid A

$\left( P\begin{matrix} 0 \\ A \end{matrix} \right)$ =100 torr, Vapour pr. of pure liquid B

$\left( P\begin{matrix} 0 \\ B \end{matrix} \right)$ =300 torr]

0%
200 torr
0%
140 torr
0%
180 torr
0%
None of these

A certain amount of ideal gas (P = 5 atm, V= 2L, T = 500K ) in state A is compressed to state B (P = 2 atm, V = "V"L, T = 100K) .The final volume V of gas in state B in litre is:

0%
1
0%
2
0%
3
0%
4

1.2 L of oxygen at a constant pressure of 2.00 atm was kept in a cylinder and provided 10.00 K cal of heat. The volume of oxygen increases to 1.8 L. The value of

$\Delta E$ is

0%
9.970 Kcal
0%
0.9970 Kcal
0%
10.100 Kcal
0%
10.970 Kcal

Find the slope of the curve plotted between

$\mathrm { P }$ Vs

$T$ for closed container of volume

$2$

$\mathrm { L }$ having same moles of gas:

0%
$\dfrac { e } { 2000 }$
0%
$2000$ $\mathrm { e }$
0%
$500$ $\mathrm { e }$
0%
$\dfrac { 2} { 1000 e }$

The law which suggests

$n_1 = n_2$ for two solutions at same temperature and pressure is

0%
van't Haff - avogadro's law
0%
van't Hoff Boyle's law
0%
van't Hoff's law
0%
Henry's law

An ideal solution contains two volatile liquids

$A (P^0=100 \ torr)$ . If the mixture contains 1 mole of A and 3 moles of B, the total vapour pressure of distillate is:

0%
150 torr
0%
188.88 torr
0%
185.72 torr
0%
198.88 torr

$Fc$ and

$Fa$ denote cohesive and adhesive force on a liquid molecule near the surface of a solid. then the surface of request is concave , when;

0%
$F_{A}<\frac{F_{0}}{\sqrt{2}}$
0%
$F_{A}=\frac{F_{c}}{\sqrt{2}}$
0%
$F_{n}>\frac{F_{c}}{\sqrt{2}}$
0%
$F_{A}>F_{C}$

A square frame of side L is dipped in a liquid. On taking out, a membrane is formed. If the surface tension of the liquid is T, then acting on one side of the frame will be :

0%
$2 TL$
0%
$8 TL$
0%
$4 TL$
0%
$\dfrac { TL }{ 2 }$

What is the total pressure (atm) in the chamber ?

0%
83.14
0%
831.4
0%
8.21
0%
None

For which of the following reactions, Gay Lussac's law is not applicable:

0%
Formation of $HI$ from its constituents
0%
Formation of $N H_{3}$ from its constituents
0%
Formation of $C O_{2}$ from its constituents
0%
Formation of $S O_{3}$ from $S O_{2}$ and $O_{2}$

According to kinetic theory of gases:

0%
Collisions are always elastic.
0%
Heavier molecules transfer more momentum to the wall of the container.
0%
Only a small number of molecules have very high velocity.
0%
Between collisison the molecules move in straight lines with constant velocities.

Both stopcocks are opened and the system is again allowed to come at equilibrium. The pressure throughout the system is

$33.5$ mm Hg. What do bulbs A, B and C contain?

0%
$A:N_2(g), B:N_2(g), H_2O(s), C:N_2(g), H_2O(s), CO_2(s)$
0%
$A:N_2(g), B:N_2(g), H_2O(s), C:N_2(g), CO_2(s)$
0%
$A:N_2(g), B:N_2(g), H_2O(s), CO_2(s), C:N_2(g), H_2O(s), CO_2(s)$
0%
$A:N_2(g), B:N_2(g), H_2O(s), CO_2(g), C:N_2(g)m H_2O(s), CO_2(s)$

The stopcock between A and B is opened and the system is allowed to come to equilibrium. The pressure in A and B is now

$0.21$ atm.

How many moles of

$H_2O$ are in the system?

0%
$0.0075$
0%
$0.015$
0%
$0.05$
0%
$0.035$

Explanation

No. of moles in flask A

Given:
P= 570 mm of Hg = 570/ 760 atm, V=1.642 L, T = 300 K

$n=\dfrac{PV}{RT}=\dfrac{570\times 1.642}{760\times 0.082\times 300}$

$n=0.05$ moles

$n_{N_2}+n_{CO_2}+n_{H_2O}=0.05$ moles

Upon opening the stop cock, the gases will dispose

In flask A:

$CO_2+N_2=(n_{CO_2}+n_{N_2})$ ……

$(1)$

In flask B:

$CO_2+N_2+H_2O=(n_{CO_2}+n_{H_2}+n_{H_2O})$ ……….

$(2)$

$(2) - (1)$

$\Rightarrow n_{CO_2}+n_{N_2}+n_{H_2O}-(n_{CO_2}+n_{N_2})=n_{H_2O}$

in flask A

no. of moles

$=\dfrac{0.21\times 1.642}{0.082\times 300}=0.014$
in flask B

no. of moles

$=\dfrac{0.21\times 1.642}{0.082\times 200}=0.021$

Total moles

$=(0.014+0.021)=n_{H_2O}$

$n_{H_2O}=0.05-0.035$

$=0.015$ Option B.

In a mixture of

$A$ and

$B$ , having vapour pressure of pure

$A$ and pure

$B$ as

$400mm\, Hg$ and

$600mm\, Hg$ respectively, mole fraction of

$B$ in liquid phases is

$0.5$ . Calculate total vapour pressure and mole fraction of

$A$ and

$B$ in vapour phases.

0%
$500, 0.4, 0.6$
0%
$500, 0.5, 0.5$
0%
$450, 0.4, 0.6$
0%
$450, 0.5, 0.5$

Explanation

$P_T = X_AP_{A^o} + X_BP_{B^o}$

$= 0.5\times 400+0.5\times 600$

$=500nm$ of

$Hg$

$\dfrac{1}{P_T} = \dfrac{y_A}{P^o_A} + \dfrac{1-y_B}{P^o_B}$

$y_A = 0.6$ &

$y_B = 1 -0.6 = 0.4$

In the case of positive deviation from an ideal gas?

0%
Interactions in molecules, $\dfrac{PV}{nRT} > 1$
0%
Interactions in molecules, $\dfrac{PV}{nRT} < 1$
0%
Finite size of molecules, $\dfrac{PV}{nRT} > 1$
0%
Finite size of molecule, $\dfrac{PV}{nRT} < 1$

Explanation

Condition for positive deviation of ideal gas

$Z=\dfrac{PV}{nRT}$

$Z > 1$ for positive deviation

Z(compressibility factor is also the ratio of actual molar mass of volume of gas to theoretical molar volume)

According to kinetic theory of gases:-

In positive deviation the interactions between molecules are very less, almost negligible.

The size of the molecules is negligible, but under the condition of deviation, from ideal gas equation, the size of molecules is not negligible.

$\therefore$ Size of molecules is finite,

$Z=\dfrac{PV}{nRT} > 1$
Option C.

A vessel of volume

$10\ l$ is evacuated by means of a piston air pump. One piston stroke captures the volume

$1\ l$ . The process is assumed to be isothermal and the gas ideal. The initial pressure of gas in the vessel was

$24.2$ atm. Select the correct statement(s).

0%
The pressure of gas remained in the vessel after first stroke is $22$ atm
0%
The pressure of gas remained in the vessel after second stroke is $20$ atm
0%
The number of strokes needed to reduce the pressure in the vessel $\eta$ times in $\dfrac{1 n\eta}{1n 1.1}$
0%
The pressure of gas remained in the vessel after 'n' strokes is $24.2\times \left(\dfrac{10}{11}\right)^n$ atm

An ideal gas obeying kinetic theory of gases can be liquified, if?

0%
Its temperature is more than critical temperature $T_c$
0%
Its pressure is more than critical pressure $P_c$
0%
Its pressure is more than $P_c$ at a temperature less than $T_c$
0%
It cannot be liquified at any value of P and T

One monoatomic gas is expanded adibatically from

$2L$ to

$10L$ at

$1\ atm$ external pressure find

$\triangle U$ (in atm

$L$ )?

0%
$-8$
0%
$0$
0%
$-66.7$
0%
$58.2$

Explanation

Process is adiabatic

$\therefore Q = 0$

$\therefore \triangle U = W = -P_{ext} \triangle V$

$= -1 (10 - 2)\ atm \ L$

$= -8\ atm\ L$ .

When the air temperature is below freezing, the saturation vapor pressure over water is _______.

0%
Equal to zero
0%
Less than the saturation vapor pressure over ice
0%
Greater than the saturation vapor pressure over ice
0%
Equal to the saturation vapor pressure over ice

A mixture contains 1 mole of volatile liquid

$A (P^0_A = 100 mm Hg )$ and 3 moles of volatile liquid

$B (P^0_B = 80 mm Hg )$ if the solution behaves ideally, the total vapour pressure of the distillate is

0%
85 mm Hg
0%
85.88 mm Hg
0%
90 mm Hg
0%
92 mm Hg

Explanation

Mole fraction of A is

$X_A=\frac{1}{1+3}=0.25$

Mole fraction of B is

$X_B=1-0.25=0.75$

The expression for the total pressure of the solution is

$P=P_A^0X_A+P_B^0X_B$

Substitute values in the above expression

$P=100\times 0.25+80\times 0.75=85$ mm of Hg

Hence the correct answer is [A]

An aqueous solution of sucrose is 0.5 molal. What is the vapour pressure of water above this solution ? The vapour pressure of pure water is 25.0 mm Hg at this temperature.

0%
24.8 mm Hg
0%
0.45 mm Hg
0%
2.22 mm Hg
0%
20.3 mm Hg

Explanation

Given that

$m = 0.5\;\;molal$

V.P. (pure water) 25.0mm Hg

According to the Raoult's law for dilute solution

$\dfrac{P_0 - P_s}{P_0} = \dfrac{w}{m+W}\times 1000 = X_{sucrose}$

here the 0.5 molal of solution means 0.5 mole in 1kg of water

Henc the mole of solvent =

$\dfrac{1000}{18}=55.5$ [ the molecular mass of water molecules is 18]

then

$\dfrac{25-P_s}{25} = \dfrac{0.5}{55.57+0.5}$

$\Rightarrow\;1401.75 - 56.07P_s = 12.5$

$\Rightarrow \; 56.07P_s = 1407.75 - 12.5$

$\Rightarrow \; P_S = \dfrac{1389.25}{\;\;56.07}$

$24.8\;mm \;Hg$

Which of the following behaviour is true about the ideal binary liquid solution of liquids A and B if

$P^0_A < P^0_B$ ?

0%
plot of $P_{total}| vs X_A$ in non linear
0%
plot of $P_{total}| vs X_B$ is linear with +ve slope
0%
plot of $P_{total}| vs X_B$ is linear with slope = 0
0%
plot of $P_{total}| vs X_B$ is linear with -ve slope

Benzene and toulene from an ideal solution, the vapour pressures of the benzene and toluene are 75 mm and 25 mm respectively at

$20^0 C$ if the mole fractions of benzene and toluene in vapour are 0.75 and 0.25 respectively the vapour pressure of the ideal solution is

0%
62.5 mm
0%
50 mm
0%
30 mm
0%
40 m

Liquids A and B from an ideal solution. A certain solution of A and B contains 25 mole percent of A whereas the vapours in the equilibrium with the solution at 298 K contains 50 mole per cent of A. The ratio of vapour pressures of pure A to that of pure B at 298 K , is

0%
1 :1
0%
3 : 1
0%
1 :3
0%
2 :1

The ratio between lowering of vapour pressure of solution and mole fraction of solute is equal to

0%
relative lowering of vapour pressure
0%
vapour pressure of pure solvent
0%
vapour pressure of solution
0%
molar mass of solvent

How many moles of

$CO_2$ are in the system?

0%
$0.0125$
0%
$0.015$
0%
$0.0225$
0%
$0.0375$

Heptane and octane from ideal solution at 373 K the vapour pressure of the pure liquids are 106 kPa and 46 kpa respectively . what will be the vapour pressure, in bar of mixture of 30.0 g of heptane and 34.2 g of octane ?

0%
76 bar
0%
152 bar
0%
1.52 bar
0%
0.76 bar

For an ideal solution of A and B ,

$Y_A$ is the mole fraction of A in the vapour phase at equilibrium . which of the following plot should be linear ?

0%
$P_{total} vs Y_A$
0%
$P_{total} vs Y_B$
0%
$\frac {1}{P_{total } } vs Y_A$
0%
$\frac {1}{P_{total} } vs \frac {1}{Y_A }$

The vapour pressure of pure liquid A at

$80^0 C$ is

0%
807.4 mm
0%
511.1 mm
0%
755.6 mm
0%
533.3 mm

If the pressure over the mixture at 300 K is reduced at what pressure does the first vapour form?

0%
40 mm Hg
0%
70 mm Hg
0%
100 Mm Hg
0%
199 mmHg

A liquid solution is formed by mixing 10 moles of anline and 20 moles of phenol at a temperature where the vapour phenol are 90 and 897 mm Hg, respectively . The possible vapour pressure of solution at that temperature is ?

0%
82 mm Hg
0%
88 mm Hg
0%
90 mm Hg
0%
94 mm Hg

If the pressure is reduced further, at what presure does the trace of liquid disppear?

0%
57.14 mm Hg
0%
40 Mm Hg
0%
100 mm Hg
0%
66.67 mm Hg

The vapour pressure of pure liquids A, B and C are 75, 22 and 10 torr , respectively. Which of the following is /are possible values of vapour pressure of binary or ternary solutions having equimolar amounts of these liquids? Assume ideal behaviour for all possible solutions

0%
53.5 torr
0%
35.67 torr
0%
48.5 torr
0%
16 torr

Explanation

$P_{AB}= \cfrac{75+22}{2}=48.5$ torr

$P_{BC}= \cfrac{22+10}{2}=16$ torr

$P_{AC}= \cfrac{75+10}{2}=42.5$ torr

$P_{ABC}= \cfrac{75+22+10}{2}=35.67$ torr

Option A, B and C are correct.

The vapour pressure of the solution this composition is:

0%
$\sqrt { P^0_A .P^0_B }$
0%
$( P^0_A - P^0_B )$
0%
$( P^0_A + P^0_B )$
0%
$0.5 ( P^0_A + P^0_B )$

A liquid mixture of A and B (assume ideal solution ) is placed in a cylinder and piston arrangement . The piston is slowly pulled out isothermally so that the volume of liquid decreases and that of vapour increases. at the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of 'A' in the vapour is 0.4 .If

$P^0_A = 0.4 atm , P^0_B = 1.2 atm$ at this temperature, the total pressure at which the liquid has almost evaporated is ?

0%
0.667 atm
0%
1.5 atm
0%
0.8 atm
0%
0.545 atm

One having high vapour pressure at temperature below its melting point is?

0%
Benzoic acid
0%
Salicylic acid
0%
Citric acid
0%
All of these

Explanation

Benzoic acid appears as a white crystalline solid. Vapour pressures of benzoic acid were measured in the temperature range

$316\,K$ to

$391 \,K$ which is below its melting point

$395.1\, K$ . Hence correct answer is option A.

The vapour pressure of pure liquid A at

$27^0 C$ is

0%
60 torr
0%
48 torr
0%
49.26 torr
0%
61.58 torr

For ideal binary solution

$p=x_{A}.p_{A}^{0}+x_{B}.p_{B}^{0}$
This equation reflects

0%
none of these
0%
Boyle's law
0%
Charles's law
0%
Dalton's law of partial pressure

The reading of pressure gauge at bubble point is :

0%
500
0%
600
0%
700
0%
None

Explanation

$X_A = 0.75 \quad X_B = 0.05$ above 500 mm Hg only liquid phase exists .

$P_{bubble} point = X_A P^0_A + X_B P^0_B$

$P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B$

$0.75 +400 + 0.25 \times 800 = 500 mm$

$y_A = 0.75 \quad \quad y_B = 0.5$
at dew point

$\frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800}$

$\Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg$
Below dew point only vapor phase exists.

The reading of pressure gauge at which only vapour phase exists is

0%
501
0%
457.14
0%
425
0%
525

Explanation

$X_A = 0.75 \quad X_B = 0.05$ above 500 mm Hg only liquid phase exists .

$P_{bubble} point = X_A P^0_A + X_B P^0_B$

$P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B$

$0.75 +400 + 0.25 \times 800 = 500 mm$

$y_A = 0.75 \quad \quad y_B = 0.5$
at dew point

$\frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800}$

$\Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg$
Below dew point only vapor phase exists.

$5 \,cm^3$ of acetone is added to

$100 \,cm^3$ of water, the vapour pressure of water over the solution

0%
It will be equal to the vapour pressure of pure water
0%
It will be less than the vapour pressure of pure water
0%
It will be greater than the vapour pressure of pure water
0%
It will be very large

The vapour pressure of two micible liquid A and B are 300 and 500 mm of Hf respectively. in a flask , 2 moles of A are mixed with moles of B. further to the mixture , 32 g of an ionic non -volatile solute MCI (partially ionised, mol mass = 70 u) were also added. thus, the final vapour pressure of solution was found to be 420 mm and Hg. Then, identify the correct statement(s). (Assume the liquid mixture of A and B to behave ideally).

0%
The numerical value of relative lowering in vapour pressure upon addition of solute MCI is 1/15
0%
Th solute MCI is 25 % ionised in the above question.
0%
The solute MCI is 23.33 % ionised in the above question
0%
Upon addition of excess $Pb(NO_3)_2$ the number of moles of $PbCl_2$ precipitated is 2/ 35 .

Explanation

$P_0= X_A P^0_A + X_B P^0_B = \frac {2}{8} \times 300 + \frac {6}{7} \times 500 = 450$ mm of Hg
Now RLVP =

$\frac {P_0 -P_s}{P_0} = \frac { 450 - 420}{450} = \frac {1}{15}$
Also RLVP =

$( \frac {in}{In +N } )$

$\frac {1}{15} = \frac { \frac { 32i}{70} }{ \frac {32i}{70} + 8 }$

$\therefore i - 1.25 = ( 1+ ( 2-1 ) \alpha ).$ .
So

$\alpha$ = 0.25 ( or 25 %)

$\therefore n_{CI^- }$ produced =

$\frac {25}{100} \times \frac {32}{70} = \frac {4}{35}$

$\therefore n_{PbCl_2 }$ precipitated =

$\frac {1}{2} \times \frac {4}{35} = \frac {2}{35}$

If the pressure of the gas contained in a closed vessel is increased by

$20$ % when heated by

$273^{\circ} \,C$ then it's initial temperature must have been:

0%
$1092^{\circ} \,C$
0%
$1092\,K$
0%
$1365^{\circ}\,C$
0%
$1365\,K$

Explanation

$\dfrac{P}{t + 273} = \dfrac{1.2P}{t + 273 + 273}$

$1.2t + 273 \times 1.2 = t + 273 \times 2$

$\Rightarrow$

$0.2t = 273 \times 0.8$

$t = 273 \times 4 =1092K$

Option B is correct.

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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