MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 9

Which of the following represents the Avogadro number?

0%
Number of molecule present in 1 L of gas at N.T.P
0%
Number of molecule present in 22.4 L of gas at N.T.P
0%
Number of molecule present in 22.4 of gas at 298K and 1 atm pressure
0%
Number of molecule present in one mole of gas at any temp and pressure

Explanation

Avogadro number is defined as the number of gas molecules present in one mole of gas at any pressure and temperature. It's value is

$6.023 \times 10^{23}$ . So the correct option is D

$250^{o}C$ and

$1$ atmosphere pressure, the vapour density of

$PCI_{5}$ is

$57.9$ . What will be the dissociation of

$PCI_{5}-$

0%
$1.00$
0%
$0.90$
0%
$0.80$
0%
$0.65$

Which of the following is an anti-knocking compound, that has been phased out in many countries so far?

0%
Lead tetrachloride
0%
Tetra Ethyl Lead
0%
Ethyl tetrachloride
0%
None of these

One gram molecule of any gas at

$NTP$ occupies

$22.4\ L$ . This fact was derived from:

0%
Dalton's theory
0%
Avogardro's hypothesis
0%
Berzelius hypothesis
0%
Law of gaseous volume

Explanation

According to Avogadro's law, 1 mole of every gas occupies 22.4L at

$NTP$ .

From ideal gas equation,

$PV=nRT$ . . . . . . .(1)

$NTP$ ,

$T=273K$

$P=1atm$

$n=1$

$R=0.0821atm L/K mol$

from equation (1),

$V=\dfrac{nRT}{P}$

$V=\dfrac{1\times 0.0821\times 273}{1}=22.4L$

The correct option is B.

$300\ K$ , the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen

$(N_{2})$ at 4 bar.The molar mass of gaseous molecule is:

0%
$224 g mol^{-1}$
0%
$112 g mol^{-1}$
0%
$56 g mol^{-1}$
0%
$28 g mol^{-1}$

Thermometer

$(A)$ and

$(B)$ have ice points marked at

$15^{o}C$ &

$25^{o}C$ and steam points at

$75^{o}C$ &

$125^{o}C$ respectively. When thermometer

$A$ measures the temperature of a bath as

$60^{o}C$ the reading of

$B$ for the same bath is :

0%
$60^{o}C$
0%
$75^{o}C$
0%
$90^{o}C$
0%
$100^{o}C$

If there unreactive gases having partial pressure

${P_A},{P_B}$ and

${P_C}$ and there moles are

$1,\,2$ and

$3$ respectively then their total pressure will be :

0%
$P = {P_A} + {P_B} + {P_C}$
0%
$P = \cfrac{{{P_A} + {P_B} + {P_C}}}{6}$
0%
$P = \cfrac{{\sqrt {{P_A} + {P_B} + {P_C}} }}{3}$
0%
None of these

Explanation

According to Dalton's law of partial pressure,

Total pressure = sum of partial pressures of individual gases

$P_{total} = P_a + P_b + P_c$

Therefore, option A is correct.

A container has

$SO_{2}$ gas at

$2\ atm$ pressure in a vessel of

$V\ L$ capacity, if no. of moles of

$SO_{2}$ are doubled in the same container at the same temperature and volume. Calculate new pressure in the container.

0%
$1\ atm$
0%
$4\ atm$
0%
$2\ atm$
0%
$8\ atm$

A solution

$X$ of

$A$ and

$B$ contains

$30$ mole

$\%$ of

$A$ & is in equilibrium with its vapour that contains

$40$ mole

$\%$ of

$B$ . The ratio of

$V.P$ of pure,

$A$ and

$B$ will be?

0%
$2:7$
0%
$7:2$
0%
$3:4$
0%
$4:3$

The vapour pressure of water at

$20^0C$ is 17.54 mm. When 20 g of non-ionic, substance is dissolved in 100 g of water, the vapour pressure is lowered by 0.30 mm.What is the molecular weight of the substances.

0%
210.2
0%
208.16
0%
215.2
0%
200.8

Consider the following statements
a. Kinetic energy of a molecule is zero at

$0^0C$ .
b. A gas in a closed container will exert much higher pressure due to gravity at the bottom than at the top
c. Between collisions, the molecules move in straight lines with constant velocities.
Choose the incorrect statement(s)

0%
a, b & c
0%
a & b
0%
b & c
0%
a & c

$27^0C$ a gas is compressed to half of its volume.To what temperature it must now be heated at constant pressure so that gas occupies just its original volume?

0%
$54^0C$
0%
$600^0C$
0%
$327^0C$
0%
$327 K$

Explanation

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$V_2=V$

$V_1=\dfrac{V}{2}$

$T_1=27^{\circ}C=(273+27)=300K$

$T_2=\dfrac{V_2}{V_1}\times T_1$

$=\dfrac{V}{\frac{V}{2}}\times 300$

$\therefore T_2=600K$

$\Rightarrow T_2=(600-273)=327C$

The sample of neon gas heated from 300 K to 390 K, percentage increases in K.E is:

0%
$10\%$
0%
$20\%$
0%
$30\%$
0%
$40\%$

Explanation

Kinetic energy of gas is given by

$K.E = \dfrac{3}{2}nRT$

therefore, kinetic energy is directly proportional to temperature.

so, if we find out percentage change in temperature , we can get percentage change in kinetic energy .

percentage change in temperature

$=\dfrac{\Delta T } {T}$

$= \dfrac{390 - 300}{90} \times 100$

$= \dfrac{90}{300}\times 100$

$= 30$ %

hence, percentage change in kinetic energy of neon gas is 30%

$273 ml$ of a gas at STP was taken to

$27^oC$ and

$600 mm$ pressure. The final volume of the gas would be:

0%
273 mL
0%
300 mL
0%
380 mL
0%
586 mL

Explanation

We know that, for constant no. of moles of gas,

$\dfrac{P_1 V_1}{T_1}$ =

$\dfrac{P_2 V_2}{T_2}$

Given, initial conditions,

$P_1 = 760mm$ (STP)

$V_1 = 273mL$

$T_1 = 273K$ (STP)

Final conditions,

$P$

$= 600mm$

$T$

$= 300K$

Thus,

$V_2 = \dfrac{P_1 V_1}{T_1} \times \dfrac{T_2}{P_2}$ =

$\dfrac{760 \times 273 \times 300}{273 \times 600} = 380mL$

Hence, option C is correct.

Equal amounts of two gases of molecular mass 4 and 40 are mixed. the pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is :

0%
0.55 atm
0%
0.11 atm,
0%
1 atm
0%
o.12 atm

Explanation

Assume 40 g each of two gases is present.

The number of moles of lighter gas are

$\displaystyle \frac {40}{4} = 10$ .

The number of moles of heavier gas are

$\displaystyle \frac {40}{40} = 1$ .

The mole fraction of the lighter gas is

$\displaystyle \frac {10}{10+1} = 0.9090$ .

The partial pressure of the lighter gas is

$\displaystyle 0.9090 \times 1.1 = 1 \: atm$ .

Mole fraction of A vapours above the solution in mixture of A and B (

$X_A=0.4$ )will be
[Given :

${ P }_{ A }^{ 0 }=100mmHg\quad and\quad { P }_{ B }^{ 0 }=200mmHg$ ]

0%
0.4
0%
0.8
0%
0.25
0%
none of these

6 g of urea is dissolved in 90 g of boiling water.The vapour preasure of the solution is:

0%
$744.8 mm$
0%
$758 mm$
0%
$761 mm$
0%
$760 mm$

A constant gas temperature is based on

0%
Archimedes prinviple
0%
Pascal laW
0%
Boyle's law
0%
Charles's law

The relative lowering of vapour pressure of a solution containing

$6\ g$ of urea dissloved in

$90\ g$ of water is

0%
$0.0196$
0%
$0.05$
0%
$1.50$
0%
$0.01$

Which of the following sets of variables give a staright line with negative slope when plotted ?
( P = Vapour pressure ; T = Temperature in K)
y- axis x -axis y -axis x-axis

0%
$P\quad \quad \quad T$
0%
$log_{10}P\quad \quad \quad T$
0%
$log_{10}P\quad \quad \quad \frac{1}{T}$
0%
$log_{10}P \quad \quad \quad log_{10} \frac{1}{T}$

The vapour pressure of two liquids are

$15000$ and

$30000$ in a unit. When equimolar solution of liquids is.. The the mole fraction of A and B in vapour phase will be:

0%
$\dfrac{2}{3}, \dfrac{1}{3}$
0%
$\dfrac{1}{3}, \dfrac{2}{3}$
0%
$\dfrac{1}{2}, \dfrac{1}{2}$
0%
$\dfrac{1}{4}, \dfrac{3}{4}$

When the temperature of 23ml of dry

$CO_{2}$ gas is changed from

$10^\circ$ to

$30^\circ$ at constant pressure of 760 mm, the volume of gas becomes closest to which one of the following?

0%
7.7 ml
0%
21.5 ml
0%
24.6 ml
0%
69 ml

The reflective lowering of sapour pressure of a solution containing

$6\ g$ of urea dissolved in

$90\ g$ of water is ?

0%
$0.0196$
0%
$0.05$
0%
$1.50$
0%
$0.01$

A 15.0 L cylinder of He gas is connected to an evacuated 235.0 L tank.If the final pressure is 750 mm Hg, then what have been the original gas pressure in the cylinder?

0%
1.645 atm
0%
16.45 atm
0%
15.46 atm
0%
23 atm

Which of the following gas mixture is not applicable for Dalton's law of particle pressure?

0%
$SO_2$ and $Cl_2$
0%
$CO_2$ and $N_2$
0%
$CO$ and $CO_2$
0%
$CO$ and $N_2$

The value of gas constant per mole is approximately-

0%
$1\ cal$
0%
$2\ cal$
0%
$3\ cal$
0%
$4\ cal$

The vapour pressure of a solution at

$100^{o}C$ containing

$3\ g$ of cane sugar in

$70\ g$ of water? (vapour pressure of water at

$100^{o}C=760\ mm$ of

$Hg$ ) is)

0%
$758\ mm$ of $Hg$
0%
$730.36\ mm$ of $Hg$
0%
$400\ mm$ of $Hg$
0%
$700\ mm$ of $Hg$

A pin or a needle floats on the surface of water, the reason for this is

0%
Surface tension
0%
Less weight
0%
upthrust of liquid
0%
None of the above

Rain drops acquire spherical shape due to

0%
viscosity
0%
surface tension
0%
friction
0%
elasticity

Which is not correct in terms of kinetic theory of gases-

0%
Gases are made up of small particles called molecules
0%
The molecules are in random motion
0%
When molecules collide, they lose energy
0%
When the gas is heated, the molecules moves faster

Small liquid drops assume spherical shape because

0%
atmospheric pressure exerts a force on liquid drop
0%
volume of a spherical drop is minimum
0%
gravitational force acts upon the drop
0%
liquid tends to have the minimum surface are due to surface tension

Explanation

Small liquid drops assume spherical shape because drop wants to be in the lowest potential state.

$E=S\times A$ And

The sphere has the minimum surface area for the given volume of the liquid.

A box of

$1L$ capacity is divided into two equal compartments by a thin partition which are filled with

$2 g\ \mathrm { H } _ { 2 } \text { and } 16 \mathrm { g } \ \mathrm { CH } _ { 4 }$ respectively. The pressure in each compartment is recorded as

$P$ atm. The total pressure when partition is removed will be:

0%
$P$
0%
$2P$
0%
$P/2$
0%
$P/4$

The water proofing agents

0%
Increase $T$ and decrease $\theta$
0%
Increase both $T$ and $\theta$
0%
Decrease both $T$ and $\theta$
0%
Decrease $T$ and increase $\theta$

The vapour pressure of pure

$CHCl_{3}$ and

$CH_{2}Cl_{2}$ are

$200$ and

$41.5\ atm$ respectively. The weight of

$CHCl_{3}$ and

$CH_{2}Cl_{2}$ are respectively

$11.9\ g$ and

$17\ gm$ . The vapour pressure of solution will be

0%
$80.5$
0%
$79.5$
0%
$94.3$
0%
$105.5$

Explanation

$P_{CHCl_{3}}{0} = 200\ atm\ P_{CH_{3}Cl_{2}}{0} = 41.5\ atm$

$n = \dfrac {11.9}{119}\ n = \dfrac {17}{85}$

$= .1\ = .2$

$P_{T} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B}$

$= 200x \dfrac {0.1}{.3}\ + \dfrac {41.5\times 0.2}{.3}$

$= 94.33$ .

A pre-weighed vessel was filled with oxygen at

$NTP$ and weighed. It was then evacuated, filled with

$SO_{2}$ at the same temperature and pressure and again weighed. The weight of oxygen is ______________.

0%
the same as that of $SO_{2}$
0%
$\dfrac{1}{2}$ that of $SO_{2}$
0%
twice that of $SO_{2}$
0%
$\dfrac{1}{4}$ that of $SO_{2}$

Explanation

At constant temperature and pressure, an equal volume of gas contains an equal number of moles.

$\therefore$ Moles of

$O_{2}=$ Moles of

$SO_{2}$

$1\ mole\ O_{2}=32\ g$

$1\ mole\ SO_{2}=64\ g$

$\therefore wt$ of

$O_2=\dfrac{1}{2}$ that of

$SO_{2}$

Option

$B$ is correct.

If water vapor comprises 3.5 percent of an air parcel whose total pressure is 1,000 mb, the water vapor pressure would be:

0%
1,035 mb
0%
35 mb
0%
350 mb
0%
965 mb

If RMS velocity of gaseous molecules is

$x$ cm/s at a pressure of P atm then RMS velocity at a pressure of

$2p$ atm at constant temperature will be:

0%
x
0%
2x
0%
4x
0%
x/4

The vapour pressure of pure

$M$ and

$N$ are

$700$ mm of

$Hg$ and

$450 mm$ of

$Hg$ respectively.which of the following option is correct?
Given :

$X_N,X_M$ - mole fraction of

$N$ and

$M$ in liquid phase

$Y_N,Y_M$ = mole fraction of

$N$ and

$M$ in vapour phase.

0%
$X_M-X_N>Y_M-Y_N$
0%
$\dfrac{X_M}{X_N}>\dfrac{Y_M}{Y_N}$
0%
$\dfrac{X_M}{X_N}<\dfrac{Y_M}{Y_N}$
0%
$\dfrac{X_M}{X_N}=\dfrac{Y_M}{Y_N}$

Explanation

Solution:- (C)

$\dfrac{X_M}{X_N}<\dfrac{Y_M}{Y_N}$

Mole fraction of more voltatile component increase in vapour phase, i.e.,

$P_N^o<P_M^o$

The volume of a gas is held constant while its temperature is raised. The pressure of the gas exerts on the walls of the container increases because:

0%
The masses of the molecules increases
0%
Each molecule loses more kinetic energy when it strikes the wall
0%
The molecules collide with the wall with relatively greater momentum
0%
The molecules strikes the walls more often

Explanation

Volume is kept constant and temperature increased.
We have relation,

$PV=nRT$

$P\propto T$
Volume constant
So,

$P\propto T$ pressure increases as temperature increases.
As we know the speed or average molecular speed depend on temperature.
As temperature increases molecules collide with wall with greater force or momentum.
Also temperature increases, molecular vibrations increases. So molecule will strike with wall more often.
So option C and D.
A is incorrect because there is no change in masses of molecule when we increase temperature.

On increasing the altitude at constant temperature, vapour pressure of liquid

0%
increases
0%
decreases
0%
remains the same
0%
depends upon climate

A pre-weighted vessel was filled with oxygen at NTP and weighted. It was then evacuated, filled with

$SO_2$ at the same temperature and pressure and again weighted. The weight of oxygen will be:

0%
the same as that of $SO_2$
0%
$1/2$ that of $SO_2$
0%
twice that of $SO_2$
0%
$1/4$ that of $SO_2$

A closed vessel contains equal number of nitrogen and oxygen molecules at a pressure of

$p\ mm$ . If nitrogen is removed from the system then the pressure will be:

0%
$p$
0%
$2p$
0%
$p/2$
0%
$p^2$

Explanation

At constant

$V, P\propto n$

$\therefore$ if no. of moles is halved then pressure will also reduce to

$p/2$ .

Option C is correct.

Equal weights of methane and hydrogen are mixed in an empty container at

$25^oC$ . The fraction of the total pressure exerted by hydrogen is

0%
$1/2$
0%
$8/9$
0%
$1/9$
0%
$16/17$

Explanation

$\dfrac{n_{CH_2}}{n_{H_2}}=\dfrac{M_{H_2}}{M_{CH_2}}$ ( for equal not )

$=\dfrac{2}{16}=\dfrac{1}{8}$

$\dfrac{n_{H_2}}{n_{CH_4}}=\dfrac{8}{1}\Rightarrow \dfrac{n_{H_2}}{n_{total}}=\dfrac{8}{9}=X_{H_2}$

$\therefore P_{H_2}=X_{H_2}P_{total}\Rightarrow X_2 =\dfrac{P_{H_2}}{P_{total}}=\dfrac{8}{9}$

Option B

With regard to the gaseous state of matter which of the following statements are correct?

0%
Complete order of molecules
0%
Complete disorder of molecules
0%
Random motion of molecules
0%
Fixed position of molecules

In the ideal gas equation, the gas constant

$R$ has the dimensions of:

0%
$mole-atm \ K$
0%
$litre \ mole$
0%
$litre-atm \ K- mole$
0%
$erg \ K$

Explanation

$PV = nRT \ R = \dfrac {PV} {nT} = litre . atm. K- mole$

The translational kinetic energy of an ideal gas depends only on its:

0%
Pressure
0%
Force
0%
Temperature
0%
Molar mass

Explanation

$K.E. = \dfrac {3RT}{2}$ it means that the Translational Kinetic energy of Ideal gas depends upon temperature only.

Kinetic energy of a gas depends upon its:

0%
Molecular mass
0%
Atomic mass
0%
Equivalent mass
0%
None of these

Explanation

Kinetic energy

$=\dfrac {3}{2} RT$

Hence, option D is correct.

For a solution formed by mixing liquids

$L$ and

$M$ , the vapour pressure of

$L$ plotted against the mole fraction of

$M$ in solution is shown in the following figure. Here

$x_L$ and

$x_M$ represent mole fractions of

$L$ and

$M$ respectively in the solution. The correct statement

$\left(s\right)$ applicable to this system is

$\left(are\right)$

0%
The point $Z$ represents vapour pressure of pure liquid $M$ and Raoults law is obeyed from $X_{L}=0$ to $X_{L}=1$ .
0%
Attractive intermolecular interactions between $L-L$ in pure liquid $L$ and $M - M$ in pure liquid $M$ are stronger than those between $L - M$ when mixed in solution.
0%
The point $Z$ represents vapour pressure of pure liquid $M$ and Raoults law is obeyed when $X_L \to 0$ .
0%
The point $Z$ represents vapour pressure of pure liquid L and Raoults law is obeyed when $X_L\to 1$

Explanation

We know from Raoult's law,

$P=P_L^ox_L+{P_M}^ox_M$

Vapour Pressure of liquid,

$P_L={P_L}^ox_L=P_L^o(1-x_M)$ at

$x_M\rightarrow 0$ or

$x_L \rightarrow 1$

$\therefore P_L={P_L}^o$ This follows the graph at point Z

Option D is correct.

The statement B is also correct because the attraction in like molecules are more than that in unlike molecules.

Option B is also correct.

Ans- B, D.

$\Delta_f G^{\circ}$ at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcal

$mol^{-1}$ respectively. Vapour pressure of liquid 'S'' at 500 K is approximately equal to:

$R = 2 cal K^{-1} mol^{-1}$

0%
10 atm
0%
0.1 atm
0%
1 atm
0%
100 atm

The type of molecular force of attraction present in the following compound is :

0%
Intermolecular H-bonding
0%
Intramolecular H-bonding
0%
van der WaaLs' force
0%
All of these

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 9 - 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 9 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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