have low masses and are spread far apart

have forces of attraction between them

never travel with a straight line motion

have molecules that are close together

Has no intermolecular attraction

Molecules do not collide with each other

The product of P and V is constant at a fixed temperature for definite mass

MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 10

Select the incorrect order of boiling point between the following compounds :

0%
$N_3H < CH_3N_3$
0%
$Me_2SO_4 < H_2SO_4$
0%
$Me_3BO_3 < B(OH)_3$
0%
$BF_3 < BI_3$

Explanation

$\rightarrow$ Boiling point of

$HN_3 > CH_3N_3$ as in hydrazoic acid (

$HN_3$ ) intermolecular H-bonding occurs and its bond energy is higher than weak dipole-dipole interactions present in

$CH_3N_3$ .

$\rightarrow$ Boiling point of

$BI_3$ is greater than that of

$BF_3$ as molecular weight of

$BI_3$ is higher than that of

$BF_3$ .

$\rightarrow$ Due to intermolecular H-bonding in

$H_2SO_4$ its boiling point is higher than

$Me_2SO_4$ .

$\rightarrow$ Similarly due to intermolecular H-bonding in

$B(OH)_3$ its boiling point is higher. than that of

$Me_3BO_3$ .

Which of the following statements are false? Rewrite them correctly.

Volume remaining constant, the pressure is inversely proportional to temperature in degrees kelvin.

0%
True
0%
False

Explanation

The above-given statement is false.

According to Gay-Lussac law, the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

$P\propto T$ ------ keeping the temperature constant.

Or,

$\dfrac{P}{T}=C$

Or,

$\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}$

where two different temperature and their corresponding pressures are given.

At a certain pressure, volume of the gas at

$27^\circ C$ is

$20$ litre. If the pressure and temperature are doubled, the volume will be

0%
$40\,litre$
0%
$20\,litre$
0%
$10.9\,litre$
0%
$8.2\,litre$

Explanation

According to question:

$P_1 =P$ ,

$V_1 =20L$ ,

$T_1 =T$

$P_2=2P$ ,

$V_2=?$ ,

$T_2= 2T$

Now,

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$\dfrac{P \times 20}{T}=\dfrac{2P \times V_2}{2T}$

$\Rightarrow V_2=20\,L$

Which of the following curve does not represent Gay lusacc's law?

Explanation

Gay Lussac's Law states that

$\mathrm{P = kT ; }$ where T is in Kelvin and k is not necessarily 1.

Hence, Options "A" & "D" are correct answers.

"Equal volumes of all gases at the same temperature and pressure contain equal number of particles." This statement is a direct consequence of:

0%
Avogadro's law
0%
Charles law
0%
Ideal gas equation
0%
Law of partial pressure

The equation of state for

$5$ g of oxygen at a pressure

$P$ and temperature

$T$ , when occupying a volume

$V$ , will be:

0%
$PV = 5 RT$
0%
$PV = \left (\dfrac{5} {2} \right ) RT$
0%
$PV = \left (\dfrac{5} {16} \right ) RT$
0%
$PV = \left (\dfrac{5} {32} \right ) RT$

Explanation

According to the ideal gas equation,

$PV=nRT$

$n=\cfrac{\text{Mass given}}{\text{Molecular mass of }O_2}$

$=\cfrac{5}{32}$

$PV= \left(\cfrac{5}{32} \right)RT$

Hence, the correct answer is option

$\text{D}$ .

According to the kinetic theory of gases, in an ideal gas, between two successive collisions, a gas molecule travels:

0%
in a wavy path
0%
in a straight line path
0%
with an accelerated velocity
0%
in a circular path

Match the following :

List -I	List-II
A. Boyle’s law	$P_{wet}=P_{dry}+P_{water\ vapour}$
B. Avogadro law	$V_{1}=n_{1}\left ( \dfrac{V_{2}}{n_{2}} \right )$
C. Charles law	$V_{1}=V_{0}\left (1+ \dfrac{t}{273} \right )$
D. Dalton’s law	$V_{1}=P_{2}\left ( \dfrac{V_{2}}{P_{1}} \right )$

The correct match is:

0%
A-1, B-2, C-3, D-4
0%
A-2, B-3, C-4, D-1
0%
A-4, B-3, C-2, D-1
0%
A-4, B-2, C-3, D-1

Explanation

Boyle's law: At constant temperature,

$PV=$ constant

$\therefore P_1V_1 = P_2V_2$

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

$\therefore \dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}$

Charles law states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature if the pressure remains constant.

$= V_{1}=V_{0}\left (1+ \dfrac{t}{273} \right )$

Dalton's law: Net vapour pressure is the addition of vapour pressure of individual components.

$\therefore P_{wet} = P_{dry} + P_{water\ vapour}$

Hence, the correct answer is option

$\text{D}$ .

2.5 L of a sample of a gas at

$27^\circ C$ and 1 bar pressure is compressed to a volume of 500 mL keeping the temperature constant, the percentage increase in the pressure is :

0%
100%
0%
400%
0%
500%
0%
80%

Dalton's law of partial pressures is not applicable to which one of the following?

0%
$H_{2}+Cl_{2}$
0%
$SO_{2}+Cl_{2}$
0%
$NH_{3}+HCl$
0%
All the above

Explanation

According to Dalton's law of partial pressures, for a mixture of

$\text{non-reacting}$ gases, the total pressure is equal to the sum of the partial pressures of the individual gases.

This law is not applicable to the mixtures

$H_{2}+Cl_{2}, NH_{3}+HCl$ and

$SO_{2}+Cl_{2}$ as they react with each other.

For example, hydrogen and chlorine react to form hydrogen chloride, ammonia and hydrochloric acid forms ammonium chloride while sulphur dioxide reacts with chlorine to form

$SO_2Cl_2$ .

Hence, the correct answer is option

$\text{D}$ .

Which of the following statement(s) is/are correct ?

0%
At $273^{o}C$ , the volume of a given mass of a gas will be twice its volume at $0^{o}C$ and $1$ atm pressure
0%
At $-136.5^{o}C$ , the volume of a given mass of a gas will be half its volume at $0^{o}C$ and 1 atm pressure
0%
The mass ratio of equal volumes of $NH_{3}$ and $H_{2}S$ under identical conditions of temperature and pressure is $1:2$
0%
The molar ratio of equal masses of $CH_{4}$ and $SO_{2}$ is $4:1$

Explanation

For options

$A$ and

$B$ , apply Charles's law.

At constant pressure,

$V_{1}T_{2}=V_{2}T_{1}$

$V_{ 273^{o}C}\times 273=V_{0^{o}C}\times 546$

Thus, at

$273^{o}$ C the volume of a given mass of gas will be twice its volume at

$0^{o}$ C and

$1$ atm pressure.

Similarly,

$V_{0^{o}C} \times 273=V_{-136.5^{o}C}\times 136.5$ .

Thus,

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

Mass

$\propto$ Molecular weight keeping other variables constant.

$\therefore$ Mass ratio of

$NH_{3}$ and

$H_{2}S$ is

$\dfrac{17}{34}= \ 1:2$

Mole

$\propto$

$\dfrac {1}{Molecular \ weight}$

$\therefore$ Molar ratio

$= \dfrac{1}{16}:\dfrac{1}{64}=4:1$

Hence, options A, B, C and D are correct.

A gas is collected by downward displacement of water. Select the correct expression for

$P_{gas}$ according to the diagram. [Density of Hg (l)

$= 13.6 \ g/cm^3]$

0%
$P_{gas}=P_{atmp}-\left [ aq.tension+\dfrac{h}{13.6} \right ]$
0%
$P_{gas}=P_{atm}-h \rho g$
0%
$P_{gas}=P_{atmp}-aq.tension+\dfrac{h}{13.6}$
0%
None of these

Explanation

The atmospheric pressure is balanced by the sum of the pressures of gases, water column of height and surface tension of water.
The pressure exerted by a column of water having height h is equal to

$\dfrac {h} {13.6}$ , where, 13.6 is the density of mercury.
Thus,

$P_{atmp}=P_{gas}+\left [ aq.Tension+\dfrac{h}{13.6} \right ]$

$P_{gas}=P_{atmp}-\left [ aq.Tension+\dfrac{h}{13.6} \right ]$

Which of the following is the only incorrect informarion regarding the composition of the system at 0.51 bar pressure ?

0%
$X_A = 0.45$
0%
$Y_A = \frac {6}{17}$
0%
$n_{A(liquid) } = \frac {12}{11}$
0%
$n_{A(vapour) } = \frac {12}{11}$

At the same temperature, HCl gas and NH

$_{3}$ gas are present in two vessels of same volume at a pressure of 'P' atmospheres each. When one jar is inverted over the other so that the two will mix, after some time the pressure in the vessels will become:

0%
$\cfrac{P}{2}$
0%
$\cfrac{P}{4}$
0%
Zero
0%
P

Explanation

$NH_3 + HCl \rightarrow NH_4Cl(solid)$

Due to the formation of a solid, final pressure in the vessel will become zero.

Hence, the correct answer is option

$\text{C}$ .

An open vessel at

$27^oC$ is heated until the three-fourths mass of the air in it has been expelled. Neglecting the expansion of the vessel, find the temperature to which the vessel has been heated.

0%
$T_2=723^oC$
0%
$T_2=958^oC$
0%
$T_2=927^oC$
0%
None of these

Explanation

No. of moles of air

$=n$ .

$\cfrac{3}{4}$ of original moles has expelled.

Remaining moles

$=n(1-\cfrac{3}{4})$

$=\cfrac{n}{4}$ .

At constant pressure & volume,

$n_{1}T_{1}=n_{2}T_{2}$ .

$\Rightarrow (n)(300)=(\cfrac{n}{4})(T_{2})$

$\Rightarrow T_{2}=(300)(4)$

$=1200\,K$

$\Rightarrow T_{2}=927^{\circ}C$ .

$\therefore$ Temperature of vessel is

$927^{\circ}C$ .

Hence, (C) is the correct option.

$45.4$ L of dinitrogen reacted with

$22.7$ L of dioxygen and

$45.4$ L of nitrous oxide was formed. The reaction is given below:

$\displaystyle 2N_{2}\left ( g \right )+O_{2}\left ( g \right )\rightarrow 2N_{2}O\left ( g \right )$
Which law is being obeyed in this experiment?

0%
Gay Lussac's law of gaseous volumes
0%
Henry's Law
0%
Raoult's Law
0%
None

Explanation

The ratio of the volumes of dinitrogen, oxygen and nitrous oxide is

$45.4$ :

$22.7$ :

$45.4$ . This is in the simple whole number ratio

$2:1:2$ .
This satisfies the Gay Lussac's law of combining volume of gases.

Gay-Lussac's law i.e, dependence of pressure of ideal gas with temperature in

$^oC$ can be expressed as

$P_t=a+bt$ where

0%
$a=P_0$ and t in $^oC$
0%
$b=\left (\frac {\partial P}{\partial t}\right )_V$
0%
$b=\left [\frac {P_0}{273.15}\right ]$
0%
$P_T=KT$ , where T is in K

Explanation

Gay Lussac's law:
It states "at constant volume, the pressure of a given mass of a gas is directly proportional to the absolute temperature of the gas".

$P\propto T$ .

$\dfrac{P}{T}=constant(K)$

Two identical vessels are filled with 44 g of hydrogen and 44 g of carbon dioxide at the same temperature. If pressure of

$CO_2$ is 2 atm. Find the pressure of hydrogen.

0%
$P_{H_2}=56\ atm$
0%
$P_{H_2}=44\ atm$
0%
$P_{H_2}=32\ atm$
0%
None of these

Explanation

Two identical vessels,

$44\,g$ of

$CO_{2}=1\,mol$ and exerts

$2\,atm$ pressure.

Now,

$44\,g$ of

$H_{2}=\cfrac{44}{2}=22\,mol$ and hence, will exert

$={22}\times{2}\,atm$ pressure.

$\therefore$ Pressure of

$H_{2}=44\,atm$ .

A zinc metal sample containing zinc chloride as impurity was made to react with an excess of dilute hydrochloric acid at

$27^{o}C$ . Liberated hydrogen gas is collected at 760 mm Hg pressure that occupies

$780.0 cm^{3}$ volume. If the vapour pressure of water at

$27^{o}C$ is 14 mm Hg, what is the volume of

$H_{2}$ at STP? The Standard pressure is 760 mm Hg (molar volume of gas at standard temperature and pressure,

$STP = 22.4 dm^{3}$ )

0%
$746 cm^{3}$
0%
$697 cm^{3}$
0%
$750 cm^{3}$
0%
$300 cm^{3}$

At constant temperature, the pressure of a gas is __________ to its volume.

0%
inversely proportional
0%
directly proportional
0%
not proportional
0%
none of the above

Explanation

At constant temperature the pressure of a gas is inversely proportional to its volume. This law is called Boyle's law.

$P \propto \cfrac { 1 }{ V }$

So answer is A.

In the reaction

$N_{2}+3H_{2}\rightarrow 2NH_{3}$ , the ratio by volume of

$N_{2},\ H_{2} \:$ and

$\: NH_{3}$ is

$1 : 3 : 2$ .

This illustrates the law of:

0%
definite proportion
0%
multiple proportion
0%
reciprocal proportion
0%
gaseous volumes

Explanation

In the reaction,

$N_{2}+3H_{2}\rightarrow 2NH_{3}$ , the ratio by volume of

$N_{2},\ H_{2}$ and

$NH_{3}$ is

$1 : 3 : 2$ . This illustrates the law of Gaseous volumes or Gay Lussac's law of combining volumes of gases.

According to this law, when gases react together to produce gaseous products, the volume of reactants and products bear a simple whole-number ratio with each other, provided volumes are measured at the same temperature and pressure.

So, the correct option is

$D$ .

To which of the following mixtures Dalton's law is not applicable?

0%
$CO_2$ and CO at room temperature
0%
Ammonia and hydrogen chloride at room temperature
0%
$NH_3$ and steam at room temperature
0%
He and $H_2$

Explanation

Dalton's Law of Partial Pressure, and it states that the pressure of a gas mixture is the sum of the partial pressures of the individual components of the gas mixture.

Ammonia and hydrogen chloride at room temperature and

$NH_3$ and steam at room temperature react with each other.

So Dalton's law is not applicable Ammonia and hydrogen chloride at room temperature and

$NH_3$ and steam at room temperature.

Hence options B & C are correct.

A balloon containing 1 mole air at 1

$atm$ initially is filled further with air till pressure increases to 4

$atm$ . The initial diameter of the balloon is 1

$m$ and the pressure at each stage is proportion to diameter of the balloon. How many no.of moles of air added to change the pressure from 1

$atm$ to 4

$atm$ .

0%
80
0%
257
0%
255
0%
256

Explanation

Pis directly proportional to d; P=kd

and k=1atm/1metre

$PV=nRT;kd(\dfrac{1}{6}\pi d^3)=nRT$

$\dfrac{d_1^4}{d_2^4}=\dfrac{n_1}{n_2}$

$\dfrac{1}{4^4}=\dfrac{n_1}{n_2}$

$n_2=256$

No.of moles added=256−1=255.

The number of effusion steps required to convert a mixture of

$H_2$ and

$O_2$ from 240 : 1600 (by mass) to 3072 : 20 (by mass) is:

0%
2
0%
4
0%
5
0%
6

Explanation

Overall separation factor

$f=\dfrac {\dfrac {n'_1}{n_2'}}{\dfrac {n_1}{n_2}}$

$f=\dfrac {\dfrac {240}{1600}}{\dfrac {3072}{20}}$

$f=\dfrac {240 \times 20}{1600 \times 3072}$

$f=9.7656 \times 10^{-4}$

$\text{log f} = \text{log}\ 9.7656 \times 10^{-4} = -3.01$

The number of effusion steps required

$x=\dfrac {\text{2 logf}}{\text{log}\dfrac {M_2}{M_1}}$

$x=\dfrac {-6.02}{-1.2}$

$x=4.95 \simeq 5$

The number of effusion steps required is 5.

Hence, the correct option is

$\text{C}$

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

Ideal gas has a volume of 10 litres at

$20^oC$ and a pressure of 750

$mm$

$Hg$ . Find the expressions which is needed to determine the volume of the same amount of gas at

$STP$ ?

0%
$\displaystyle 10 \times \frac{750}{760} \times \frac{0}{20}L$
0%
$\displaystyle 10 \times \frac{750}{760} \times \frac{293}{273}L$
0%
$\displaystyle 10 \times \frac{760}{750} \times \frac{0}{20}L$
0%
$\displaystyle 10 \times \frac{760}{750} \times \frac{273}{293}L$
0%
$\displaystyle 10 \times \frac{750}{760} \times \frac{273}{293}L$

Explanation

Initial volume

$V_1 = 10 \:L$
Final volume

$V_2 = ?$

Initial pressure

$P_1 = 750 \:mm \:Hg$
Final pressure

$P_2 = 760 \:mm \:Hg$

Initial temperature

$T_1 =20^oC = 293 \:K$
Final temperature

$T_2 = 273 \:K$

$\displaystyle \dfrac {P_1V_1}{T_1}=\dfrac {P_2V_2}{T_2} \\ \dfrac {750 \times 10}{293}=\dfrac {760 \times V2}{273} \\ V_2 = 10 \times \dfrac {750}{760} \times \dfrac {273}{293}$

Hence, the option (E) is the correct answer.

"All gases have the same number of moles in the same volume at constant temperature and pressure:. This statments belongs to :

0%
Boyle's law
0%
Charles's law
0%
Avogadro's principle
0%
ideal gas law
0%
Dalton's law

Statement 1 : In the kinetic theory of gases, collisions between gas particles and the walls of the container are considered elastic.
Statement 2 : Gas molecules are considered point like, volume less particles with no inter molecular forces and in constant, random motion.

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%

Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%

Both the Statement 1 and Statement 2 are not correct.

''Total pressure of a gaseous mix is equal to the sum of the partial pressures'', this statement is from :

0%
Boyle's law
0%
Charles's law
0%
Avogadro's principle
0%
ideal gas law
0%
Dalton's law

Explanation

Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Each gas is assumed to be an ideal gas.

$P_total = P_1 + P_2 + P_3 + . .$
Where,

$P_1$ and

$P_2$ are the partial pressure of gas 1 and gas 2 in the mixture. Since, each gas behaves independently, the ideal gas law can be used to calculate the pressure of that ga,s if we know the number of moles of the gas, the total volume of the container, and the temperature of the gas.

Inelastic collisions occur in :
I. Real gases
II. Ideal gases
III. Fusion reactions
Which of the following is correct option about the above problem?

0%
I and II
0%
II and III
0%
I and III
0%
I only
0%
II only

Statement 1: At the same temperature and pressure, 1 L of hydrogen gas and 1 L of neon gas have the same mass.
Statement 2: Equal volumes of ideal gases at the same temperature and pressure contain the same number of moles.

0%

Both statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1
0%

Both statement 1 and statement 2 are correct but Statement 2 is not the correct explanation of Statement 1
0%

Statement 1 is correct but statement 2 is incorrect
0%

Statement 1 is incorrect but statement 2 is correct
0%

Both the statement 1 and statement 2 are incorrect

Among the following which one shows that the total pressure of a mixture of gases is equal to the sum of the partial pressure of the component gases.

0%
V/T $=$ k
0%
P/T $=$ k
0%
PV $=$ k
0%
$P_T=P_1+P_2+P_3$
0%
PT $=$ k

Explanation

The total pressure of a mixture of gases is equal to the sum of the partial pressure of the component gases.

$\displaystyle P_T=P_1+P_2+P_3$

$\displaystyle P_T=$ total pressure of a mixture of gases.

$\displaystyle P_1, P_2, P_3 =$ partial pressures of the component gases.

Equal weights of

$CH_{4}$ and

$H_{2}$ are mixed in an empty container at

$25^{\circ}C$ . The fraction of the total pressure exerted by

$H_{2}$ is:

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

Explanation

Ratio of number of moles of

$CH_{4}$ and

$H_{2}$ are

$CH_{4} : H_{2} = \dfrac {x}{16} : \dfrac {x}{2}$

$= 1 : 8$

Hence, partial pressure exerted by

$H_{2} = \dfrac {8}{1 + 8}\times p_{total} = \dfrac {8}{9}\times p_{total}$

All of the following statements are related to the kinetic molecule theory of gases except:

0%
gas molecules have no inter-molecular forces
0%
gas particles are in random motion
0%
the collisions between gas particles are elastic
0%
gas particles have no volume
0%
the average kinetic energy is proportional to the celsius temperature of the gas

Statement I : At STP, 22.4 liters of He will have the same volume as one mole of

$\displaystyle { H }_{ 2 }$ (assume ideal gases).
Statement II : One mole or 22.4 liters of any gas at STP will have the same mass.

0%
true, false
0%
false, true
0%
true, true, correct explanation
0%
true, true, not correct explanation

$CuSO_4$

$\cdot$

$5H_2O(s)$

$\rightleftharpoons$

$CuSO_4\cdot$ 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.

What is the effect on the pressure of a gas if its temperature is increased at constant volume?

0%
The pressure of the gas increases
0%
The pressure of the gas decreases
0%
The pressure of the gas remains same
0%
The pressure of the gas becomes double

Explanation

According to Gay Lussac's law, at constant volume, the pressure of the gas is directly proportional to the temperature.

$\therefore P\propto T$

Thus, with an increase in temperature at constant volume the pressure of the gas increases.

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

$C$ .

The vapour pressures of two liquids A and B in their pure states are in the ratio of 1 :A binary solution of A and B contains A and B in the mole proportion of 1 :The mole fraction of A in the vapour phase of the solution will be:

0%
$0.33$
0%
$0.2$
0%
$0.25$
0%
$0.52$

Explanation

$\dfrac {P^o_A}{P^o_B} = \dfrac{1}{2}$

$\dfrac {x_A}{x_B} = \dfrac{1}{2}$

$P_T \gamma _A =\ P^o_Ax_A$

$\therefore \dfrac{ \gamma _A}{ \gamma _B}= \dfrac {P^o_A}{P^o_B} \dfrac {x_A}{x_B}$

We know

$\gamma _A + \gamma _B =1$

$\therefore \gamma _A = 0.2$

Hence, the correct option is

$\text{B}$

Calculate the temperature at which the root mean square velocity of chlorine gas molecules is twice the root mean square velocity of carbon dioxide gas molecules at

$40^0C$ .

0%
$500 K$
0%
$1000 K$
0%
$1500 K$
0%
$2000 K$

Explanation

The expression for the root mean square velocity is

$\displaystyle u = \sqrt {\dfrac {3RT}{M}}$

For carbon dioxide gas

$\displaystyle u = \sqrt {\dfrac {3R \times (40+273)}{44}}$

$\displaystyle u = \sqrt {3R \times7.114}$ ...... (1)

For chlorine gas

$\displaystyle u = \sqrt {\dfrac {3RT}{71}}$ ..... (2)

The root mean square velocity of chlorine gas molecules is twice The root mean square velocity of carbon dioxide gas molecules.

From (1) and (2)

$\displaystyle \sqrt {\dfrac {3RT}{71}} =2 \times \sqrt {3R \times7.114}$

$\displaystyle \dfrac {3RT}{71} =4 \times 3R \times7.114$

$\displaystyle \dfrac {T}{71}$

$=4 \times 7.114$

$\displaystyle T =71 \times 4 \times 7.114$

$T=2020 \: K \simeq 2000$ K

Ideal gases:

0%
have forces of attraction between them
0%
are always linear in shape
0%
never travel with a straight line motion
0%
have molecules that are close together
0%
have low masses and are spread far apart

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.

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

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

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,

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

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

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

To use Gay-Lussac's Law, which of the following needs to remain constant?

0%
Volume and the number of moles of a gas
0%
Pressure and temperature
0%
Temperature and the number of moles of a gas
0%
Pressure and the number of moles of a gas
0%
Temperature and volume

Explanation

$P \propto T$
At constant volume and number of moles of a gas, the pressure is directly proportional to the absolute temperature.

Hence, option

$A$ is correct.

Avogadro's law shows the relationship between which two variables?

0%
Volume and number of moles
0%
Pressure and number of moles
0%
Volume and pressure
0%
Temperature and pressure
0%
Temperature and number of moles

When

$100\ ml$ of a

$O_{2} - O_{3}$ mixture was passed through turpentine, there was reduction of volume by

$20\ mL$ . If

$100\ ml$ of such a mixture is heated, what will be the increase in volume?

0%
10%
0%
20%
0%
40%
0%
30%

Explanation

Given,

Total volume of ${O_2} - {O_3}$ mixture $= 100mL$

Reduction in volume $= 20mL$

We know that ${O_3}$ gets absorbed in turpentine oil, therefore

Volume of ${O_3}$ that gets absorbed $= 20mL$

Volume of ${O_2} = (100 - 80)mL = 20mL$

On heating, the following reaction takes place between ${O_3}$ and ${O_2}$

${20_3} \rightleftharpoons 3{O_2}$

Since volume is proportional to moles, therefore

${20_3} \rightleftharpoons 3{O_2}$

$2mol$ $3mol$

$20mL$ $30mL$

Therefore, increase in volume will be $= (30 - 20)mL = 10mL$

Hence, percentage increase in volume will be = Increase in volume/Total volume *100

$\Rightarrow \dfrac{{10mL}}{{100mL}} \times 100$

$\Rightarrow 10$ %

The correct answer will be option A.

An ideal gas:

0%
Has no intermolecular attraction
0%
Molecules do not collide with each other
0%
The product of P and V is constant at a fixed temperature for definite mass
0%
Can be liquefied easily

If the vapour are compressed slowly and isothermally , at what pressure , the first drop of liquid will appear ?

0%
0.4 bar
0%
0. 5 bar
0%
0.52 bar
0%
0.6 bar

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

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

At certain temperature (

$T$ ) for the gas phase reaction

$2{ H }_{ 2 }O(g)+2{ Cl }_{ 2 }(g)\rightleftharpoons 4HCl(g)+{ O }_{ 2 }(g);{ K }_{ P }=12\times { 10 }^{ 8 }atm$
If

${Cl}_{2}$ ,

$HCl$ and

${O}_{2}$ are mixed in such a manner that the partial pressure of each is

$2atm$ and the mixture is brough into contact with excess of liquid water. What would be approximate partial pressure of

${Cl}_{2}$ when equilibrium is attained at temperature (

$T$ )?
[Given:Vapour pressure of water is

$380mm$

$Hg$ at temperature (

$T$ )]

0%
$3.6\times { 10 }^{ -5 }atm$
0%
${ 10 }^{ -4 }atm$
0%
$3.6\times { 10 }^{ -3 }atm$
0%
$0.01$ atm

Explanation

$2H_2O + 2Cl_2 \rightleftharpoons 4HCl + O_2$

$K_P = \dfrac{[HCl]^4[O_2]}{[H_2O]^2[Cl_2]^2}$

Given, partial pressure is

$2\space atm$ for

$Cl_2, HCl, O_2.$

$\Rightarrow 12\times 10^8 = \dfrac{(2)^4(2)}{(2)^2[Cl]^2}$

Hence,

$P_{Cl_2}$ is

$3.6\times 10^{-5}\space atm$

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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