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 Medical Chemistry States Of Matter Gases And Liquids Quiz 10

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,

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

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

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.

Calculate the volume of

$CO_{2}$ evolved by the combustion of

$50$ ml of a mixture containing

$40$ per

$C_{2}H_{4}$ and

$60$ per

$CH_{4}$ (by volume).

0%
$70$ ml
0%
$75$ ml
0%
$80$ ml
0%
$82$ ml

Explanation

The volume of

$C_2H_4$ in

$50$ ml of sample is

$50 ml \times \dfrac {40} {100} =20 ml$ .

The volume of

$CH_4$ in

$50$ ml of sample is

$50 ml \times \dfrac {60} {100} =30 ml$ .

$\underset {30 ml}{CH_4} \rightarrow \underset {30 ml}{CO_2} + \underset {60 ml} {2H_2O}$

$\underset {20 ml}{C_2H_4} \rightarrow \underset {40}{2CO_2} + \underset {40 ml} {2H_2O}$
The volume of carbon dioxide formed is

$30 ml + 40 ml=70 ml$ .

The pressure exerted by saturated water vapour is called aqueous tension. The pressure of the dry gas can be calculated as?

0%
$\displaystyle P_{Dry \: gas}=P_{Total}-$ aqueous tension
0%
$\displaystyle P_{Dry \: gas}=P_{Total}+$ aqueous tension
0%
$\displaystyle P_{Dry \: gas}=P_{Total}/$ aqueous tension
0%
$\displaystyle P_{Dry \: gas}=P_{Total}*$ aqueous tension

Explanation

Pressure exerted by saturated water vapour is called aqueous tension. Pressure of dry gas can be calculated as

$\displaystyle P_{Dry \: gas}=P_{Total}-$ aqueous tension.
Thus, the pressure of dry gas is the difference between the total pressure and aqueous tension.

$45.4$ 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%
Henry's Law
0%
Gay Lussac's law of gaseous volumes
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.

Calculate the mean free path in

$\displaystyle CO_{2}$ at

$\displaystyle 27^{\circ}C$ and a pressure of

$\displaystyle 10^{-9}$ bar. Molecular diameter of

$CO_2$ is

$500$ pm.

[ Given : R

$\displaystyle =\frac{25}{3}$

$\displaystyle J.mol^{-1}K^{-1}, \sqrt{2}=1.4, \pi =\frac{22}{7}, N_{A}=6\times 10^{23}$ ]

0%
$\displaystyle 3.788\times 10^{3}$ cm
0%
$\displaystyle 7.47\times 10^{3}$ cm
0%
$\displaystyle 6.43\times 10^{5}$ cm
0%
None of these

Explanation

The number of molecules per unit volume

$\displaystyle N = \frac {n}{V} \times N_A= \frac {P}{RT} \times N_A= \frac {10^{-9}}{1.01325 \times 0.0821 \times 300} \times 6.023 \times 10^{23}$

$= 2.4 \times 10^{13} \: \text{molecules/dm}^3 = 2.4 \times 10^{10} \: \text{molecules/cm}^3$

The mean free path is

$\displaystyle \frac {1}{\sqrt {2} \times \pi \times \sigma ^2 \times N} = \frac {1}{1.414 \times 3.1416 \times (500 \times 10^{-10})^2 \times 2.4 \times 10^{+10}} = 3.788 \times 10^3 \: cm$

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%
$300 cm^{3}$

According to kinetic theory of gases:

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

Explanation

As per the postulates of the kinetic theory of gases, the molecules of gas collide with each other and with the walls of the container. These collisions are perfectly elastic in nature.

The gas molecule with mass

$m$ and speed

$u$ collides with the walls of the container to transfer the momentum

$\Delta p=-2mu$ . Thus heavier molecules transfer more momentum to the walls of the container. But this is not the postulate of the kinetic theory of gases.

As per Maxwell -Boltzmann distribution of molecular speed, very few molecules move with very high or very low speed. Most of the molecules move with average speed. Hence, only a small number of molecules have very high velocity.

According to the kinetic theory of gases, a gas molecule moves in a straight line with constant velocity. When it collides with another molecule, it changes direction. Between collisions, the molecules move in straight lines with constant velocities.

The compound A undergoes reaction as shown above.

Find out the reduction product of

$'A'$ .

0%
Isoamyl amine
0%
Secondary butyl amine
0%
Neopentyl amine
0%
Tertiary amyl amine

A mixture of hydrogen and helium is prepared such that the number of wall collisions per unit time by molecules of each gas is the same. Which gas has the higher concentration?

0%
Helium
0%
Hydrogen
0%
Both have same concentration
0%
None of these

Explanation

The number of wall collisions per unit time by molecules of each gas is the same. Hence, the pressure is the same.

$\displaystyle P=CRT$

Since pressure is directly proportional to concentration, the two gases have equal concentration.

Hence, the correct answer is option

$\text{C}$ .

The gas constant has units:

0%
$L \: atm \: K^{-1} \: mol^{-1}$
0%
$L \: atm^{-1} \: K^{-1} \: mol^{-1}$
0%
$atm \:cm^3 \:K^{-1} \: mol^{-1}$
0%
$erg\: K^{-1}$

Explanation

$\displaystyle R=\frac{PV}{nT} \therefore R \:(unit) =L \: atm \: K^{-1} \: mol^{-1}$ and

$atm \:cm^3 \:K^{-1} \: mol^{-1}$

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 statements are true about the law of chemical combination?

0%
Potassium combines with two isotopes of chlorine ( $^{35}Cl$ and $^{37}Cl$ ) to form two samples of $KCl$ . Their formation follows the law of definite composition.
0%
Different proportions of oxygen in the various oxides of sulphur proves the law of multiple proportions.
0%
$H_2O$ and $H_2S$ contain $11.11\%$ hydrogen and $5.88\%$ hydrogen respectively whereas $SO_2$ contains $50\%$ sulphur. The above data prove the law of reciprocal proportions.
0%
In the decomposition of $NH_3\ (2NH_3\xrightarrow {\Delta}N_2+3H_2)$ , the ratio of volumes of $NH_3, N_2$ and $H_2$ is $2:1:3$ . The above data proves the Gay Lussac's law.

Explanation

(A) Potassium combines with two isotopes of chlorine (

$^{35}Cl$ and

$^{37}Cl$ ) to form two samples of

$KCl$ . Their formation does not follow the law of definite composition as the proportion of chlorine in two samples are different.
Hence, the statement A is false.

(B) Different proportions of oxygen in the various oxides of sulphur prove the law of multiple proportions.
This is because the ratio of masses of oxygen that combine with same mass of sulphur (to form

$SO_2$ and

$SO_3$ ) is

$2:3$ , a simple ratio.
Hence, the statement B is true.

(C)

$H_2O$ and

$H_2S$ contain

$11.11\%$ hydrogen and

$5.88\%$ hydrogen respectively whereas

$SO_2$ contains

$50\%$ sulphur. The above data proves the law of reciprocal proportions.
According to the law of reciprocal proportions, "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine."
Hence, the statement C is true.

(D) In the decomposition of

$NH_3\ (2NH_2\xrightarrow {\Delta}N_2+3H_2)$ , the ratio of volumes of

$NH_3,\ N_2$ and

$H_2$ is

$2:1:3$ . The above data proves the Gay Lussac law.
According to the Gay Lussac law, "The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers."
Hence, the statement D is true.

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".

$P\propto T$ .

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

3000 cc of oxygen was burnt with 600 cc of ethane(

$C_2H_6$ ). Calculate the volume of unused oxygen.

0%
300 cc
0%
600 cc
0%
900 cc
0%
1200 cc

Explanation

The reaction is as follows:

$C_2H_6+\dfrac72O_2\longrightarrow2CO_2+3H_2O$

600 cc. of ethane reacts with

$600\times\dfrac72$ , i.e., 2100 cc volume of oxygen.

Remaining volume of

$O_2$ is

$3000-2100=900\ cc.$

60 cc of oxygen was added to 24 cc of carbon monoxide and the mixture ignited. Calculate the volume of

$O_2$ used up and the volume of

$CO_2$ formed.

0%
15 cc, 30 cc
0%
12 cc, 24 cc
0%
20 cc, 40 cc
0%
10 cc, 20 cc

Which one of the following is the wrong statement about the liquid?

0%
It has intermolecular force of attraction
0%
Evaporation of liquids increases with the decrease of surface area
0%
It resembles a gas near the critical temperature
0%
It is an intermediate state between gaseous and solid state

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.

For an ideal gas, number of mol per litre in terms of its pressure p, temperature T and gas constant R is:

0%
p/RT
0%
p/R
0%
pT/R
0%
RT/p

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

$D$ .

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

$\text{C}$

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.

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

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

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

$\text{B}$

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.

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

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

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

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

$P_{gas}$ according to the diagram

$[d_{Hg} = 13.6 g/cm^3$ ]:

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

Explanation

The atmospheric pressure is balanced by the sum of the pressures of gases, the water column of height

$h$ , and the 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,

$P_{atm}$ =

$P_{gas}$ + [

$aq. tension$ +

$\dfrac{h}{13.6}$ ]

⇒

$P_{gas}$ =

$P_{atm}$ - [

$aq. tension$ +

$\dfrac{h}{13.6}$ ]

Hence, option A is correct.

The density of a liquid is 1.2 g/mL. That are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70):

0%
$\frac{1.2}{35} N_A$
0%
$\left (\frac{1}{35} N_A \right )^2$
0%
$\frac{1.2}{(35)^2} N_A$
0%
$1.2 N_A$

Explanation

Volume of one drop

$=\frac{2}{35}\, mL$

Moles in one drop

$=\frac{2 \times 1.2}{35 \times 70}=\frac{1.2}{(35)^2}\ mol$

Number of molecules in one drop

$=\frac{1.2}{(35)^2}\times N_A$

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,

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

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

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

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.

$18$ 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

$C$ .

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

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

The reaction,

$ZnO(s)+CO(g) \rightleftharpoons Zn(g)+{CO}_{2}(g)$ , has an equilibrium constant of

$1$ atm at

$1500K$ . The equilibrium partial pressure of zinc vapour in a reaction vessel if an equimolar mixture of

$CO$ and

${CO}_{2}$ is brought into contact with solid

$ZnO$ at

$1500K$ and the equilibrium is achieved at

$1$ atm is?

0%
$0.68$ atm
0%
$0.76$ atm
0%
$0.24$ atm
0%
$0.5$ atm

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

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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