MCQ Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 5

An ideal gas undergoes a process in which

$T = T_{0} + aV^{3}$ , where

$T_{0}$ and

$'a'$ are positive constants and

$V$ is molar volume. The molar volume for which pressure will be minimum is :

0%
$\left (\dfrac {T_{0}}{2a}\right )^{1/3}$
0%
$\left (\dfrac {T_{0}}{3a}\right )^{1/3}$
0%
$\left (\dfrac {a}{2T_{0}}\right )^{1/3}$
0%
None of these

Explanation

Given,

$T = T_0 + aV^3$

We know that

$PV = nRT$

$\Rightarrow T =\dfrac{PV}{nR}$

$\Rightarrow \dfrac{PV}{nR} = T_0 + aV^3$

$\Rightarrow \dfrac{P}{nR} = \dfrac{T_0}{V} + aV^2$

Now, for pressure to be minimum

$\dfrac{dP}{dV} = 0$

$-\dfrac{nRT_0}{aV^2} + nRa2V = 0$

$V = (\dfrac{T_0}{2a})^{\dfrac{1}{3}}$

A balloon contains

$14.0\ L$ of air at

$760\ torr.$ . What will be the volume of the balloon when it is taken to a depth of

$10\ ft$ in a swimming pool? Assume that the temperature of the air and water are equal (density:

$Hg=13.6\ g/mL$ )

0%
$11.0$
0%
$11.3$
0%
$10$
0%
$10.8$

Explanation

we know,

$P_1V_1=P_2V_2$

$=$ pressure

$=$ volume

$P_1=$ 760 torr

$V_1=$ 14 L

$10ft = 3.048m$

Density of water (

$\rho$ )= 1000

$g = 9.8\ m/s^2$

We know,

$p = \rho gh$

So,

$P_2 = 1000 \times 9.8\times 3.048+101325$

(we will get the value in Pascal)

In torr, we will get

$984$

So,

$V_2= \dfrac{P_1V_1} {P_2}$

$V_2=760 \times \dfrac{14}{984}=10.8\ L$

The vapour pressure of a solvent decreased by

$10\ mm$ of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is

$0.2$ . What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be

$20\ mm$ of mercury?

0%
$0.8$
0%
$0.6$
0%
$0.4$
0%
$0.2$

Explanation

$\text{Mole fraction of solute}=\cfrac { \text{lowering in vapour pressure} }{\text{ vapour pressure of solvent} }$

comparing under the two conditions

$\cfrac { 0.2 }{ \text{mole fraction of solute} } =\cfrac { 10 }{ 20 } \quad \quad$

or, the mole fraction of solute

$=0.4$

mole fraction of solvent

$=(1-0.4)=0.6$

Hence, the correct option is

$\text{B}$

If the mean free-path of gaseous molecule if 60 cm at a pressure of

$1\times 10^{-4}mm$ mercury what will be its mean free-path when the pressure is increased to

$1\times 10^{-2}$ mm mercury.

0%
$.0 \times 10^{-1} cm$
0%
0.6 cm
0%
$.0 \times 10^{-2} cm$
0%
$.0 \times 10^{3} cm$

Explanation

$\begin{array}{l} \lambda =\frac { { RT } }{ { \sqrt { 2 } \pi { d^{ 2 } }NaP } } & \lambda P=\frac { { RT } }{ { \sqrt { 2 } \pi { d^{ 2 } }Na } } \\ { \lambda _{ 1 } }{ P_{ 1 } }={ \lambda _{ 2 } }{ P_{ 2 } } & \\ 60\times { 10^{ -4 } }={ \lambda _{ 2 } }\times { 10^{ -2 } } & \\ { \lambda _{ 2 } }=60\times { 10^{ -2 } }\, \, cm & \\ { \lambda _{ 2 } }=0.6\, \, cm & \end{array}$

Which of the following mixtures of gases does not obey Dalton's law of partial pressure?

0%
$O_{2}$ and $CO_{2}$
0%
$N_{2}$ and $O_{2}$
0%
$Cl_{2}$ and $O_{2}$
0%
$NH_{3}$ and $HCl$

Explanation

When a mixture of

$NH_3$ and

$HCl$ react chemically with the spontaneous reaction it would lead in the formation of a chemical called

$NH_4Cl$ .

By this, the volume of the gas is automatically changed and from this reaction, there is no formation of pressure.

Hence the two gases do not obey Dalton's law of partial pressure.

Hence, the correct option is

$\text{D}$

Which mixture of gases at room temp does not obey daltons law of partial pressure?

0%
$He + Ne + O_{2}$
0%
$CO_{2} + N_{2} + O_{2}$
0%
$NH_{3} + HCl + O_{2}$
0%
$N_{2} + O_{2} + CH_{4}$

Explanation

The mixture of

$NH_3$ and

$HCl$ at room temperature does not obey Dalton's law of partial pressure.

$NH_3$ and

$HCl$ react spontaneously when putting together within the same apparatus. If they react, then the volume obviously changes. Which will not result in the pressure that you would predict if the two gases don't react and remained as separate entities.

Hence, the correct option is

$\text{C}$

If K.E. of an electron,

$\alpha-particle$ and a proton are represented by

$E_{1}, E_{2}$ and

$E_{3}$ respectively (provided all these have same de Broglie wavelength) then______________.

0%
$E_{1} = E_{2} = E_{3}$
0%
$E_{1} > E_{2} > E_{3}$
0%
$E_{1} > E_{3} > E_{2}$
0%
$E_{2} > E_{3} > E_{1}$

Explanation

$KE=\dfrac{1}{2}mu^2$ and

$\lambda =\displaystyle \dfrac{h}{mu}$

$KE=\displaystyle\frac{1}{2}\cdot \displaystyle\frac{h^2}{m\lambda ^2}=\frac{h^2}{2m\lambda^2}$

Given, all these have the same de Broglie wavelength

$E_1=\cfrac{h^2}{2m_e\lambda^2}$ ,

$E_2=\cfrac{h^2}{2m_p\lambda^2}$

$E_3=\cfrac{h^2}{2m_{\alpha}\lambda^2}$

$E\propto \cfrac{1}{m}$

$m_e<m_{\alpha}<m_p$

So lesser is the mass, higher is the kinetic energy so the order is

$E_1 > E_3 > E_2$

A open ended mercury manometer is used to measure the pressure exerted by a trapped gas as shown in the figure. Initially manometer shows no difference in mercury level in both columns as shown in figure.
After sparking

$A$ dissociates according to following reaction

$A(g)\longrightarrow B(g)+3C(g)$
If pressure of Gas

$A$ decreases to

$0.9atm$ . Then
(Assume temperature to be constant and is

$300K$ )

0%
total pressure increased to $1.3$ atm
0%
total pressure increased by $0.3$ atm
0%
total pressure increased by $22.3cm$ of $Hg$
0%
difference in mercury level is $228mm$

Explanation

Given

Initial pressure of A gas is 1 atm

After sparking A dissociates into B and C

Pressure of A decreases to 0.9 atm

Solution

Let x be the decrease in Pressure of A

$A(g)=B(g)+3C(g)$

t=0 1 0 0

t=t 1-x x 3x

We know that

1-x=0.9

x= 0.1 atm

Total pressure of system at equilibrium is

=1 - x + x + 3x

=1.3 atm

The correct option is A

If the pressure of gas contained in a closed vessel is increased by 0.4% when heated by 1 degree C its initial temperature must be _____

0%
$20K$
0%
$250K$
0%
$25^oC$
0%
$300K$

Explanation

$PV=nRT$ .....(i)

$P\left(1+\dfrac{0.4}{100}\right)V=nRT(T+1)$ ....(ii)

eqn. (ii) divided by (i)

$1+\dfrac{0.4}{100}=\dfrac{T+1}{T}=1+\dfrac{1}{T}$

$T=\dfrac{100}{0.4}=250K$

Hence, the correct option is

$\text{B}$

The blotting paper absorbs liquids because

0%
It has got small pores.
0%
It is white.
0%
An absorbing chemical is mixed.
0%
None of these.

Which one of the following gases is lighter than air?

0%
Carbon dioxide
0%
Chlorine
0%
Oxygen
0%
Hydrogen

Explanation

Hydrogen (density

$0.090$ g/L at STP, average molecular mass

$2.016$ g/mol) and helium (density

$0.179$ g/L at STP, average molecular mass

$4.003$ g/mol) are the most commonly used lift gases. Although helium is twice as heavy as (diatomic) hydrogen, they are both so much lighter than air that this difference only results in hydrogen having

$8\%$ more buoyancy than helium.

The

$K_p$ for

$2SO_{2(g)+O_{2(g)}}\rightleftharpoons 2SO_{3(g)}$ is

$5.0\,atm^{-1}$ .What is the equilibrium partial pressure of

$O_2$ if the equilibrium pressure of

$SO_2$ and

$SO_3$ are equal?

0%
0.2
0%
0.3
0%
0.4
0%
0.1

Explanation

$2SO_2 + O_2 \rightarrow2SO_3$

$K_p$ can be calculated as:

$K_p=\dfrac{[P_{SO_3^{2-}}]}{[P_{SO_2}^2]\times [P_{O_2}]}$

As given:

$[P_{SO_3}]=[P_{SO_2}]$

So,

$5=\dfrac{1}{[O_2]}$

$P_{O_2}=\dfrac{1}{5}$

$P_{O_2}=0.2$

So, partial pressure of oxygen will be 0.2 atm.

0.2 atm will be the partial pressure of $O_2$ .

At same temperature

$N_{2}O_{4}$ is dissociated to

$40$ % and

$50$ % at total pressure

$P_{1}$ and

$P_{2}$ atm respectively in

$NO_{2}$ . Then the ratio of

$P_{1}$ and

$P_{2}$ is?

0%
$\dfrac {4}{5}$
0%
$\dfrac {7}{4}$
0%
$\dfrac {4}{7}$
0%
None of these

Explanation

Given

$N_{2}O_{4}$ dissociated

$P_{1}=40 percent =\alpha_{1}$

$P_{2}=50 percent=\alpha_{2}$

Solution

As temperature is constant

$K_{P_{1}}=K_{P_{2}}$

$K_{P}=\dfrac{\alpha_{1}^{2}}{1-\alpha_{1}^{2}}*P$

Hence

$\dfrac{\alpha_{1}^{2}}{1-\alpha_{1}^{2}}*P_{1}=\dfrac{\alpha_{2}^{2}}{1-\alpha_{2}^{2}}*P_{2}$

Putting Values

$\dfrac{P_{1}}{P_{2}}=\dfrac{7}{4}$

The correct option is B

The property utilized in the manufacture of lead shots is

0%
Specific weight of liquid lead
0%
Specific gravity of liquid lead
0%
Compressibility of liquid lead
0%
Surface tension of liquid lead

The given diagram shows percentage composition of gases

$P, Q, R$ and other gases present in air.
Which of the following statements is/ are correct regarding gases

$P, Q$ and

$R$ ?
(i)

$P$ can be fixed in soil by certain bacteria present in roots of leguminous plants.
(ii)

$P$ supports combustion whereas

$Q$ and

$R$ help in extinguishing fire.
(iii) Plants and animals release

$R$ in respiration whereas plants use

$Q$ in photosynthesis.
(iv)

$R$ entraps sun's heat and makes earth warm and hospitable.

0%
(i) and (iv) only
0%
(ii) and (iii) only
0%
(i), (ii) and (iv) only
0%
(iii) only

The molecular weights of

$O_2$ and

$N_2$ are

$32$ and

$28$ respectively. At

$15^0$ C, the pressure of

$1$ gm

$O_2$ will be the same as that of

$1$ gm

$N_2$ in the same bottle at the temperature:

0%
$-21^0$ C
0%
$-13^0$ C
0%
$15^0$ C
0%
$56.4^0$ V

Explanation

Temperature of

$O_{2}$

$=273+15=298 K$

Temperature of

$N_{2}$

$=x$

$n$

$\propto$

$\dfrac 1T$ , as pressure is constant.

$\Longrightarrow { n }_{ 1 }{ T }_{ 1 }={ n }_{ 2 }{ T }_{ 2 }\Longrightarrow \left( \cfrac { 1 }{ 32 } \right) \left( 298 \right) =\left( \cfrac { 1 }{ 28 } \right) \left( x \right) \\ \Longrightarrow x=\cfrac { 28 }{ 32 } \times 298=260.75K$

Temperature in

$^oC=260.75-273=-13 ^oC$

Hence, option

$B$ is correct.

Lowering of vapour pressure in

$2$ molal aqueous solution at

$373K$ is____________.

0%
$0.35$ bar
0%
$0.022$ bar
0%
$0.22$ bar
0%
$0.035$ bar

For the reaction

$A(g)+B(g)\rightleftharpoons C(g)$ , the equilibrium partial pressures are

${P}_{A}=0.15atm$ ,

${P}_{B}=0.10atm$ and

${P}_{C}=0.30atm$ . The value of the pressure was so reduced that after attainment of equilibrium again, the partial pressures of

$A$ and

$B$ were doubled. The partial pressure of

$C$ would be:

0%
$0.30atm$
0%
$0.60atm$
0%
$1.20atm$
0%
$1.80atm$

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

The rms speed of hydrogen is

$\sqrt { 7 }$ times the rms speed of nitrogen. If T is the temperature of the gas, then:

0%
${ T }_{ { H }_{ 2 } }={ T }_{ { N }_{ 2 } }$
0%
${ T }_{ { H }_{ 2 } }>{ T }_{ { N }_{ 2 } }$
0%
${ T }_{ { H }_{ 2 } }<{ T }_{ { N }_{ 2 } }$
0%
${ T }_{ { H }_{ 2 } }={ \sqrt { 7 } T }_{ { N }_{ 2 } }$

Explanation

$U_{rms}H_2=\sqrt{\frac{3RT_1}{2}}$

$U_{rms}N_2=\sqrt{\frac{3RT_2}{28}}$

$U_{rms}H_2=\sqrt7\times U_{rms}N_2$

$\sqrt{\frac{3RT_1}{2}}=\sqrt7\times\sqrt{\frac{3RT_2}{28}}$

$\therefore\frac{T_1}{2}=\frac{T_2}{4}$

$T_2=2T_1$

$\therefore T_{N_2}>T_{H_2}$

The ratio of the vapour pressure of a solution to the vapour pressure of the solvent is:

0%
equal to the mole fraction of the solvent
0%
equal to the mole fraction of the solute
0%
directly proportional to mole fraction of the solute
0%
None of the above

At a given temperature, total vapour pressure in Torr of a mixture volatile components A and B is given by

$P_{Total}=120-75X_B$
Hence, vapour pressure of pure A and B respectively (in Torr) are ?

0%
120, 75
0%
120, 195
0%
120, 45
0%
75, 45

Explanation

$\displaystyle P_{Total}=P_AX_A+P_BX_B$
But

$\displaystyle X_A=1-X_B$

$\displaystyle ( \because X_A+X_B=1)$

Hence,

$\displaystyle P_{Total}=P_A(1-X_B)+P_BX_B$

$\displaystyle P_{Total}=P_A-P_AX_B+P_BX_B$

$\displaystyle P_{Total}=P_A-(P_A-P_B)X_B$

But,

$\displaystyle P_{Total}=120-75X_B$

Hence,

$\displaystyle P_A=120 \ torr$

$\displaystyle P_A-P_B = 75$

$\displaystyle P_B=P_A-75$

$\displaystyle P_B=120-75$

$\displaystyle P_B=45 \ torr$

Pressure exerted by one mole of an ideal gas kept in vessel of '

$V$ ' L having root mean square molecule '

$V$ ' and '

$m$ ' mass of each molecule is correctly given by the equation?

0%
$P =$ $\dfrac {1}{2}$ $\dfrac {N}{V}$ mv $^{2}$
0%
$P =$ $\dfrac {1}{3}$ $\dfrac {N}{V}$ mv $^{2}$
0%
$P =$ $\dfrac {2}{3}$ $\dfrac {N}{V}$ mv $^{2}$
0%
$P =$ $\dfrac {3}{2}$ $\dfrac {N}{V}$ mv $^{2}$

Explanation

According to ideal gas and kinetic theory of gases, pressure exerted by one mole of ideal gas in a vessel of volume

$V$ , having mean square molecule

$'V'$ and

$'m'$ mass of each molecule is given by

$P=\cfrac { 1 }{ 3 } \cfrac { N }{ V } { mv }^{ 2 }$

Water and chlorobenzene are immiscible liquids. Their mixture boils at

$89^o$ C under a reduced pressure of

$7.7 \times 10^4$ Pa. The vapour pressure of pure water at

$89^o$ C is

$7 \times 10^4$ Pa. Weight percent of chlorobenzene in the distillate is:

0%
50
0%
60
0%
79
0%
38.46

Explanation

Vapour pressure =

$7.7 \times 10^4$ Pa

Total pressure =

$7 \times 10^4$ Pa

Partial pressure = Vapour pressure = Mole fraction x Total pressure

Mole fraction of

$H_2O$ =

$\dfrac{Vapour pressure}{Total pressure}$ =

$\dfrac{7.7 \times 10^4}{7 \times 10^4}$ = 0.909

Mole fraction of

$CHCl_3$ = 1 - 0.909 = 0.091

Molar mass of

$H_2O$ = 18g/mol

Molar mass of

$CHCl_3$ = 119.5g/mol

Weight percent of chloroform = Mole fraction of

$CHCl_3$ x Molar mass of

$CHCl_3$ / [Molar mass of

$CHCl_3$ +Molar mass of

$H_2O$ }

Weight percent of chloroform = 79%

Answer is 79%.

Two glass bulbs A and B are connected by a very small tube having a stop cock. Bulb A has a volume of 100

$cm^{3}$ and contained the gas, while bulb B was empty. On opening the stop cock, the pressure fell down 40%. The volume of the bulb B must be ?

0%
75 $cm^{3}$
0%
125 $cm^{3}$
0%
150 $cm^{3}$
0%
250 $cm^{3}$

Explanation

Let the initial pressure in bulb A be P.

Given that the pressure fell down

$40 \%$ on opening the stop cock

$\therefore$ Final pressure

$= \cfrac{40}{100} P = \cfrac{2}{5} P$

Initial voume of bulb A

$= 100{cm}^{3}$

Let initial volume of bulb B be V.

$\therefore$ total volume after opening the stopcock

$= 100 + V$

By using Boyle's law,

${P}_{i}{V}_{i} = {P}_{f}{V}_{f}$

$P \times 100 = \cfrac{2}{5} P \times \left( 100 + V \right)$

$\Rightarrow \; V = 150 {cm}^{3}$

Hence, the volume of bulb B must be

$150{cm}^{3}$ .

A pre-weighed vessel was filled with

$O_{2}$ at STP and weighed. It was then evacuated, filled with

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

$O_{2}$ will be

$[O = 16, S = 32]$ .

0%
The same as that of the $SO_{2}$
0%
Twice as that of the $SO_{2}$
0%
Half of that of the $SO_{2}$
0%
One-fourth of that of the $SO_{2}$

Explanation

Half of that of the

$S{O}_{2}$

Given:-

Wt. of

${O}_{2} = 32$

Wt. of

$S{O}_{2} = 64$

$\therefore \; \cfrac{\text{Mol. wt. of } {O}_{2}}{\text{Mol. wt. of } {O}_{2}} = \cfrac{32}{64} = \cfrac{1}{2}$

$\Rightarrow \; \text{Mol. wt. of } {O}_{2} = \cfrac{1}{2} \text{Mol. wt. of } S{O}_{2}$

Hence, The weight of

${O}_{2}$ will be half of that of the

$S{O}_{2}$ .

$80^0$ C, the vapour pressure of pure liquid A is

$250$ mm of Hg and that of pure liquid B is

$1000$ mm of Hg. If a solution of A and B boils at

$80^0$ C and

$1$ atm pressure, the amount of A in the mixture is :

$(1atm = 760mm$ Hg)

0%
$50$ mole percent
0%
$52$ mole percent
0%
$32$ mole percent
0%
$48$ mole percent

Explanation

1 atm = 760 mm Hg =

$P_T$

$P_T = P^0_A X_A \times P^0_BX_B$

$760 = 250X_A + 1000(1 - X_A)$

$240 = 750X_A$

$X_A = 0.32$

Hence mole

$\%$ of A =

$32\%$

Which of the following statement is wrong ?

0%
For gas A, a= 0 and z will linearly depend on pressure
0%
For gas B, b = 0 and z will linearly depend on pressure
0%
Gas C is real gas and we can find "a" and 'b' if intersection data is given
0%
All vander waal gases will behave like gas C and give positive slope at high pressure

Explanation

According to the real gas equation, and characteristic of constants 'a' and 'b'

$(P-\cfrac{n^2a}{V^2}) (V-nb)=RT$

$Z=\cfrac{PV}{RT}$

For gas A, a= 0 and z will linearly depend on pressure as per the below relation-

$Z=\cfrac{P(V-nb)}{RT}$

For gas B, b= 0 and z will linearly depend on the volume, not pressure as per the below relation-

$Z=\cfrac{(P-\cfrac{n^2a}{V^2})V}{RT}$

The dotted plot given by gas C is for real gases.

At higher pressure Z>1.

Hence, statement B is incorrect.

$80^o$ C, the vapour pressure of pure liquid

$A$ is

$250$ mm of

$Hg$ and that of pure liquid

$B$ is

$1000$ mm of

$Hg$ . If a solution of

$A$ and

$B$ boils at

$80^o$ and

$1$ atm pressure, the amount of

$A$ in the mixture is?

(

$1$ atm

$=760$ mm Hg).

0%
$50$ mole percent
0%
$52$ mole percent
0%
$32$ mole percent
0%
$48$ mole percent

Explanation

1 atm = 760 mm Hg =

$P_T$

$P_T = P^0_A X_A \times P^0_BX_B$

$760 = 250X_A + 1000(1 - X_A)$

$240 = 750X_A$

$X_A = 0.32$

Hence mole

$\%$ of A

$= 32\%$

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.

On a humid day in summer, the mole fraction of gaseous

$H_{2}O$ (water vapour) in the air at

$25^0$ C can be as high as

$0.0287$ . Assuming a total pressure of

$0.977$ atm. What is the partial pressure of dry air?

0%
$94.9$ atm
0%
$0.949$ atm
0%
$949$ atm
0%
$0.648$ atm

Explanation

$p_{H_2O}=X_{H_2O} p_{\text{total}}$

$=0.0287\times 0.977 =0.028atm$

$p_{\text{total}}=p_{\text{dryair}}+p_{H_2O}$

$p_{\text{dryair}}=p_{\text{total}}−p_{H_2O} =0.977−0.028=0.949 atm$

Which of the following relationships for various gas laws is not correct?

0%
$V_{t} = V_{0} + \dfrac {V_{0}}{273}\times t$
0%
$\dfrac {V_{1}}{T_{1}} = \dfrac {V_{2}}{T_{1}} (constant\ P)$
0%
$\dfrac {P_{1}}{T_{1}} = \dfrac {P_{2}}{T_{1}} (constant\ V)$
0%
$\dfrac {P_{1}T_{1}}{V_{1}} = \dfrac {P_{2}T_{2}}{V_{2}}$

Explanation

According to ideal gas equation

$PV=nRT$

$\therefore { P }_{ 1 }{ V }_{ 1 }=nR{ T }_{ 1 }$ or

${ P }_{ 1 }{ V }_{ 1 }/{ T }_{ 1 }=nR$

${ P }_{ 2 }{ V }_{ 2 }=nR{ T }_{ 2 }$ or

${ P }_{ 2 }{ V }_{ 2 }/{ T }_{ 2 }=nR$

$\Rightarrow \cfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\cfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } }$

Hence

$\cfrac { { P }_{ 1 }{ T }_{ 1 } }{ { V }_{ 1 } } =\cfrac { { P }_{ 2 }{ T }_{ 2 } }{ { V }_{ 2 } }$ is incorrect

Two flasks of equal volume have been joined by a narrow tube of negligible volume. Initially, both flasks are at 300K containing 0.6 mole of

$O_2$ at 0.5 atm pressure. one of the flasks is placed in the thermostat of 600K. FInd the number of mole of

$O_2$ gas in each flask.

0%
0.5mol and 0.3 mol
0%
0.5mol and 0.2 mol
0%
0.4mol and 0.2 mol
0%
0.6mol and 0.3 mol

Explanation

$0.5 V_1 =0.6R\times 300$

$V = 360R$

in 2nd case the pressure will be same for both flask and sum of moles is 0.6

Flask-1

$P\times 360R = n_1R\times 300$

$n_1 =1.2P$

Flask-2

$P\times 360R = n_2R\times 600$

$n_2 = 0.6P$

$n_1+n_2 = 0.6$

$1.2P+0.6P =0.6$

$1.8P = 0.6$

Final pressure-

$P = \dfrac{1}{3}$

Number of moles of oxygen in flasks-

$n_1 = 1.2\times \dfrac{1}{3}= 0.4$

$n_2 = 0.6\times \dfrac{1}{3} = 0.2$

Which of the following values does not represent the correct value of

$R$ ?

0%
$8.314\ Pa\ m^{2} K^{-1} mol^{-1}$
0%
$8.314\times 10^{-2} bar\ L\ K^{-1} mol^{-1}$
0%
$0.0821\ J\ K^{-1} mol^{-1}$
0%
$8.314\ J\ K^{-1} mol^{-1}$

Explanation

The value of ideal gas constant in vaious units can be given as follow:

$R = 8.314\ Pa\ m^{3}\ K^{-1} mol^{-1}$

$R = 8.314\times 10^{-2} bar\ L\ K^{-1} mol^{-1}$

$R = 8.314\ J\ K^{-1} mol^{-1}$

$R = 0.0821\ L\ atm\ K^{-1} mol^{-1}$ .

Thus value given in option C is incorrect.

The statements for laws of chemical combinations are given below. Mark the statement which is not correct.

0%
Matter can neither be created nor destroyed: Law of conservation of mass
0%
A compound always contains exactly the same proportion of elements by weight: Law of definite proportions
0%
When gases combine they do so in a simple ratio by weight: Gay Lussac's Law
0%
Equal volumes of gases at same temperature and pressure contain same number of molecules: Avogadro's Law

$6 g$ . of the area of dissolved in

$90 g$ . of boiling water. The vapour pressure of the solution is:

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

Explanation

Vapour pressure

$P=P°x,\quad P°=$ Boiling point of pure solvent

$P=P°\cfrac { { n }_{ 1 } }{ { n }_{ 1 }+{ n }_{ 2 } }$

$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right) \quad \quad 90g{ H }_{ 2 }O=5mol={ n }_{ 2 };\quad 6g$ urea

$=0.1mol={ n }_{ 1 }$

$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right)$

$=760\times 0.0196=14.89$ Torr=lowered pressure

Now, 1 torr = 1 mm of Hg

Vapour pressure of solution

$=760-14.89\approx 744.8 mm\ of\ Hg$

Option A is correct.

Which of the following statement about Avogadro's hypothesis is correct?

0%
Under similar conditions of temperature and pressure, gases react with each other in simple ratio.
0%
Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules
0%
At NTP all gases contain same number of molecules
0%
Gases always react with gases only at the given temperature and pressure

Two moles of pure liquid 'A' (

$P^o_A$ = 80mm of Hg) and 3 moles of pure liquid 'B' (

$P^o_B$ =

$120$ mm of

$Hg$ ) are mixed. Assuming ideal behaviour.

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

Explanation

Formula of Ammonia =

$NH_{3}$

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

$n_{A}=2$ mole

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

$n_{B}=3$ mole

$x_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{2}{2+3}=0.4$

$x_{B}=\frac{n_{B}}{n_{A}+n_{B}}=\frac{3}{2+3}=0.6$

Vapour pressure

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

$P_{A}=80\times 0.4+120\times 0.6$

$=104 mmHg$

Option A is correct

At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole percent which at

$80$ is in equilibrium with vapour containing 30 mole percent of benzene.

0%
24, 56
0%
26 ,54
0%
56 24
0%
34 ,76

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

0%
$H_{2}$ and $He$
0%
$NH_{3}$ and $HCl$
0%
$N_{2}$ and $H_{2}$
0%
$N_{2}$ and $O_{2}$

Explanation

Daltons' law of partial pressure is applicable to non-reacting gases,

$NH_{3}$ and

$HCl$ are reacting gases, so Dalton's law will not be applicable for them.

$10\ L$ flask contains a gaseous mixture of

$CO$ and

$CO_{2}$ at a total pressure of

$2\ atm$ and

$298\ K$ . If

$0.20\ mole$ of

$CO$ is present, find its partial pressure.

0%
$0.49\ atm$
0%
$1.51\ atm$
0%
$1\ atm$
0%
$2\ atm$

Explanation

Here,

$V=10L$ ,

$n=0.20mole$ , and

$T=298K$

By Ideal gas equation

$PV=nRT$

$\therefore P = \dfrac {nRT}{V} = \dfrac {0.20\times 0.0821\times 298}{10} = 0.49\ atm$ .

How many number of moles of nitrogen will be present in

$2.24\ L$ of nitrogen gas at

$STP?$

0%
$9.9$
0%
$0.099$
0%
$0.001$
0%
$1.00$

Explanation

$PV = nRT$

$P = 1\ atm, T = 273\ K, V = 2.24\ L$

$R = 0.0821\ L\ atm\ K^{-1} mol^{-1}$

$n = \dfrac {1\times 2.24}{0.0821\times 273} = 0.099$ moles.

Equal masses of helium and oxygen are mixed in a container at

$25^{\circ}C$ . The fraction of the total pressure exerted by oxygen in the mixture of gases is :

0%
$1/3$
0%
$2/3$
0%
$1/9$
0%
$4/9$

Explanation

Mole fraction of

$O_{2} = \dfrac {\dfrac {w}{32}}{\dfrac {w}{32} + \dfrac {w}{4}} = \dfrac {w}{32}\times \dfrac {32}{9w} = \dfrac {1}{9}$
Partial pressure of oxygen

$= P \times x_{O_{2}}$

$= P_{total}\times \dfrac {1}{9}$ or

$\dfrac {1}{9}P_{total}$ .

A gas occupies a volume of

$300\ cm^{3}$ at

$27^{\circ}C$ and

$620\ mm$ pressure. The volume of gas at

$47^{\circ}C$ and

$640\ mm$ pressure is:

0%
$260\ cm^{3}$
0%
$310\ cm^{3}$
0%
$390\ cm^{3}$
0%
$450\ cm^{3}$

Explanation

Here,

$V_1=300cm^3$ ,

$T=27^{\circ}C=300K$ ,

$P_1=620mm$

$T_2=47^{\circ}C=320K$ ,

$P_2=640mm$

From Ideal gas equation,

$\dfrac {P_{1}V_{1}}{T_{1}} = \dfrac {P_{2}V_{2}}{T_{2}} \Rightarrow \dfrac {620\times 300}{300} = \dfrac {640\times V_{2}}{320}$

$V_{2} = \dfrac {620\times 300\times 320}{640\times 300} = 310\ cm^{3}$ .

A closed container contains equal number of moles of two gases

$X$ and

$Y$ at a total pressure of

$710\ mm$ of

$Hg$ . If gas

$X$ is removed from the mixture, the pressure will :

0%
become double
0%
become half
0%
remain same
0%
become one-fourth

Explanation

Here in the closed container, the equal number of moles of two gases are taken.

Thus, when gas

$X$ is removed, the number of moles of the gases in the container becomes half.

According to an Ideal gas equation,

$PV = nRT$ .

$\therefore P \propto n$

Thus, if all the other conditions are kept constant, the pressure is directly proportional to the number of moles.

Thus, if the number of moles of the gas is halved, the pressure will also become half of the original pressure.

Thus, B is the correct answer.

What will be the volume of

$2.8\ g$ of carbon monoxide at

$27^{\circ}C$ and

$0.821$ atmospheric pressure?

0%
$2.5\ L$
0%
$4\ L$
0%
$3.5\ L$
0%
$3\ L$

Explanation

Given,

$m=2.8\ g$ ,

$T=27^{\circ}C=300K$ , and

$P=0.821\ atm$

$n = \dfrac {m}{M} = \dfrac {2.8}{28} = 0.1$

By Ideal gas equation;

$PV=nRT$

$\therefore V = \dfrac {nRT}{P} = \dfrac {0.1\times 0.0821\times 300}{0.821} = 3\ L$ .

Density of a gas is found to be

$5.46\ g/ dm^{3}$ at

$27^{\circ}C$ and

$2\ bar$ pressure. What will be its density at STP?

0%
$3.0\ g\ dm^{-3}$
0%
$5.0\ g\ dm^{-3}$
0%
$6.0\ g\ dm^{-3}$
0%
$10.82\ g\ dm^{-3}$

Explanation

$P = \dfrac {n}{V}RT = \dfrac {m}{V} \times \dfrac {RT}{M} = \dfrac {dRT}{M}$

$d = \dfrac {PM}{RT}$ or

$d\propto \dfrac {P}{T}$

$\dfrac {d_{1}}{d_{2}} = \dfrac {P_{1}}{P_{2}}\times \dfrac {T_{2}}{T_{1}}$ or

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

$d_{2} = \dfrac {5.46\times 1\times 300}{2\times 273} = 3.0\ g \ dm^{-3}$

Therefore, the correct option is A.

What volume in litres will be occupied by

$4.4$ g of

$CO_2$ at STP?

0%
$22.4\ L$
0%
$44.8\ L$
0%
$12.2\ L$
0%
$2.24\ L$

Explanation

$44\ g$ of

$CO_{2}$ occupies

$22.4\ L$

$4.4\ g$ of

$CO_{2}$ occupies

$\dfrac {22.4}{44}\times 4.4 = 2.24\ L$ .

What will be the pressure of the gas mixture of

$3.2\ g$ methane and

$4.4\ g$ carbon dioxide contained in at

$9\ dm^{3}$ flask at

$27^{\circ}C$ ?

0%
$0.82\ atm$
0%
$8.314\times 10^{4}\ atm$
0%
$1\ atm$
0%
$1.8\ atm$

Explanation

From Ideal gas law-

$P = \dfrac {n}{V}RT$

$n$ for

$CH_{4} = \dfrac {3.2}{16} = 0.2;$

$n$ for

$CO_{2} = \dfrac {4.4}{44} = 0.1$

$n_{total} = 0.2 + 0.1 = 0.3$

$P = \dfrac {0.3\times 0.0821\times 300}{9} = 0.82\ atm$ .

Hence, option A is correct.

$34.05\ mL$ of phosphorus vapours weight

$0.0625\ g$ at

$546^{\circ}C$ and

$0.1\ bar$ pressure. What is the molar mass of phosphorus?

0%
$124.77\ g\ mol^{-1}$
0%
$1247.74\ g\ mol^{-1}$
0%
$12.47\ g\ mol^{-1}$
0%
$30\ g\ mol^{-1}$

Explanation

Given,

$V=34.05mL$ ,

$m=0.0625g$ ,

$T=546^{\circ}C=819K$

and

$P=0.1\ bar$

Thus, molar mass of the phosphorus can be given as;

$M = \dfrac {mRT}{PV} = \dfrac {0.0625\times 0.083\times 819}{0.1\times (34.05/ 1000)}$

$= 1247/74\ g\ mol^{-1}$ .

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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