MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 4

Mole fraction of the component A in vapour phase is

$x_1$ and mole fraction of component A in liquid mixture is

$x_2$ , then (

$p^0_A =$ vapour pressure of pure A;

$p^0_B =$ vapour pressure of pure B), the total vapour pressure of liquid mixture is :

0%
$p^0_A\dfrac{x_2}{x_1}$
0%
$p^0_A\dfrac{x_1}{x_2}$
0%
$p^0_B\dfrac{x_1}{x_2}$
0%
$p^0_B\dfrac{x_2}{x_1}$

Explanation

Mole fraction of component A in vapour phase is

$x_1$

that means

$P_A = P^0_A * x_1$

and mole fraction of compound A in liquid mistureis

$x_2$

that means we have,

$P_A = x_2 * P^0_A$

so, from both equation we get

$x_2 * P^0_A = x_1 * P_t$

so,

$P_t = \dfrac{x_2}{x_1}* P^0_A$

What is the vapour pressure of a solution of glucose which has an osmotic pressure of 3 atmospheres at

$20^C?$ The vapour pressure of water at

$20^oC$ is

$17.39 mm$ . Consider the density of solution equal to that of the solvent.

0%
$12.35 mm$
0%
$14.35 mm$
0%
$16.35 mm$
0%
$17.35 mm$

Which of the following are correct for an ideal gas?

0%
${ \left( \cfrac { \partial U }{ \partial V } \right) }_{ T }=0\quad$
0%
${ \left( \cfrac { \partial H }{ \partial P } \right) }_{ T }=0$
0%
${ \left( \cfrac { \partial T }{ \partial P } \right) }_{ H }=0\quad$
0%
${ \left( \cfrac { \partial P }{ \partial T } \right) }_{ H }=0\quad$

$25^oC$ , the vapour pressure of pure benzene is 100 torr, while that of pure ethyl alcohol is 44 torr. Assuming ideal behaviour, calculate the vapour pressure at

$25^oC$ of a solution which contains 10 g of each substance.

0%
33.775 torr
0%
54.775 torr
0%
64.775 torr
0%
60.775 torr

Explanation

We have,

moles of benzene =

$\dfrac{10}{78}$ = 0.128

moles of ethyl alcohol =

$\dfrac{10}{46}$ = 0.217

$P_t = P^0_A * x_A + P^0_B * x_B$

$P_t = 100 * \dfrac{0.128}{0.345} + 44*\dfrac{0.217}{.345}$

$P_t = 64.77 torr$

5 gram each of the following gases at

$87^{\circ}C$ and

$750\ mm$ pressure is taken. Which of them will have the least volume?

0%
$HF$
0%
$HCl$
0%
$HBr$
0%
$HI$

Explanation

From Ideal gas law,

$PV = \dfrac{Weight}{Molar\ mass}\times RT$

The gas with maximum molecular weight will have minimum volume.

$HI$ has the maximum molecular weight, hence has the least volume.

For 1 mole of a real gas. graph between P and V is given as at

${27^ \circ }$ C

0%
2 atm
0%
3 atm
0%
30 atm
0%
20 atm

Which of the following do not pertain to the postulates of kinetic theory of gases?

0%
The gas molecules are perfectly elastic
0%
Speed of gas molecules are every changing
0%
Pressure exerted by the gas is due to the collision of molecules with the walls of the container
0%
Kinetic energy of a gas is given by the sum of $273$ and temperature in Celsius scale

Plot of Maxwell's distribution of velocities is given below:
which of following is correct?

0%
$T_{1} > T_{2}$
0%
$T_{1} < T_{2}$
0%
$v_{1} < v_{2}$
0%
$f_{1} > f_{2}$

Explanation

On increasing temperature, the molecular velocity increases but fraction of the molecules possessing that velocity decreases. Therefore,

$T_1>T_2$ .

Which of the following property indicates weak intermolecular forces of attraction in liquid?

0%
High heat of vaporization
0%
High vapour pressure
0%
High critical temperature
0%
High boiling point

Explanation

The molecules that are having weak intermolecular forces of attraction, they can easily dissociate and convert into vapour state,

so, molecules with weak intractions have high vapour pressure.

Hence, option

$(B)$ is correct.

$27^{\circ}C$ a sample of ammonia gas exerts a pressure of

$5.3\ atm$ . What is the pressure when the volume of the gas is reduced to one-tenth of the original value at the same temperature?

0%
$0.53\ atm$
0%
$5.3\ atm$
0%
$53\ atm$
0%
None of these

Explanation

$\begin{array}{l} Pv=nRT \\ { p_{ 1 } }{ v_{ 1 } }={ p_{ 2 } }{ v_{ 2 } } \\ \Rightarrow 5.3\lambda \frac { v }{ 1 } ={ p_{ 2 } }\times \frac { { \frac { v }{ 1 } } }{ { 10 } } \\ \Rightarrow { p_{ 2 } }=53\, atm. \\ Hence,\, the\, option\, C\, is\, the\, required\, answer. \end{array}$

Soap bubbles can be formed floating in air by blowing soap solution in air, with the help of a glass tube, but not water bubbles. It is because

0%
The excess pressure inside water bubble being more due to large surface tension
0%
The excess pressure inside water bubble being less due to large surface tension
0%
The excess pressure inside water bubble being more due to large viscosity
0%
The excess pressure inside water bubble being less due to less viscosity

The following graphs illustrate:

0%
Dalton's law
0%
Boyle's law
0%
Charles' law
0%
Gay-Lussac's law

Explanation

Chrle's law-

It states that at constant pressure, the volume of a fixed mass of a gas is directly proportional to the temperature.

$PV=nRT\longrightarrow$ Constant pressure

$V=(\cfrac{nR}{P})T\Longrightarrow y=mx$

$V\propto T$ straight line,

1105193_667491_ans_19a78eeb52d44ce68c8b4c1ee0ad1df2.png

The temperature, at which the density of

$O_{2}$ at

$1\ atm$ , is the same as that of

$CH_{4}$ at S.T.P is:

0%
$273^{\circ}C$
0%
$100^{\circ}C$
0%
$546^{\circ}C$
0%
$150^{\circ}C$

Explanation

$\begin{array}{l} density\alpha \frac { { pressure } }{ { \sqrt { temperature } } } \\ \frac { { { d_{ { o_{ 2 } } } } } }{ { { d_{ C{ H_{ 4 } } } } } } =\frac { { P{ o_{ 2 } }\times \sqrt { { T_{ C{ H_{ 4 } } } } } } }{ { \sqrt { { T_{ { o_{ 2 } } } } } \times { P_{ C{ H_{ 4 } } } } } } \\ \sqrt { { T_{ { o_{ 2 } } } } } =\frac { { 1\times \sqrt { 373 } } }{ { 0.99 } } \\ { T_{ { o_{ 2 } } } }=\frac { { 373 } }{ { { { \left( { 0.99 } \right) }^{ 2 } } } } =380.57=\sim { 100^{ \circ } }C \\ Hence,\, option\, B\, \, is\, the\, correct\, answer. \end{array}$

Equal weights of two gases of molecular weight

$4$ 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%
$0.15$ atm
0%
$1$ atm
0%
$0.11$ atm

Explanation

Number of moles of lighter gas

$=\dfrac{m}{4}$

Number of moles of heavier gas

$=\dfrac{m}{40}$

Total Number of moles

$=\dfrac{m}{4} + \dfrac{m}{40} = \dfrac{11m}{40}$

Mole fraction of lighter gas

$=\dfrac{m}{40} \times \dfrac{40}{11m} = \dfrac{10}{11}$

Partial pressure due to lighter gas

$= P_o \times \dfrac{10}{11}$

$=1.1 \times \dfrac{10}{11} = 1 atm$

Assume that you take a flask, evacuate it to remove all the air and find it's mass to be 478.1 g .You then fill the flask with argon to a pressure of 2.15 atm and reweigh it. What would the balance read (in grams) of the flask has a volume of 7.35 L and the temperature is

$20^0 C$ ?

0%
203.6 g
0%
504.3 g
0%
471.1 g
0%
None of these

Explanation

Given that,

$P=2.15atm$

$V=7.35L$

$T=20.0$

$R=8.3145$ (gas constant )

$n=?$

$PV=nRT$

$2.15 \times 7.35= n\times 8.31 \times 20$

$n=0.0950$

therefore mass =

$0.0950 \times 39.95$

$3.80g$

But original mass =

$478.1+3.80g$

$=481.90g$

Thus, option

$D$ is correct.

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

$\text{B}$

At a constant temperature, which of the following aqueous solutions will have the maximum vapour pressure?

[Mol weight:

$NaCl=58.5,{ H }_{ 2 }{ SO }_{ 4 }=98.0g.{ mol }^{ -1 }$ ]

0%
$1$ molal $NaCl$
0%
$1$ molar $NaCl$
0%
$1$ molal ${H}_{2}{SO}_{4}$
0%
$1$ molar ${H}_{2}{SO}_{4}$

Explanation

One molar (1M) aqueous solution is more concentrated than one molal aqueous solution of the same solute.

In solution,

${H}_{2}{SO}_{4}$ provides three ions, while

$NaCl$ provides two ions. Hence, the vapour pressure of a solution of

$NaCl$ is higher (as it gives less number of ions).

Therefore, 1 molal

$NaCl$ will have the maximum vapour pressure.

Hence, option

$A$ is correct.

A spherical soap bubble of radius

$1\ cm$ is formed inside another of radius

$3\ cm$ . The radius of single soap bubble which maintains the same pressure differences as inside the smaller and outside the larger soap bubble is ________

$cm$ .

0%
$1$
0%
$0.75$
0%
$0.5$
0%
$0.25$

'a' moles of

$PCl_5$ , undergoes, the thermal dissociation as:

$PCl_5\rightleftharpoons PCl_3+Cl_2$ , the mole fraction of

$PCl_3$ at equilibrium is

$0.25$ and the total pressure is

$2.0$ atmosphere. The partial pressure of

$Cl_2$ at equilibrium is?

0%
$2.5$
0%
$1.0$
0%
$0.5$
0%
None

Explanation

$PCl_5\rightarrow PCl_3+Cl_2$

Thus mole fraction of

$PCl_3$ and

$Cl_2$ are equal

so,

$xCl_2=0.25$

$P\ of\ Cl_2=x\times P_{total}=0.25\times2=0.5\ atm$

On the surface of the earth at

$1\ atm$ pressure, a balloon filled with

$H_{2}$ gas occupies

$500\ mL$ . This volume is

$5/6$ of its maximum capacity. The balloon is left in air. It starts rising. The height above which the balloon will burst if temperature of the atmosphere remains constant and the pressure decreases

$1\ mm$ for every

$100\ cm$ rise of height is:

0%
$120\ m$
0%
$136.67\ m$
0%
$126\ m$
0%
$100\ m$

Explanation

Given:

$\ P_1 = 1 atm = 760 mm of Hg$

$\ V_1 = 500 ml$

Solution:

Let Maximum capacity

$\ V_2$ = x

According to question ,

$500 = \dfrac {5}{6} \times x$

$\Rightarrow x = 500 × \dfrac{6}{5}$

$\Rightarrow x = 600$

From the ideal gas law ,

$\ P_1V_1 = P_2V_2$

$\Rightarrow 760 × 500 = \ P_2 × 600$

$\Rightarrow \ P_2 = 760 × \dfrac{500}{600}$

$\Rightarrow \ P_2 = 633.33 mm of Hg$

Change in pressure =

$\ P_2 - \ P_1$

$\Rightarrow Change\ in\ pressure\ = 760 - 633.33$

$\Rightarrow Change\ in\ pressure\ = 126.67 mm of Hg$

It is given that for each 1 mm decrease, rise = 100 cm

Therefore, Height = 100 × 126.67 cm

$\Rightarrow$ Height = 126.67 m

Hence, option C is correct and the answer is 126.67m.

The two bulbs of volume

$5$ litre and

$10$ litre containing an ideal gas at

$9\ atm$ and

$6\ atm$ respectively are connected. What is the final pressure in the two bulbs if the temperature remains constant ?

0%
$12\ atm$
0%
$21\ atm$
0%
$15\ atm$
0%
$7\ atm$

Explanation

Explanation

Step 1: Preparing data

For ideal gases, the equation $PV= nRT$ is always valid …(1)
where , $P =$ Pressure
$V=$ Volume
$T =$ Temperature
$n =$ Number of gaseous moles
̣ $R =$ Ideal gas constant
Let the two bulbs be A and B. The diffusion of gases will continue to take place until $P_{1} = P_{2}$
Before connecting, $P_{1} = 9 atm$ and $P_{2} = 6 atm$ … (2)
Also, $V_{1} = 5 L$ and $V_{2} = 10 L$ . Volumes of bulbs are fixed and do not change. …(3)

Step 2: Solving using equations

Total moles of gas in the two bulbs do not change. Therefore, $n_{total} (Initial) = n_{total} (Final)$
Or, $n_{1} + n_{2} = n_{3} + n_{4}$
Since the final pressure after diffusion will be equal in both bulbs, let it be $P^{o}$
Also, R and T are constant in equation (1), we can write $n \propto PV$
Therefore, $P_{1}V_{1} + P_{2}V_{2} = P^{o}V_{1} + P^{o}V_{2}$ …(4)
Putting values from (2) and (3) in (4), we get:
$9 \times 5 + 6 \times 10 = P^{o} (V_{1} + V_{2})$
Or, $45 + 60 = P^{o} (15)$
Therefore, $P^{o} = \dfrac{105}{15} = 7$

Correct Option: $B$

The P-T graph as given below was observed for a process of an ideal gas, which of the following statement is true?

0%
$w = +ve, \triangle H = +ve$
0%
$w = -ve, \triangle H = -ve$
0%
$w = -ve, \triangle H = +ve$
0%
$w = +ve, \triangle H = -ve$

Explanation

Hint

Change in enthalpy, $\Delta H=n{{C}_{P}}\Delta T$

Explanation

Enthalpy change will be positive as temperature changes can never be negative.

And both the constant term is positive,

Thus,

$\Delta H=\left( + \right)ve$

When the temperature fluctuates on this graph, the enthalpy changes as well.

Again, from the gas equation it is clear that,

$~V\propto T$

When the temperature rises, the volume of the gas rises with it.

$W=-P\left( {{V}_{2}}-{{V}_{1}} \right)$

As volume increases so $\left( {{V}_{2}}-{{V}_{1}} \right)$ is positive.

So, amount of work done is positive.

Final answer

The correct answer is option (A).

What will be the pressure of the gaseous mixture when

$0.5\ L$ of

$H_{2}$ at

$0.8$ bar and

$2.0\ L$ of dioxygen at

$0.7$ bar are introduced in a

$1\ L$ vessel at

$27^{\circ}C$ ?

0%
$2.5 \ bar$
0%
$2.1 \ bar$
0%
$1.8 \ bar$
0%
None of these

Explanation

Use gaseous formula ,

$PV = nRT$

$n = \dfrac{PV}{RT}$

now,

moles of

$H2 = \dfrac{PV}{RT}$

for

$H_2$ , pressure ( P ) = 0.8 bar and volume ( V ) = 0.5 L ,

moles of

$H_2 = 0.8 \times \dfrac{0.5}{RT} = \dfrac{0.4}{RT}$

similarly for

$O_2$ ,

pressure ( P ) = 0.7 bar , volume ( V ) = 2 L

moles of

$O_2 = \dfrac{PV}{RT}$

$= 0.7 \times \dfrac{2}{RT} = \dfrac{1.4}{RT}$

hence, total number of moles

$= \dfrac{0.4}{RT} + \dfrac{1.4}{RT}$

$= \dfrac{1.8}{RT}$

now, Total pressure

$= = \dfrac{n'RT}{V'}$ [ by gas formula]

here,

$n'$ is Total moles

$= \cfrac{1.8}{RT}$

$V'$ is total volume

$= 1 L$

now, total pressure

$= \dfrac{1.8}{RT} \times \dfrac{RT}1 = 1.8$

$bar$

Hence, the correct option is

$\text{C}$

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 $O_2$ .

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

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

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

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

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

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

Assuming the volume is held constant, what happens to the pressure of gas when the absolute temperature doubles?

0%
It is doubled
0%
It is quadrupled
0%
It is halved
0%
None of these

Explanation

The ideal gas equation is,

$PV = nRT$ .

When

$V, R$ and

$n$ are constants, Pressure becomes directly proportional to the Temperature.

⇒

$P ∝ T$

⇒

$\dfrac{P_1}{T_1}$ =

$\dfrac{P_2}{T_2}$ (Gay-Lussac's law)

Now, when temperature is doubled i.e.

$T_2$ = 2

$T_1$

$P_2$ =

$\dfrac{T_2}{T_1}$ x

$P_1$

⇒

$P_2$ =

$2P_1$

Thus, option A is correct.

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

$\text{D}$

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

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

$\text{C}$

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

$\text{B}$

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 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 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.

The molecular weight of

$O_2$ and

$N_2$ are 32 and 28 respectively. At

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

Explanation

Temperature of

$O_{2}$

$=273+15=298 K$

Temperature of

$N_{2}$

$=x$

$P$

$\propto$

$nT$ , as pressure is same.

Temperature in

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

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

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

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

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

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

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

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

$T_2=2T_1$

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

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%
$1.80atm$

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

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

$=0.0287\times 0.977 =0.028atm$

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

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

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,

$\displaystyle P_A=120 \ torr$

$\displaystyle P_A-P_B = 75$

$\displaystyle P_B=P_A-75$

$\displaystyle P_B=120-75$

$\displaystyle P_B=45 \ torr$

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-

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

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

Among the following substances, the lowest vapour pressure is exerted by:

0%
water
0%
alcohol
0%
ether
0%
mercury

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%.

$80^0$ 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 =

$32\%$

$80^o$ 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

$= 32\%$

If the concentration of water vapour in the air is

$1\%$ and the total atmospheric pressure equals

$1$ atm then the partial pressure of water vapour is?

0%
$0.1$ atm
0%
$1$ mm Hg
0%
$7.6$ mm Hg
0%
$100$ atm

Explanation

According to Dalton's law of partial pressures,

pressure

$\propto \Lambda_{\text{water vapour}}$

Pressure of water vapour=

$\left( \cfrac { 1 }{ 100 } \right) 1$ atm=0.01 atm

1 atm

$\Longrightarrow$ 760 mm of Hg

0.01 atm

$\Longrightarrow$ 7.6 mm of Hg

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

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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