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

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

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

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

Dalton's law of partial pressure will not apply to which of the following mixture of gases?

0%
$H_{2}$ and $SO_{2}$
0%
$H_{2}$ and $Cl_{2}$
0%
$H_{2}$ and $CO_{2}$
0%
$CO_{2}$ and $Cl_{2}$

Equal mass of

$H_2$ ,

$He$ and

$CH_4$ are mixed in empty container at

$300$ K, when total pressure is

$2.6$ atm. The partial pressure of

$H_2$ in the mixture is:

0%
$2.1$ atm.
0%
$1.6$ atm.
0%
$0.8$ atm.
0%
$0.2$ atm.

Explanation

According to Dalton's law, the partial pressure of a gas is equals to the product of total pressure and mole fraction.

We have

$H_2,He\:and\:CH_4$ gas with equal mass, m.

Total Pressure = 2.6 atm

$Partial\:Pressure of {H_2}=Total\:Pressure\times\chi_{H_2}$

$=2.6\times\dfrac{n_{H_2}}{n_{H_2}+{n_{He}}+{n_{CH_4}}}$

$=2.6\times\dfrac{\frac{m}{2}}{\frac{m}{2}+\frac{m}{4}+\frac{m}{16}}$

$=2.6\times\dfrac{8}{13}$

$Partial\:Pressure_{H_2}=1.6\:atm$

$18$ 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

$C$ .

4 g argon (Atomic mass=40) in a bulb is at a temperature of T K has a pressure P atm. When the bulb was placed in hot bath at a temperature

$50^0 C$ more than the first one, 0.8 g of gas had to be removed to get the original pressure. T is equal to :

0%
510 k
0%
200 k
0%
100 k
0%
73 k

Explanation

$\begin{array}{l} PV=nRT \\ P=\frac { { nRT } }{ V } \\ So,\, \, \frac { 4 }{ { 40V } } RT=\frac { { 3.2 } }{ { 40V } } R\left( { T+50 } \right) \Rightarrow 40T=32\left( { T+50 } \right) \\ 40T-32T=50\times 32 \\ T=\frac { { 50\times 32 } }{ 8 } \Rightarrow 200K \end{array}$

Option B is correct.

A sample of an ideal gas with initial pressure P and volume V is taken through an isothermal process during which entropy change is found to be

$\Delta$ S. The universal gas constant is R. Then, the work done by the gas is given by:

0%
$\displaystyle\frac{PV\Delta S}{nR}$
0%
$nR\Delta S$
0%
PV
0%
$\displaystyle\frac{P\Delta S}{nRT}$

Explanation

Dimensional analysis,

work done

$=$ Energy

$=\cfrac{PV\triangle S}{nR}$

We know,

$PV=nRT.$

$\Longrightarrow Energy=\cfrac{PVT\triangle S}{nRT}=T\triangle S,$

Also,

$\triangle S=\cfrac{\triangle H}{T}$

Energy

$=\cfrac{\triangle H}{T}\times T$

Energy

$=$ Energy.

$\therefore R.H.S=L.H.S\Longrightarrow$ Work done by gas is

$\cfrac{PV\triangle S}{nR}.$

A gaseous mixture contains

$56\ g\ N_{2}. 44\ g\ CO_{2}$ and

$16\ g\ CH_{4}$ . The total pressure of the mixture is

$720\ mm\ Hg$ . The partial pressure of

$CH_4$ in mm Hg is:

0%
$620\ mm$ of $Hg$
0%
$180\ mm$ of $Hg$
0%
$160\ mm$ of $Hg$
0%
$200\ mm$ of $Hg$

Explanation

Given: Mass of $N_{2}$ = 56g

Mass of $CO_{2}$ = 44g

Mass of $CH_{4}$ = 16g

Solution: Moles of $N_{2}$ = $\dfrac{56}{28}$ = 2

Moles of $CO_{2}$ = $\dfrac{44}{44}$ = 1

Moles of $CH_{4}$ = $\dfrac{16}{16}$ = 1

Now, as we know, partial pressure of a gaseous component = mole fraction of the component * total pressure

Thus, partial pressure of $CH_{4}$ = ( $\dfrac{1}{2 + 1 +1}$ * 720) mm Hg

= 180 mm Hg

Thus, the correct option is (B).

The vapour pressure of pure benzene at a certain temperature is

$640\ mm\ Hg$ . A non-volatile solute weighing

$2.175\ g$ is added to

$39.0\ g$ of benzene. The vapour pressure of solution is

$600\ mm\ Hg$ . What is the molar mass of the solute?

0%
$72.1\ g\ mol^{-1}$ .
0%
$70.6\ g\ mol^{-1}$ .
0%
$69.4\ g\ mol^{-1}$ .
0%
$56.2\ g\ mol^{-1}$ .

Explanation

$\begin{aligned} P_{A}^{0} &=640 \mathrm{~mm}{\mathrm{Hg}} \\ P_{S} &=600 \mathrm{~mm} {\mathrm{Hg}} \\ & \omega_{A}=39 \mathrm{~g} ; \quad \omega_{B}=2.175 \mathrm{~g} \end{aligned}$

$\begin{array}{l}\text{According to Rault's law (solution }\\ \text { is ideal or dilute), } \frac{P_{A}^{\circ}-P_{S}}{P_{S}}=\frac{n_{B}}{n_{A}}=\frac{\omega_{B}}{M_{B}} \times \frac{M_{n}}{\omega_{A}}\end{array}$

$\begin{array}{l} \Rightarrow \frac{640-600}{600}=\frac{2.175}{M _B} \times \frac{78}{39} \\ \Rightarrow \quad M_{B}=\frac{2.175 \times 78}{39} \times \frac{600}{40}=65.25 \\ \therefore \text { Molar mass of solute }=65.25 \mathrm{~g/mol} \end{array}$

The coefficient of thermal expansion,

$\alpha$ , is nearly constant for a liquid

$\alpha =\cfrac { 1 }{ V } { \left( \cfrac { dV }{ dT } \right) }_{ P }$
At

$293K$ and

$1atm$ for water

$\alpha =2.1{ K }^{ -1 }$
What is the approximate work done when 1 mole of liquid is heated from

$288K$ to

$298K$ at

$1atm$ (Molar volume of liquid

$=18ml$ )?

0%
$-78.25kJ$
0%
$78.15J$
0%
$-37.8J$
0%
$-83.14J$

Explanation

$dV=\alpha VdT$

$\Delta W=−\int{PdV}=−\alpha V\times P\overset {T_2} {\underset {T_1} \int}dT$

$\Delta W=−2.1\times 18\times 10^{−6}\times 10^5\times 10J$

$\Delta W=−2.1\times 18J$

$\Delta W =−37.8J$

${ PCl }_{ 5 }$ is 80% dissociated at

$250$ then its vapour density at room temperature will be:

0%
56.5
0%
104.25
0%
101.2
0%
52.7

The quantity

$\dfrac { PV }{ { k }_{ B }T }$ represents the:

$({ k }_{ B }$ : Boltzmann constant)

0%
number of particle of the gas
0%
mass of the gas
0%
number of the moles of the gas
0%
translation energy of the gas

Explanation

$PV=nRT$

n = no. of moles

$PV= (N/N_A)\times RT$

Where

$N =$ no. of molecules

$N_A =$ Avagadro number

$\therefore PV= NT \times (R/N_A)$

$R/N_A = k_B =$ Boltzmann's constant

Hence

$PV = Nk_B T$

$(PV/k_BT) = N =$ no. of moecules

Two containers, X and Y at 300K and 350K with water vapour pressures 22 mm and 40 mm respectively a are connected, initially closed with a valve. If the valves opened.

0%
The final pressure in each container is 31 mm
0%
The final pressure in each container is 40 mm
0%
Mass of $H_{2}O$ (l) in X increases
0%
Mass of $H_{2}O$ (l) in Y decreases roar.

Explanation

$\begin{array}{l} \text { - The vapour pressur in the cointainer } y \text { is higher } \\ \text { than the vapour pressure in the container } x \text { . } \\ \text { - When the value is opened, the vapour from } \\ \text { container y diffuse in the container } x \text { } \\ \text { and condense there. } \\ \text { - When the equilibrium is re-established, the } \\ \text { mass of liquid water in the contaier y } \\ \text { decreases and container } x \text { increases. } \\ \text { Hence the answer is, option ( } A \text { ) } \\ \text { i.e. The final pressure in each container is } 31 \mathrm{~mm} \text { . } \end{array}$

The vapour pressure of

$C_6H_6$ and

$C_7H_8$ mixture at

$50^{\circ}C$ is given by

$p=179X_B+92$ where

$X_B$ is the mole fraction of

$C_6H_6$ .

Calculate (in mm) Vapour pressure of liquid mixture obtained by mixing

$936 \, g\ C_6H_6 \,$ and

$736\ g$ toluene is:

0%
300 mm Hg
0%
250 mm Hg
0%
199.4 mm Hg
0%
180.6 mm Hg

Explanation

Given,PM = 179

$X_B$ +92

For pure

$C_{6} h_{6}$ ,

$x_{B},$ i.e. mole fraction of benzene

$=1$

$\Rightarrow \quad P_{B}^{0}=179+92=271\mathrm{~mm}$

$x_{B}=\frac{\frac{936}{78}}{\frac{936}{78}+\frac{736}{92}}=\frac{12}{12+8}=0.60$

$\begin{aligned} x_{T} &=\frac{\frac{736}{92}}{\frac{736}{92}+\frac{936}{78}}=\frac{8}{12+8}=0.40 \\ P_{M}=& 179 x_{B}+92 \\=& 179 \times 0.6+92 \\=& 107.4+92 \\=& 199.4 \mathrm{~mm} \text { of } \mathrm{Hg} \\ \text { Hence, }(\mathrm{C}) \text { is correct option. } \end{aligned}$

Pressure of the gas in column (1) is :

0%
60 cm of Hg
0%
55 cm of Hg
0%
50 cm of Hg
0%
45 cm of Hg

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 k Calculate vapour pressure of water at 298K?

0%
$4.53kPa$
0%
$3.53kPa$
0%
$5.53kPa$
0%
$6.53kPa$

The vapour pressure of two pure liquids A and B which form an ideal solution are 1000 and 1600 torr respectively at 400 K. A liquid solution of A and B for which the mole fraction of A is 0.60 is contained in a cylinder by a piston on which the pressure can be varied. The solution use slowly vapourised at 400 K by decreasing the applied pressure. What is the composition of last droplet of liquid remaining in equilibrium with vapour?

0%
$X_A= 0.30, X_B= 0.70$
0%
$X_A= 0.40, X_B= 0.60$
0%
$X_A= 0.70, X_B= 0.30$
0%
$X_A= 0.50, X_B= 0.50$

Explanation

$\begin{array}{l}\text{Here,} Y_A =\dfrac{ P_A^0 X_A }{P_A^0 X_A+ P^0_B(1-X_A)}\implies 0.6 =\dfrac{1000 X_A}{1000X_A+1600(1-X_A)}\\ \quad \therefore X_A=0.70 \quad X_B = 1-X_A =0.30 \end{array}$

2004004_878189_ans_b792462a43ab4499a61d477b0f33aea7.jpeg

$P_A \, = \, X_AP_A$ and

$P_B \, = \, X_BP_B$

$P_T \, = \, X_AP_A \, + \, X_BP_B$
Vapour pressure of mixtures of Benzene

$(C_6H_6)$ and toluene

$(C_7H_8)$ at

$50^{\circ}C$ are given by

$P_M \, = \, 179X_B \, + \, 92$ where

$X_B$ is mole fraction of

$C_6H_6$ .
What is the vapour pressure of pure liquids?

0%
$P_B \, = \, 92 mm, \, P_T \, = \, 179 mm$
0%
$P_B \, = \, 271 mm, \, P_T \, = \, 92 mm$
0%
$P_B \, = \, 180 mm, \, P_T \, = \, 91 mm$
0%
None of these

Explanation

$x_A+x_B=1$

$PT=x_AP_A+x_BP_B= (1-xB)P_A+x_BP_B$

$P_T=P_A+x_B(P_B-P_A) \longrightarrow (1)$

Given,

$P_M= 92+179x_B\longrightarrow (2)$

on comparing equation

$(1)$ &

$(2)$ we have

$P_A= 92 mm, P_B-P_A= 179mm$

$P_B= 179+P_A=179+92$

$P_A=92 mm, P_B=271mm$

A sample of pure

$NO_{2}$ gas healed lo 1000 K decomposes.

$NO_{2}(g) \leftrightharpoons 2NO(g) + O_{2}(g)$

The equillibrium constant

$K_{P}$ is 100atm. Analysis shows that the partial pressure of

$O_{2}$ is 0.25 atm at equillibrium. The partial pressure of

$NO_{2}$ at equillibrium is:

0%
0.03
0%
0.25
0%
0.0245
0%
0.04

Explanation

$2NO_{2}(g) \leftrightharpoons 2NO(g) + O_{2}(g)$
Analysis shows that the partial pressure of

$O_{2}$ is 0.25 atm at equilibrium. The partial pressure of

$NO$ at equilibrium

$\displaystyle 2 \times$ the partial pressure of

$O_{2}$

$\displaystyle 2 \times 0.25=0.50$ atm.
The equillibrium constant

$K_{P}$ is 100 atm.

$\displaystyle K_p=\dfrac { P_{ NO}^2 \times P_{ O2} }{ P_{ NO2} ^2}$

$\displaystyle 100=\dfrac {(0.50)^2 \times 0.25}{ P_{ NO2} ^2 }$

$\displaystyle P_{ NO2} ^2=0.000625$

$\displaystyle P_{ NO2} =0.025 \simeq 0.03$
The partial pressure of

$NO_{2}$ at equilibrium is 0.03 atm.

The vapour pressure of two pure liquids

$A$ and

$B$ , that form an ideal solution are

$100$ and

$900$ torr respectively at temperature

$T$ . This liquid solution of

$A$ and

$B$ is composed of

$1\ mole$ of

$A$ and

$1\ mole$ of

$B$ . What will be the pressure, when

$1$ mole of mixture has been vapourized?

0%
$800\ torr$
0%
$500\ torr$
0%
$300\ torr$
0%
None of these

Explanation

$P_A^o = 100 \, torr \,\,\,\, P_B^o = 900 \, torr$

$n_A = 1 \, mole \,\,\,\, n_B = 1 \, mol$

$x_A = \dfrac{1}{1 + 1} = 0.5 mole$

$x_B = \dfrac{1}{1 + 1} = 0.5 mole$

Pressure

$= P_A^o x_A + P_B^o x_B$

$= 100 \times 0.5 + 900 \times 0.5 = 50 +450 = 500 \, torr$

A nitrogen-hydrogen mixture initially in the molar ratio 1:3 reached equilibrium to form ammonia when 25% of the

$H_{2}$ and

$N_{2}$ had reacted. If the total pressure of the system was 21 atm, the partial pressure of ammonia at the equilibrium was:

0%
4.5 atm
0%
3.0 atm
0%
2.0 atm
0%
1.5 atm

Which of the following is a correct graph?

Explanation

According to kinetic theory of gas kinetic energy only depending on temperature

$K=\dfrac{3}{2}K_B T$

Hence option A is correct

1064050_878409_ans_f3e53c944e834031844a531d1694efb9.png

The vapour pressure of two pure liquids A and B, that form an ideal solution are

$100$ and

$900$ torr respectively at temperature T. This liquid solution of A and B is composed of

$1$ mole of A and

$1$ mole of B. What will be the pressure, when

$1$ mole of mixture has been vapourized?

0%
$800$ torr
0%
$500$ torr
0%
$300$ torr
0%
None of these

Explanation

Vapour pressure of pure A

$P_{A}^{\circ}=100 torr$

Vapour presuure of pure B

$P_{B}=900 torr$

$n_{A}=1 mole$

$n_{B}= 1 mole$

$x_{A}=\dfrac{1}{1+1}=0.5$

$x_B=\dfrac{1}{1+1}=0.5$

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

$=100\times 0.5+900\times 0.5=500torr$

Two moles of pure liquid 'A'

$(P_{A}^{0} =80mm$ of Hg) and

$3$ moles of pure liquid 'B' (

$P_{B}^{0} = 120mm$ 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

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

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

$n_{A}=2 moles$

$n_{B}=3 moles$

$x_{A}=\dfrac{2}{2+3}=\dfrac{2}{5}=0.4$

$x_{B}=\dfrac{3}{5}=0.6$

$P_{A}=P_{A}^{\circ}x_{A}=80\times 0.4=32 mm Hg$

$P_{B}=P_{B}^{\circ}x_{B}=120\times 0.6=72 mm Hg$

Total vapour pressure

$P=P_{A}+P_{B}$

$=32+72=104 mm Hg$

Two glass bulbs

$A$ (of

$100\ mL$ capacity), and

$B$ (of

$150\ mL$ capacity) containing same gas are connected by a small tube of negligible volume. At particular temperature, the pressure in

$A$ was found to be

$20$ times more than that in bulb

$B$ . The stopcock is opened without changing the temperature. The pressure in

$A$ will :

0%
Drop by $75$ %
0%
Drop $57$ %
0%
Drop by $25$ %
0%
Will remain same

Explanation

Let the original pressure in B be P.

$\therefore$ Pressure in A = 20 P.

Given that :

Volume in A = 100 mL

Volume in B = 150 mL

After opening the stopcock,

total volume

$= \left( 100 + 150 \right) mL = 250 mL$

According to Boyle's law :

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

Let the partial pressure of gas in A be

${P}_{A}$ :

${P}_{A} \times 250 = 100 \times 20 P$

$\Rightarrow \; {P}_{A} = \cfrac{100 \times 20P}{250} = 8 P$

Let Partial pressure of gas in B be

${P}_{B}$ :

${P}_{B} \times 250 = 150 \times P$

${P}_{B} = \cfrac{150 P}{250} = 0.6 P$

$\therefore$ Total pressure

$= 8 P + 0.6 P = 8.6 P$

Now,

Drop in pressure in A

$= 20P - 8.6P = 11.4 P$

$\therefore \; \%$ drop in pressure

$= \cfrac{11.4 P}{20 P} \times 100 = 57 \%$

Therefore, the correct option is B.

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

Two liquids

$X$ and

$Y$ are perfectly immiscible. If

$X$ and

$Y$ have molecular masses in ration

$1 : 2$ , the total vapour pressure of a mixture of

$X$ and

$Y$ prepared in weight ratio

$2 : 3$ should be:

$(P_{X}^{0} = 400\ torr, P_{Y}^{0} = 200\ torr)$ .

0%
$314\ torr$
0%
$466.7\ torr$
0%
$600\ torr$
0%
$700\ torr$

Explanation

Let molecular mass of

$X = M$ and molecular mass of

$Y = 2M$

vapour pressure of

$X \, P_{x_0}^0 = 400$ torr

vapour pressure of

$Y \, P_y = 200$ torr

Let weight of

$X = 2w$

weight of

$Y = 3w$

Mole fraction

$X_x = \dfrac{\dfrac{2w}{m}}{\dfrac{2w}{m} + \dfrac{3w}{2m}} = \dfrac{4}{7}$

Mole fraction =

$Y_x = 1 - \dfrac{4}{7} = \dfrac{3}{7}$

$P = P_x^0 X_x + P_y^o Y_x$

$= 400 \times \dfrac{4}{7} + 200 \times \dfrac{3}{7} = \dfrac{2200}{7}$

$= 314 \, torr$

A rigid and insulated tank of 3

$m^3$ volume is divided into two components. One compartment of volume of 2

$m^3$ contains an ideal gas at 0.8314 MPa and 400 K and while the second compartment of volume 1

$m^3$ contains the same gas 8.314 MPa and 500 k. If the partition between the two compartments is ruptured, the final temperature of the gas is:

0%
420 K
0%
450 K
0%
480 K
0%
None of these

Explanation

First compartment

$P_1=0.8314\ M\ Pa\ V_1=2\ xm^2\ T_1=400\ K$

Second compartment

$P_2=8.314\ M\ Pa\ V_2=1\ m^3\ T_2=500\ K$

$n_1=\dfrac {P_1 V_1}{RT_1}=\dfrac {0.8314\times 10^6 \times 2}{8.314 \times 400}=500$

$n_2=\dfrac {P_2 V_2}{RT_2}=\dfrac {8.314\times 1\times 10^6 }{8.314 \times 500}=200$

After Ruptewe of compartment

$n=500+200=700\ V=2\ P_1=3\ m^3$

$P=0.8314\ M\ Pq$

$T=\dfrac {PV}{nR}=\dfrac {3\times 8.314\times 10^6}{700\times 8.314}$

$=\dfrac {3000}{7}$

$\simeq 420\ K$

$100$ grams of oxygen

$(O_{2})$ gas and

$100$ grams of helium

$(He)$ gas are in separate containers of equal volume at

$100^{\circ}C$ . Which one of the following statement is correct?

0%
Both gases would have the same pressure
0%
The average kinetic energy of $O_{2}$ molecules is greater than that of $He$ molecules
0%
The pressure of $He$ gas would be greater than that of the $O_{2}$ gas
0%
The average kinetic energy of $He$ and $O_{2}$ molecules is same

Explanation

According to ideal gas equation,

$PV = nRT$
For

$100\ gm\ O_{2}$ ,

$P_{1}\times V = \dfrac {100}{32} \times R\times T ..... (1)$
For

$100\ gm\ He$ ,

$P_{2}\times V = \dfrac {100}{4}\times R\times T .... (2)$
Dividing eq

$(1)$ by

$(2)$

$\dfrac {P_{1}\times V}{P_{2}\times V} = \dfrac {100\times R\times T}{32}\times \dfrac {4}{100}\times \dfrac {1}{R\times T}$

$\dfrac {P_{1}}{P_{2}} = \dfrac {1}{8}, P_{2} = 8P_{1}$

$\therefore P_{He} > P_{O_{2}}$ .

A reaction at 300 K with

$\Delta G^o$ = -1743 J/mol consists of 3 mole of

$A(g)$ , 6 mole of

$B(g)$ and 3 mole of

$C(g)$ . If given

$A,\ B$ and

$C$ are in equilibrium in 1 litre container then the reaction may be:
[Given : 2 =

$e^{0.7}$ , R = 8.3 J/K - mol]

0%
$A\,+\,B \rightleftharpoons C$
0%
$A \rightleftharpoons B\,+\,2C$
0%
$2A \rightleftharpoons B\,+\,C$
0%
$2C \rightleftharpoons A\,+\,B$

Explanation

$\Delta G = -RT \, \ln K$

$-1743 = -8.3 \times 300 \, \ln K$

$\dfrac{-1743}{-8.3 \times 300} = \ln k$

$0.7 = \ln \, k$

$k = e^{0.7} = 2$

Let reaction x

$A + y \, B \rightarrow C$

$k = \dfrac{[C]^2}{[A]^x [B]^y}$

$2 = \dfrac{3^2}{3^x 6^y}$

$2 = 3^{z - x - y} 2^{-y}$

$y = -1$

$z - x = 1$

$z = 2 \,\,\,\, x = 1$

Reaction would be

since y becomes

$-1$ so it be in product

$A \rightleftharpoons B + 2C$

Two liquids X and Y are perfectly immiscible. If X and Y have molecular masses in ratio 1 : 2, the total vapour pressure of a mixture of X and Y prepared in weight ratio 2 : 3 should be (

$Px^0 = 400 torr, Py^0 = 200 torr$ )

0%
314 torr
0%
466.7 torr
0%
600 torr
0%
700 torr

Explanation

Molecular mass of

$X = M$

Molecular mass of

$Y = 2M$

weight of

$X = 2W$

weight of

$Y = 3w$

no.of mole of

$X = \dfrac{2w}{M}$

no. of mole of

$Y = \dfrac{3w}{2m}$

mole fraction of

$X = Xx = \dfrac{\dfrac{2w}{m}}{\dfrac{2w}{M} + \dfrac{3w}{2M}}$

$X_x = \dfrac{4}{7}$

Mole fraction of

$Y = Xy = \dfrac{3}{7}$

$P_x^o = 400 \, torr$

$P^o_y = 200 \, torr$

$P = \dfrac{4}{7} \times 400 + 200 \times \dfrac{3}{7}$

$= \dfrac{2200}{7} = 314 torr$

option

$A$ may be correct approximately

A mixture (by weight) of hydrogen and helium is enclosed in a two-litre flask at

$27^o C$ . Assuming ideal kept behaviour, the partial pressure of helium is found to be 0.2atm. Then the concentration of hydrogen would be:

0%
0.045
0%
8.26
0%
0.0162
0%
1.62

Explanation

$\text { Given, } \mathrm{H}_{2} \text { and He gas mixture is enclosed in a flask. }$

$\begin{array}{l} V=2 L \\ T=300 K \end{array}$

$\text { It is given, Partial pressure of He is } 0.2 \mathrm{~atm} \text { . }$

$\text { so, } n_{\text {He }}=\dfrac{P_{\text {He }} V}{R T}=\dfrac{0.2 \times 2}{0.0821 \times 300} \text { moles }$

$=0.01624 \text { moles. }$

$\text { Since, mixture constitutes } 1: 1 \quad H_{2} \text { - He ratio by weight. }$

$\text { Thus, weight of } H_{2}=0.01624 \times 4 \mathrm{gm}=0.06496 \mathrm{gm}$

$\therefore \text { moles of } H_{2}=0.03248 \mathrm{mol}$

$\text { so, concentration of } H_{2}=\dfrac{\text { moles }}{\text { volume }}=\dfrac{0.03248}{2}=0.01624 M$

$0.2\ mol$ of

$H_{2}(g)$ and

$2\ mol$ of

$S(s)$ are mixed in a

$1\ dm^{3}$ vessel at

$90^{\circ}C$ , the partial pressure of

$H_{2}S(g)$ formed according to the reaction,

$H_{2}(g) + S(s)\rightleftharpoons H_{2}S, K_{P} = 6.8\times 10^{-2}$ would be:

0%
$0.19\ atm$
0%
$0.38\ atm$
0%
$0.6\ atm$
0%
$0.072\ atm$

Explanation

$H_2 (g) + S(s) \rightleftharpoons H_2S$

Number mole of

$H_2 \,\ \ n = 0.2\ mol$

temp

$t = 90^o C + 273 = 363\ K$

volume =

$1\ dm^3 = 1 litre$

$P = \dfrac{nRT}{V} = \dfrac{0.2 \times 0.082 \times 363}{1}$

$=5.96\ atm$

$H_2 (g) + S(s) \rightleftharpoons H_2 (s)$

$5.96 - P_0$

$P_0$

$K_p = \dfrac{P_0}{5.96 - P_0}$

$6.8 \times 10^{-2} = \dfrac{P_0}{5.96 - P_0}$

$40.53 \times 10^{-2} - 0.068 P_0 = P_0$

$1.068 P_0 = 40.53 \times 10^{-2}$

$P_0 = \dfrac{40.53 \times 10^{-2}}{1.068} = 0.3795$

$\approx 0.38\ atm$

CuSO

$_4.$ 5

$H_2O$ (s)

$\rightleftharpoons$ CuSO

$_4$ .3H

$_2O$ (s) + 2H

$_2O$ (g);

$k_p$ = 4

$10^{4}$ 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

$\begin{aligned} K_ P &=\left(P_{ H_{2} O}\right)^{2} \\ P_{ H_{2} O }&=\sqrt{K_ P} \\ &=\sqrt{4 \times 10^{-4}} \\ &=2 \times 10^{-2} \end{aligned}$

$\text { % Relative Humidity }=\dfrac{P_{H_2O} \times 760}{\text { v.p of water}}$

$\begin{aligned} =& \dfrac{2 \times 10^{-2} \times 760 \times 100}{38} \\ &=40 \end{aligned}$

Vapour Pressure of a mixture of benzene and toluene is given by

$P= 179X_{B} + 92$ , Where

$X_{B}$ is mole fraction of benzene.
This condensed liquid again brought to the same temperature then what will be the mole fraction of benzene in vapour phase :

0%
0.007
0%
0.93
0%
0.65
0%
4.5

Explanation

Mole fraction of

${ C }_{ 6 }{ H }_{ 6 }$ in vapour phase

$={ X }_{ B }^{ 1 }$ of initial mixture

${ X }_{ B }^{ 1 }=\dfrac { { P }_{ B }^{ 1 } }{ { P }_{ M } } =\dfrac { 162.6 }{ 199.4 } =0.815$

then mole fraction of

${ C }_{ 7 }{ H }_{ 8 }=0.185$

(mole fraction of

${ C }_{ 6 }{ H }_{ 6 }$ in initial mixture)

$=$ (mole fraction of

${ C }_{ 6 }{ H }_{ 6 }$ in liquid phase on

$II$ mixture

${ X }_{ B }^{ 1 }$ )

${ P }_{ M }={ P }_{ B }^{ 1 }+{ P }_{ T }^{ 1 }$

now,

${ P }_{ M }={ P }_{ B }^{ 0 }{ X }_{ B }^{ 1 }+{ P }_{ T }^{ 0 }{ X }_{ T }^{ 1 }$

$=\left[ \left( 271 \right) \left( 0.815 \right) +\left( 92 \right) \left( 0.185 \right) \right]$ mm

$=237.885$ mm

new mole fraction of

${ C }_{ 6 }{ H }_{ 6 }=\dfrac { 220.865 }{ 237.885 } =0.928\cong 0.93$

$\therefore$ new mole fraction of Benzene in vapour phase is

$0.93$ .

The pressure of a mixture of equal weights of two gases

$X$ and

$Y$ with molecular weight

$4$ and

$40$ respectively is

$1.1\ atm$ . The partial pressure of the gas

$X$ in the mixture is :

0%
$1\ atm$
0%
$0.1\ atm$
0%
$0.15\ atm$
0%
$0.5\ atm$

Explanation

Let

$w\ g$ of each gas is taken.
Mole fraction of

$X = \dfrac {\dfrac {w}{4}}{\dfrac {w}{4} + \dfrac {w}{40}} = \dfrac {10}{11}$
Partial pressure of

$X = P_{total}\times Mole\ fraction$

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

A container of

$1\ L$ capacity contains a mixture of

$4\ g$ of

$O_{2}$ and

$2\ g$ of

$H_{2}$ at

$0^{\circ}C$ . What will be the total pressure of the mixture?

0%
$50.42\ atm$
0%
$25.21\ atm$
0%
$15.2\ atm$
0%
$12.5\ atm$

Explanation

Applying

$PV = \dfrac {w}{M}RT$

$p_{O_{2}} = \dfrac {4}{32}\times \dfrac {0.0821\times 273}{1} = 2.81\ atm$

$p_{H_{2}} = \dfrac {2}{2}\times \dfrac {0.0821\times 273}{1} = 22.4\ atm$

$p_{total} = p_{O_{2}} + p_{H_{2}} = 2.81 + 22.4 = 25.21\ atm$ .

An open flask contains air at

$27^{\circ}C$ . At what temperature should it be heated so that

$1/3rd$ of air present in it goes out?

0%
$177^{\circ}C$
0%
$100^{\circ}C$
0%
$300^{\circ}C$
0%
$150^{\circ}C$

Explanation

Let initial number of moles of air at

$27^{\circ}C (300\ K) = n$
At temperature

$T\ K$ , the no. of moles left

$= n - \dfrac {n}{3} = \dfrac {2n}{3}$
At constant pressure and volume,

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

$n\times 300 = \dfrac {2n}{3}\times T\Rightarrow T = 450\ K$ or

$177^{\circ}C$ .

The certain volume of a gas exerts on its walls some pressure at a particular temperature. It has been found that by reducing the volume of the gas to half of its original value the pressure becomes twice that of the initial value at a constant temperature. This happens because:

0%
mass of the gas increases with pressure
0%
speed of the gas molecules decreases
0%
more number of gas molecules strikes the surface per second
0%
gas molecules attract each other

Value of gas constant

$R$ in the ideal gas equation

$PV = nRT$ depends upon:

0%
temperature of the gas
0%
pressure of the gas
0%
units in the $P, V$ and $T$ are measured
0%
nature of the gas

Explanation

$R$ is universal gas constant

It is has fixed value which would depend upon the units in which

$P,V$ and

$T$ expressed in the equation

$PV=nRT$

$R=\dfrac { PV }{ nT }$

What is a weight of $CO_2$ in a $10L$ cylinder at $5atm$ and $27^0C$ ?

0%
$200\ g$
0%
$224\ g$
0%
$44\ g$
0%
$89.3\ g$

Explanation

$PV = nRT = \dfrac {m}{M}RT$

$P = 5\ atm, V = 10\ L, T = 300\ K$ ,

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

$m = \dfrac {PVM}{RT} = \dfrac {5\times 10\times 44}{0.0821\times 300} = 89.3\ g$ .

The main reason for deviation of gases from ideal behaviour is few assumptions of kinetic theory. These are
(i) there is no force of attraction between the molecules of a gas
(ii) volume of the molecules of a gas is negligibly small in comparison to the volume of the gas
(iii) Particles of a gas are always in constant random motion.

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

Explanation

The main reasons for deviation of gases from ideal behavior is few assumptions of Kinetic theory, they are-

$\Longrightarrow$ The kinetic theory of gases assume that gas molecules are point particles without any volume and any force acting between them; so that their collisions could be treated as perfectly elastic.

$\Longrightarrow$ But in reality, neither can one assure gas molecules to be with nil volume nor free from a mild attractive force.

$X,Y$ and

$Z$ in the given graph are?

0%
$X={p}_{1}+{p}_{2},Y=1,Z=0$
0%
$X={p}_{1}+{p}_{2},Y=0,Z=0$
0%
$X={p}_{1}\times {p}_{2},Y=0,Z=0$
0%
$X={p}_{1}-{p}_{2},Y=1,Z=0$

Explanation

Here,

$X =P_T=p_1+p_2=$ Total pressure of the solution.

$p_1$ and

$p_2$ is the partial pressure of component 1 and 2 respectively.

$Y=x_2=$ mole fraction of component 2 in the liquid phase.

$Z=y_2=$ mole fraction of component 2 in the vapour phase.

When mole fraction of component 2 in the liquid phase is 0 then mole fraction of component 2 in the vapour phase is also 0 which means the liquid is pure of component 1 only.

So, the correct option is B.

$P_o$ and

$P_s$ are the vapour pressure of solvent and its solution respectively.

$x_1$ and

$x_2$ are the mole fraction of solvent and solute respectively then:

0%
$P_s = \dfrac{P_o}{x_2}$
0%
$P_o - P_s = P_ox_2$
0%
$P_s = P_ox_2$
0%
$\dfrac{(P_o-P_s)}{P_s} = \dfrac{x_1}{(x_1+x_2)}$

A plot of volume

$(V)$ versus temperature

$(T)$ for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order of pressure is correct for this gas?

0%
$P_1>P_2>P_3>P_4$
0%
$P_1=P_2=P_3=P_4$
0%
$P_1 < P_2 < P_3 < P_4$
0%
$P_1 < P_2 = P_3 < P_4$

Explanation

$\bf{Explanation-}$

From ideal gas equation:

$PV = nRT$

Where

$n$ is the number of moles.

$R$ is redberg's constant.

$P$ is pressure

$V$ is Volume

$T$ is Absolute Temperature

Here,

$n$ and

$R$ is constant.

So,

$PV = T$

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

That means

$P$ is directly proportional to

$T$ and inversely proportional to

$V$ .

According to graph:

$V_1>V_2>V_3>V_4$

Hence,

$P_1<P_2<P_3<P_4$

$\left(\because P=\dfrac{1}{V}\right)$

$\bf{Conclusion-}$ Hence, option C is correct.

In three beakers labeled as (A), (B) and (C),

$100mL$ of water,

$100mL$ of

$1M$ solution of glucose in water and

$100mL$ of

$0.5M$ solution of glucose in water are taken respectively and kept at same temperature.
Which of the following statements is correct?

0%
Vapour pressure in all the three beakers is same
0%
Vapour pressure of beaker B is highest
0%
Vapour pressure of beaker C is highest
0%
Vapour pressure of beaker B is lower than that of C and vapour pressure of beaker C is lower than that of A

5 litre of

$N_2$ under a pressure of 2 atm, 2 litre of

$O_2$ at 5.5 atm and 3 litre of

$CO_2$ at 5 atm are mixed. The resultant volume of the mixture is 15 litre. Calculate the total pressure of the mixture and partial pressure of each constituent.

0%
2.06 atm
0%
1.75 atm
0%
3.06 atm
0%
4.50 atm

Explanation

From the equation

$Pv = n RT$ for the two gases. We can write

$nN_2 = \dfrac{5 \times 2} {RT}$

$nO_2 = \dfrac{2 \times 5.5} {RT}$

$nCO_2 = \dfrac{3\times 5}{RT}$

When introduced in

$1 L$ vessel, then

$P\times15L = (nN_2 + nO_2+nCO_2) RT$

Putting the values, we get

$P = 2.06 bar$

Hence, the total pressure of the gaseous mixture in the vessel is

$1.8$ bar

What is the total pressure exerted by the mixture of 7.0 g of

$N_2$ , 2g of hydrogen and 8.0 g of sulphur dioxide gases in a vessel of 6.1L capacity that has been kept in a reservoir at

$27^oC$ ?

0%
2,5 bar
0%
4.5 bar
0%
10 atm
0%
5.7 bar

Explanation

Total moles of gases

$=\left(\dfrac{7}{28}+\dfrac{2}{2}+\dfrac{8}{64}\right)=\left(\dfrac{1}{4}+\dfrac{1}{8}+1\right)=\dfrac{11}{8}$ moles.

volume of vessel

$=6.1 \mathrm{~L}$

and, temperature

$=27^{\circ}\mathrm{C}=300\mathrm{k}$ .

By ideal gas equation,

$\quad P V=n R T$ .

$\begin{aligned}\Rightarrow P=\dfrac{n R T}{V} &=\dfrac{11 \times 0.0831 . x 300}{6.1} \mathrm{bar} .\\& \approx 5.7 \mathrm{bar} .\end{aligned}$

Thus, total pressure exerted by the mixture

$=5.7$ bar.

so, option (D) is correct.

In a gaseous mixture at

$20^o C$ the partial pressure of the components are:

${ H }_{ 2 }$ : 150 Torr

${ CH }_{ 4 }$ : 300 Torr

${ CO }_{ 2 }$ : 200 Torr

${ C }_{ 2 }{ H }_{ 4 }$ : 100 Torr

The volume % of

${ H }_{ 2 }$ in mixture:

0%
26.67
0%
73.33
0%
80.00
0%
20

Calculate the change in pressure when

$1.04$ mole of

$NO$ and

$20 \,g \,O_2$ in

$20$ litre vessel originally at

$27^o C$ react to produce the maximum quantity of

$NO_2$ possible according to the equation.

$2NO(g) + O_2(g) \longrightarrow 2NO_2(g)$

0%
0.2113 atm
0%
1.2321 atm
0%
0.6396 atm
0%
0.5687 atm

Explanation

According to equation, 2 moles of

$NO$ reacts with one mole of oxygen.

Number of moles of

$NO$ = 1.04 mole

Number of moles of oxygen =

$\frac { 20 }{ 32 } \quad =\quad 0.625\quad mol$

So, according to reaction,

$1.04$ mol of

$NO$ reacts with

$0.52$ mol of oxygen

So, leftover oxygen =

$0.625-0.52=0.105$

Now for calculation of pressure

(1). Before reaction, 1.04 moles of

$NO$ and 0.625 moles of oxygen are present.

So, total number of moles =

$1.04+0.625=1.665$

Since,

$PV=nRT$

So,

${ P }_{ initial }=\frac { nRT }{ V } \\ { P }_{ initial }=\frac { 1.665RT }{ V }$

(2). After reaction, 1.04 mole of

${ NO }_{ 2 }$ is formed and 0.105 moles of

${ O }_{ 2 }$ is left.

So, total number of moles =

$1.04+0.105=1.145$

So,

${ P }_{ final }=\frac { 1.145RT }{ V }$ .

Now, change in pressure

$=\quad { P }_{ initial }-{ P }_{ final }\\ =\quad \frac { (1.665-1.145)RT }{ V } \\ =\quad \frac { 0.52\times 0.082\times 300 }{ 20 } \\ =\quad 0.6396\quad atm$

So, correct answer is option C.

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

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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