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

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).

$CuSO_4$

$\cdot$

$5H_2O(s)$

$\rightleftharpoons$

$CuSO_4\cdot$ 3

$H_2O(s)$ +

$2H_2O(g)$ ;

$K_p$ = 4

$\times10^{-4}$

$atm^2$ if the vapour pressure of water is 38 torr then percentage of relative humidity is: ( Assume all data at constant temperature)

0%
$4$
0%
$10$
0%
$40$
0%
$none\ of\ these$

Explanation

${K_p} = P_{{H_2}O}^2 = 4 \times {10^{ - 4}}$

${P_{{H_2}O}} = 2 \times {10^{ - 2}}\,\,atm \Rightarrow 0.02\,\,atm$

$P_{{H_2}O}^1 = 38\,\,atm \Rightarrow \frac{{38}}{{760}}\,\,atm \Rightarrow \frac{1}{{20}}\,\,atm \Rightarrow 0.05\,\,atm$

$\% \,\,{\text{Relative}}\,\,humidity = \frac{{{P_{{H_2}O}}}}{{P_{{H_2}O}^1}} \Rightarrow \frac{{0.02}}{{0.05}} = \frac{2}{5} \times 100 \Rightarrow 40\,\,\%$

Option C is correct.

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 $Cl_{2}$

A gas obeys the equation of state

$P(V-b)=RT$ (The parameter

$b$ is a constant). The slope for an isochore will be:

0%
negative
0%
zero
0%
$R/(V-b)$
0%
$R/P$

Explanation

Given that :

$P(V-b)=RT$

$P=\dfrac{RT}{(V-b)}$

$P=\dfrac{R}{(V-b)}T+O$

$y=mx+c$ (

$Y=P$ and

$X=T$ for isochore)

Slope

$=\left ( \dfrac{R}{V-b} \right )$

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

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

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

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

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

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

In a closed vessel, an ideal gas at

$1\ atm$ is heated from

$27^{\circ}C$ to

$327^{\circ}C$ . The final pressure of the gas will approximately be :

0%
$3\ atm$
0%
$0.5\ atm$
0%
$2\ atm$
0%
$12\ atm$

Explanation

For an ideal gas we have :

$PV=nRT$

And at constant volume:

$P\propto T$

Using the relation we get :

$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { T }_{ 1 } }{ { T }_{ 2 } }$

Given :

$P_1=1\ atm$

$T_1=27+273=300\ K$

$P_2=x\ atm$

$T_2=327+273=600\ K$

$\dfrac { 1 }{ P_2 } =\dfrac { 300 }{ 600 }$

$\Rightarrow P_2=2$

Therefore the final pressure is 2 atm.

Option C is correct.

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

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

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

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

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

$C_6H_6$ 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

$P_A \, = \, X_AP_A$ 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.

Vapour pressure of

$CC{l}_{4}$ at

$25^{O}C$ is 143mm Hg. 0.5gm of a non-volatile solute (mol. wt. 65) is dissolved in 100ml of

$CC{l}_{4}$ . Find the vapour pressure of the solution.

(Density of

$CC{l}_{4} = 1.58 gm/{cm}^{3}$ )

0%
141.93 mm
0%
94.39 mm
0%
199.34 mm
0%
143.99 mm

${ PCl }_{ 5 }$ 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 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

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

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

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%
Mass of $H_{2}O$ (l) in X increases
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 Y decreases roar.

Explanation

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

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

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

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.

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 =

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

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

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

$= 314 \, torr$

Which of the following relationships between partial pressure, volume and temperature is correct?
(i)

$P = \dfrac {nRT}{V}$
(ii)

$P_{total} = p_{1} + p_{2} + p_{3}$
(iii)

$P_{total} = (n_{1} + n_{2} + n_{3}) \dfrac {RT}{V}$ .

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

Explanation

$i.$ Ideal gas equation

$\Longrightarrow PV=nRT\Longrightarrow P=\cfrac{nRT}{V}$

$ii.$ Dalton's law of partial pressure,

$\Longrightarrow P_{total}=P_1+P_2+P_3$

$iii.$ Partial pressure ,

$P_1,P_2,P_3\dots P_K$

According to ideal gas equation,

$P=\cfrac{nRT}{V}\\ \Longrightarrow P_1=n_1(\cfrac{RT}{V})$

$P_{total}=(n_1+n_2+n_3)(\cfrac{RT}{V})$

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

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

$P_{B}=900 torr$

$n_{A}=1 mole$

$n_{B}= 1 mole$

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

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

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

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

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 =

$1\ dm^3 = 1 litre$

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

$=5.96\ atm$

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

$5.96 - P_0$

$P_0$

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

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

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

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

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

$\approx 0.38\ atm$

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

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

$=32+72=104 mm Hg$

The virial equation for 1 mole of a real gas is written as:
PV=RT

[1+

$\frac{A}{V}+\frac{B}{V^2}+\frac{C}{V^3}+$ ................to the higher power of n]

Where A, B and C are known as virial coefficients. If Vander Waal's equation is written in virial form, then what will be the value of B?

0%
$a-\cfrac{b}{RT}$
0%
$b^3$
0%
$b-\cfrac{a}{RT}$
0%
$b^2$

Explanation

A virial equation is an equation of state of gases that has additional terms beyond that for an ideal gas, which account for the interactions between the molecules.

The equation of state for a van der Waals gas is:

$\boxed{\left(p+\dfrac{a}{V^2_m}\right)(V_m-b)=RT}$

This leads to the following virial expansion:

$\dfrac{pV_m}{RT}+1+\dfrac{B}{V_m}+\dfrac{C}{V^2_m}+-----(1)$

The van der Waals equations of state can be rewritten as:

$pV=\dfrac{RT}{V-b}+\dfrac{a}{V^2}=\dfrac{RT}{V}\left(1-\dfrac{b}{V}\right)^{-1}+\dfrac{a}{V^2}$

and using the binomial expansion, the terms in brackets can be expanded into a series, resulting in

$\dfrac{pV}{RT}=1+\dfrac{1}{V}(b-\dfrac{a}{RT})+\left(\dfrac{b}{V}\right)^2+\left(\dfrac{b}{V}\right)^3+----,$

which is in the same form as the virial expansion is equation (1) with

$B(T)=b-\dfrac{a}{RT}$

If the Van der Waal's equation is written in the virial form then the value of

$B=b-\dfrac{a}{RT}$ .

Hence, option C is correct.

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

$0.93$ .

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

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

$P=0.8314\ M\ Pq$

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

$=\dfrac {3000}{7}$

$\simeq 420\ K$

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

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

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

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

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

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

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

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

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

$2\ L$ vessel is filled with air at

$50^{\circ}C$ and a pressure of

$3\ atm$ . The temperature is now raised to

$200^{\circ}C$ . A valve is now opened so that the pressure inside drops to one atm. What will be the fraction of the total number of moles, inside escaped on opening the valve? (Assume no change in the volume of the container).

0%
$7.7$
0%
$9.9$
0%
$8.9$
0%
$0.77$

Explanation

Given that,

$V = 2\ L, T_{1} = 50 + 273 = 323\ K, P_{1} = 3\ atm$
On heating

$V = 2L, T_{2} = 200 + 273 = 473, P_{2} = ?$
Using

$P-T\ law, \dfrac {P_{1}}{T_{1}} = \dfrac {P_{2}}{T_{2}}$

$P_{2} = \dfrac {3\times 473}{323} = 4.39\ atm$
and

$n = \dfrac {PV}{RT} = \dfrac {3\times 2}{0.0821\times 323} = 0.226$
Now valve is opened till the pressure is maintained at

$1\ atm$ .
Thus, at constant

$V$ and

$T, P\propto n$

$\therefore 4.39\propto 0.226$

$\therefore 1\propto n_{left}$

$\therefore n = \dfrac {0.226}{4.39} = 0.052$

$\therefore$ Moles escaped out

$= 0.226 - 0.052 = 0.174$

$\therefore$ Fraction of moles escaped out

$= \dfrac {0.174}{0.226} = 0.77$

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:

$2N_2(g) + O_2(g) \rightarrow 2N_2O(g)$
Which law is being obeyed in this experiment?

0%
Gay Lussac's law
0%
Law of definite proportion
0%
Law of multiple proportion
0%
Avogadro's law

Explanation

$2N_2(g)+O_2(g)\longrightarrow 2N_2O(g)$

$45.4$

$L$

$22.7$

$L$

$45.4$

$L$

$V\propto$ Number of moles

$\Rightarrow$ Gay Lussac Law is obeyed

Also, Reaction happen in

$2:1$

$mole$ ratio of

$N_2$ and

$O_2$ .

Law of definite proportion is also obeyed.

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

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

Explanation

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

$\therefore P\propto T$

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

$100$ 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

$(2)$

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

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

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

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.

A mixture in which the mole ratio of

$H_{2}$ and

$O_{2}$ is

$2 : 1$ is used to prepare water by the reaction,

$2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)}$
The total pressure in the container is

$0.8\ atm$ at

$20^{\circ}C$ before the reaction. The final pressure at

$120^{\circ}C$ after reaction is: (assuming

$80$ % yield of water).

0%
$1.787\ atm$
0%
$0.878\ atm$
0%
$0.787\ atm$
0%
$1.878\ atm$

Explanation

$\underset {\underset {Final\ mode\ (2a - 2a)}{Initial\ mole\ 2a}}{2H_{2}(g)} + \underset {\underset {(a - x)}{a}}{O_{2(g)}}\rightarrow \underset {\underset {2x}{0}}{2H_{2}O_{(g)}}$
Given

$2x = 2a \times \dfrac {80}{100} = 1.6a$

$\therefore x = 0.8a$
Thus, after the reaction

$H_{2}\ left = 2a - 1.6a = 0.4a\ mole$

$O_{2}\ left = 0.2a\ mole$

$H_{2}O\ formed = 1.6a\ mole$

$\therefore$ Total mole at

$120^{\circ}C$ in gaseous phase

$= 0.4a + 0.2a + 1.6a = 2.2a$
Now, given at initial conditions

$P = 0.8\ atm, T = 293\ K$

$P\times V = nRT$

$0.8\ V = 3a \times R\times 293$

$\therefore V = \dfrac {3a\times R\times 293}{0.8}$
The volume of container remains constant.
After the reaction using,

$P\times V = nRT$

$\therefore P\times \dfrac {3a\times R\times 293}{0.8} = 2.2a \times R\times 393$

$P = \dfrac {393\times 0.8\times 2.2}{3\times 293} = 0.787\ atm$ .

On heating a liquid having coefficient of volume expansion

$\gamma$ in a container having coefficient of linear expansion

$\alpha = \gamma/2$ (numerically), the level of the liquid in the container would

0%
Rise
0%
Fall
0%
Remains almost stationary
0%
Cannot be predicted

Explanation

Here,Initially, both container and liquid have same volume, (Say V)

Given,

Coefficient of Volume expansion for container

$=3\alpha=\dfrac{3\gamma}{2}$

Coefficient of Volume expansion for liquid

$=\gamma$

After

$\Delta T$ temperature,

Final volume of container

$(V_1)=V\left(1+\dfrac{3\gamma}{2}\Delta T\right)$

Final volume of Liquid

$(V_2)=V\left(1+\gamma\Delta T\right)$

It is clear that

$V_1>V_2$ ,

In a nutshell, Container will expand more that liquid

leading to a fall in the level of liquid with respect to the container.

Hence option

$\textbf B$ is the right answer.

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

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.

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

When a non volatile solute is added to a pure solvent, the :

0%
vapour pressure of the solution becomes lower then that of the pure solvent
0%
rate of evaporation of the pure solvent is reduced
0%
solute does not effect the rate of condensation
0%
rate of the evaporation of the solution is equal to the rate of condensation of the solution at a lower vapour pressure than that in the case of the pure solvent

Explanation

If we add a non volatile solute to solvent such

as water, we decease the tendency for water

molecules to evaporate into the gase phase.

in essence, the solute particles obstract the

evaporation of water, as a result, fcream

molecules change from the liquid to the gas

phase, thus reducing the vapour pressure.

$\boxed {hence\, Ans\, - \,'A' }$

1092148_1033515_ans_2c151c7082674f35bf516337a8aa8e6b.png

An inflated balloon has a volume of

$6.0L$ at sea level (

$P=1atm$ ) and is allowed to ascend in altitude until the pressure is

$0.4atm$ . During the rise, the temperature of the gas falls from

${27}^{o}C$ to

$-{23}^{o}C$ . The volume of balloon at its final altitude is

0%
$12.5L$
0%
$15L$
0%
$10L$
0%
$3L$

When intermolecular attractive forces are neglected, isobar formed between V and T at constant p is of the type given in the figure.
Pressure at which isobar made is:

0%
2 atm
0%
0.5 atm
0%
1 atm
0%
4 atm

Explanation

$\Rightarrow PV = n RT$

$Pv \times 10^{-1} = n RT$ (volume is litre)

$P = \dfrac{n R}{tan \theta \times 10^{-1}}$

$\left \{ tan \theta = slope = \dfrac{V}{T} (in\,dm^{3}) \right \}$

putting n = 1, R = 0.082ltatm/mole x

$= \dfrac{0082}{0.1642} \times 10 = \dfrac{0.82}{0.1642} = 0.5 atm$

1170648_1035444_ans_29c40456cb8e44ca8a7696f0e62cc8a1.jpeg

A glass bulb of volume 400

${ cm }^{ 3 }$ is connected to another bulb of volume 200

${ cm }^{ 3 }$ by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of

${ 20 }^{ 0 }$ C and 1.00 atm. The larger bulb is immersed in steam at

${ 100 }^{ 0 }$ C; the smaller, in melting ice at

${ 0 }^{ 0 }$ . Find the final common pressure.

0%
2.05 atm
0%
1.563 atm
0%
2.36 atm
0%
1.134 atm

Explanation

$P_i=1atm$

$T_i=20^oC=293K$

$V_i=0.6L$

$\therefore \dfrac{PiVi}{T_i}=\dfrac{0.6}{293}$

let final press be P

$T_A=373K$

$T_B=273K$

$\therefore \dfrac{PV_A}{T_A}+\dfrac{PV_B}{T_B}=\dfrac{P_iV_i}{T_i}$

$\Rightarrow P\left(\dfrac{0.4}{373}+\dfrac{0.2}{273}\right)=\dfrac{0.6}{293}$

$\Rightarrow P\left(\dfrac{2}{373}+\dfrac{1}{273}\right)=\dfrac{3}{293}$

$\Rightarrow P=\dfrac{3\times 373 \times273}{293 \times 919}$

$\approx1.134atm$

1053736_1036655_ans_75059dfb6bd144fb9f13646a2e14e0f0.png

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

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

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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