MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 12

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

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%
1.563 atm
0%
2.05 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

Vapour pressure of a pure liquid

$X$ is

$2$ atm at

$300$ K. It is lowered to

$1$ atm on dissolving

$1$ g of

$Y$ in

$20$ g of liquid

$X$ . If molar mass of

$X$ is

$200$ , what is the molar mass of

$Y$ ?

0%
$20$
0%
$10$
0%
$100$
0%
$30$

Explanation

Given=

$P^o=2atm$

$\Rightarrow P^o-P= 1 atm$

$W_2= 1$

$W_1= 20g$

$M_1=200$

$M_2=?$

The expression for the lowering of the vapour pressure is given by,

$\cfrac {P^o-P}{P^o}=x_2$

$\Rightarrow \cfrac {1}{2}= \cfrac {W_2 \times M_1}{M_2 \times W_1}$

$\Rightarrow \cfrac {1}{2}=\cfrac {1 \times 200}{M_2 \times 20}$

$\Rightarrow M_2= \cfrac {200 \times 2}{20}=20$

Molar mass of

$Y$ is

$20$ .

Hence, the correct option is

$A$

$4.40\ g$ piece of solid

$CO_{2}$ is allowed to sublime in a balloon. The final volume of the balloon is

$1.00L$ at

$300K$ . What is the pressure of the gas?

0%
$0.122$
0%
$122$
0%
$2.46$
0%
$24.6$

Explanation

moles of

$C{ O }_{ 2 }=\dfrac { 4.40 }{ 44 } =0.1$

$P=0.1\times 0.0821\times \dfrac { 300 }{ 1.00 } =2.46\quad atm$

What is the composition of last deoplet of liquid remaining in equilibrium with vapour?

0%
$x_A = 0.6; x_B = 0.4$
0%
$x_A = 0.5; x_B = 0.5$
0%
$x_A = 0.7; x_B = 0.3$
0%
$x_A = 0.3; x_B = 0.7$

Explanation

$P_{A}=500$

$\quad P_{B}=800$

$x_{A}=0.6$

$x_{B}=1-x_{A}=0.4$

To Find

$X_{A}$ and

$X_{B}$

where

$X_{A}=$ Last drop of liquid remaining in Equilibrium in A.

$X_{B}=$ Last drop of liquid remaining in Equilibrium

$B$

We Know that

Total pressure

$(P)=\quad P_{A}^{0} x_{A}+P_{B}^{0} x_{B}$

$=500 \times(0.6)+800 \times(0.4)=620$

Now partial vapour pressure of

$A$

$\begin{aligned} y_{A}=\dfrac{P_{A}}{P} &=\dfrac{P_{A X A}^{0} }{P} \\ &=\dfrac{500 \times 0.6}{620}=0.48 \end{aligned}$

Mole fraction of

$A=\dfrac{P_{A} x_{A}}{P_{A} x_{A}+P_{B}^{\circ}\left(1-x_{A}\right)}$

$0 .6=\dfrac{500 x_{A}}{500 x_{A}+800\left(1-x_{A}\right)}$

we get

$\begin{array}{rl}x_{A}=0.7 & x_{B}=1-x_{A} \\ & =0.3\end{array}$

Two liquids X and Y form an ideal solution.
The mixture has a vapour pressure of 400mm at 300K, when mixed in the molar ratio of 1:1 and a vapour pressure of 350mm when mixed in the molar ratio of 1:2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are?

0%
250mm, 550mm
0%
350mm, 450mm
0%
350mm, 700mm
0%
550mm, 250mm

Explanation

Total pressure

$P=pA^0\dfrac{2}{A}+pB^0\dfrac{2}{B}$

for case (i)

$400p=pA^0\dfrac{1}{2}+pB^0\dfrac{1}{2}$

$\{molar\, ratio\theta=1:1\}$

$pA^0+pB^0=800$

for case (ii)

$350=pA^0\dfrac{1}{3}+pB^0\dfrac{2}{3}$

$\{molar\,ration\theta=1:2\}$

$pA^0+2pB^0=1050$ .....(2)

from (1) and (2)

$pA^0=550mm$

$pB^0=250\ mm$

An aqueous solution containing liquid

$A(M.\ wt.\ =128)\ 64\%$ by weight has a vapour pressure of

$145\ mm\ Hg$ . If the vapour pressure of water is

$155\ mm\ Hg$ then vapour of

$A$ at the same temperature will be

0%
$205\ mm\ Hg$
0%
$105\ mm\ Hg$
0%
$185\ mm\ Hg$
0%
$52.5\ mm\ Hg$

Equal weights of methane and hydrogen are mixed in an empty container at

${25}^{o}C$ . The fraction of the total pressure exerted by hydrogen is:

0%
$\cfrac{1}{2}$
0%
$\cfrac{8}{9}$
0%
$\cfrac{1}{9}$
0%
$\cfrac{16}{17}$

Explanation

Molar ratio of

$C{ H }_{ 4 }$ :

${ H }_{ 2 }$ = 16 : 2 = 8 : 1 (Mixed in equal quantities)

Pressure is directly proportional to a number of moles.
Methane is 8/9 of total no. of moles.
So,
Fraction of pressure by

$C{ H }_{ 4 }$ = 8/9 .

Hence

$\dfrac { 8 }{ 9 }$ is the answer.

For the two compounds, the vapour pressure of (2) at a particular temperature is expected to be:-

0%
Higher than (i)
0%
Lower than that of (i)
0%
Same as that of (i)
0%
Can be 'higher or lower depending upon the size of the vessel

${N}_{2}+3{H}_{2}\rightleftharpoons 2{NH}_{3}$ . Starting with one mole of nitrogen and

$3$ moles of hydrogen, at equilibrium

$50$ % of each had reacted. If the equilibrium pressure is

$P$ , the partial pressure of hydrogen at equilibrium would be:

0%
$P/2$
0%
$P/3$
0%
$P/4$
0%
$P/6$

Explanation

Initial mole ratio N2:H2 is 1:3

At equilibrium, 50% of N2, H2 has reacted and equilibrium pressure is P

${ N }_{ 2 }+3{ H }_{ 2 }\rightarrow 2N{ H }_{ 3 }\\ y\quad \quad \quad 3y\quad \quad \quad \quad 2y$

At equilibrium, 50% of each reactant reacted, the number of moles will become half.

The mole of N2=

$y-\frac { y }{ 2 }$

The moles of H2=

$3y-\frac { 3y }{ 2 }$

THE moles of NH3=

$\frac { 2y }{ 2 }$

Total moles at equilibrium=

$\frac { y }{ 2 } +\frac { 3y }{ 2 } +\frac { 2y }{ 2 } =3y$

Partial pressure of H2=

$\frac { Moles\quad of\quad H2 }{ Total\quad moles } \times \quad Total\quad pressure$ =

$\frac { 3y }{ 2 } \times \frac { 1 }{ 3y } \times \quad P$ =

$\frac { P }{ 2 }$

The K.E of

$N$ molecule of

$O_{2}$ is

$xjoules$ at__

${123}^{o}C$ . Another sample of

$O_{2}$ at

${327}^{o}C$ has a K.E of

$2x$ joules. The latter sample contains:

0%
$N$ molecules of $O_{2}$
0%
$2N$ molecules of $O_{2}$
0%
$N/2$ molecules of $O_{2}$
0%
$N/4$ molecules of $O_{2}$

Explanation

Though the individual speeds are changing, the distribution of speeds remains constant at a particular temperature.

$K.E_{ O_{ 2 } }=\dfrac { \dfrac { 3 }{ 2 } \times \dfrac { N }{ 32 } \times R\times 150 }{ \dfrac { 3 }{ 2 } \times \dfrac { { N }' }{ 32 } \times R\times 300 }$

$=\dfrac { x }{ 2x }$

$K.E_{ O_{ 2 } }=\dfrac { N\times 1 }{ N'\times 2 }$

$N=N'$

conclusion: hence the option (A) is correct.

A volume

$V$ of a gas at a temperature

$T_{1}$ and a pressure

$p$ is enclosed in a sphere. It is connected to another sphere of volume

$V/2$ by a tube and stopcock is opened the temperature of the gas in the second sphere becomes

$T_{2}$ . The first sphere is maintained at a temperature

$Y_{1}$ . What is the final pressure

$p_{1}$ within the apparatus?

0%
$\dfrac {2pT_{2}}{2T_{2}+T_{1}}$
0%
$\dfrac {2pT_{2}}{T_{2}+2T_{1}}$
0%
$\dfrac {pT_{2}}{2T_{2}+T_{1}}$
0%
$\dfrac {2pT_{2}}{T_{1}+T_{2}}$

Explanation

We have the ideal gas equation as follow:-

$PV={ nRT }$

The number of moles of gases can be written as:-

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

Number of moles present in 1st sphere before stopcock open is

$\\ { n }=\frac { pV }{ R{ T }_{ 1 } }$

Number of moles present in 1st sphere after stopcock open is

$\\ { n }_{ 1 }=\frac { { p }_{ 1 }V }{ { RT }_{ 1 } }$

Number of moles present in 2st sphere before stopcock open is

$\\ { n }_{ 1 }=\frac { { { p }_{ 2 } }\frac { V }{ 2 } }{ R{ T }_{ 2 } }$

Now we have,

$\\ { n }{ ={ n }_{ 1 } }{ +{ n }_{ 2 } }$

$\\ \frac { pV }{ { RT }_{ 1 } } =\frac { p_{ 1 }V }{ { RT }_{ 1 } } +\frac { { p }_{ 1 }V }{ { 2RT }_{ 2 } }$

After solving above expression we get final pressure as

$\\ { p }_{ 1 }=\frac { { 2pT }_{ 2 } }{ { 2T }_{ 2 }+{ T }_{ 1 } }$

Correct option is A

The average velocity of an ideal gas molecule at

${27}^{o}C$ is

$0.3m/sec$ . The average velocity at

${927}^{o}C$ will be:

0%
$0.6m/sec$
0%
$0.3m/sec$
0%
$0.9m/sec$
0%
$3.0m/sec$

Explanation

The expression for average velocity is:

$\sqrt { \frac { 8RT }{ \pi M } }$

$\\ \sqrt { \frac { T1 }{ T2 } }$ =

$\\ \sqrt { \frac { 300 }{ 1200 } }$ =1/2

(

$92{ 7 }^{ \circ }$ ) =2 x (

$2{ 7 }^{ \circ }$ ) =0.6

$m{ s }^{ -1 }$ is the answer.

${100}^{o}C$ and

$1$ atm, if the density of liquid water is

$1.0g$

${cm}^{-3}$ and that of water vapour is

$0.0006g$

${cm}^{-3}$ , then the volume occupied by water molecules in

$1$ litre of steam at that temperature is:

0%
$6{cm}^{3}$
0%
$60{cm}^{3}$
0%
$0.6{cm}^{3}$
0%
$0.06{cm}^{3}$

Explanation

Density of liquid water =

$1.0 g$

$c{ m }^{ -3 }$

Density of water vapour =

$0.0006g$

$c{ m }^{ -3 }$

mass of 1000mL stream =

$\frac { mass\quad of\quad 1000mL\quad stream }{ Density\quad of\quad liquid\quad water } =\frac { 0.6\ gm }{ 1\ gm\ c{ m }^{ -3 } }$

$\therefore$ volume of liquid water =

$0.6$

$c{ m }^{ -3 }$

Hence, option (C) is the correct answer.

Reaction of

${NO}$ takes place with hydrogen if equal molar mixture of snow and hydrogen is taken at initial total pressure of 350 mm of Mercury, total pressure reduces to half its value after 121 seconds while if initial total pressure would have been 275 mm it reduces to half of the 196 seconds calculate the order of reaction?

0%
0
0%
1
0%
2
0%
3

The brown gas Y in the mixture is?

0%
Nitric oxide
0%
Nitrous oxide
0%
Dinitrogen tetroxide
0%
Nitrogen dioxide

Explanation

Given

Colorless gas - $X$

Brown gas - $Y$

The bromine does not change its color at a given temperature, but nitrogen dioxide and dinitrogen tetraoxide change their color at a given temperature and give the same color as given in the question.

Heat dinitrogen tetroxide

${N_2}{O_4}$ $\xrightarrow{{{{160}^{\rm O}}C}}$ $2N{O_2}$ $\xrightarrow{{{{600}^{\rm O}}C}}$ $2NO$ $+$ ${O_2}$

In this reaction,

${N_2}{O_4}$ - dinitrogen tetroxide (colorless)

$2N{O_2}$ - nitrogen dioxide (brown)

$2NO$ - Nitric oxide (colorless)

Therefore, the name of $Y$ gas is dinitrogen tetroxide that is brown.

Hence, option (C) is the correct answer.

$100\space g$ of liquid

$A$ (molar mass

$140\space g { mol }^{ -1 })$ was dissolved in

$1000\space g$ of liquid

$B$ (molar mass

$180 \space g { mol }^{ -1 })$ . The vapour pressure of pure liquid

$B$ was found to be

$500 \space torr.$ Calculate the vapour of pure liquid

$A$ and its vapour pressure in the solution if the total vapour pressure of the solution is

$475 \space torr$ .

0%
$22\space torr$
0%
$32\space torr$
0%
$45 \space torr$
0%
$36 \space torr$

Explanation

Given:-

$P_{T}=475$ Torr,

$P_{B}^{\circ}=500$ Torr

$\omega_{A}=100\space g, M_{A}=140\space g ; \omega_{B}=1000\space g, M_{B}=180 g$

To find:-

$P_{A}^{0}$ and

$P_{A}$

$n_{A}=\dfrac{10 0 }{140}=\dfrac{5}{7}, n_{B}=\dfrac{1000}{180}=\dfrac{50}{9}$

Now,

$x_{A}=\dfrac{n_{A}}{h_{A}+{h_{B}}}=\dfrac{\dfrac{5}{7}}{\dfrac{5}{7}+\dfrac{50}{9}}=0.114$

$x_{B}=1-u_{A}=1-0.114=0.886$

we know, that,

$\quad P_{T}=P_{A}^{0} x_{A}+P_{B}^{0} x_{B}$

where

$p_{A} x_{A}=P_{A}$ and

$p_{B} x_{B}=P_{B}$

Now,

$\quad P^{0} A \times 0.114+500 \times 0.886=475$

$\Rightarrow p^{0} A=280 \cdot 70$ Torr

$\therefore \quad P_{A}=p_{A} x_{A}=280.7 \times 0.114$

So,

$P_A=31.9998$ Torr

$P_{A} \approx 32$ Torr

option

$B$ is correct

$1.00L$ vessel containing

$1.00g$

${H}_{2}$ gas at

${27}^{o}C$ is connected to a

$2.00L$ vessel containg

$88.0g$

${CO}_{2}$ gas, at also

${27}^{o}C$ . When the gases are completely mixed, total pressure is?

0%
$20.525$ atm
0%
$4.105$ atm
0%
$16.420$ atm
0%
$730.69$ atm

Explanation

Given mass of H2=1 g

Moles of H2=

$\frac { 1 }{ 2.016 } =0.496$

Given mass of CO2=88 g

Moles of CO2=

$\frac { 88 }{ 44 } =2$

Total moles=2+0.496=2.496

Total volume=(1.00+2.00)l=3 l

According to Ideal gas equation,

pV=nRT

${ p }_{ total }$ =

$\frac { 2.496\times 0.0821\times 300 }{ 3 } =20.5\quad$ atm

$\frac { p_{ H_{ 2 } } }{ { p }_{ total } } =\frac { No.\quad of\quad moles\quad of\quad H2 }{ Total\quad moles } \\ p_{ H_{ 2 } }=\frac { 0.496\times 20.5 }{ 2.496 } =4.07\quad atm\\$

$\frac { p_{ CO_{ 2 } } }{ { p }_{ total } } =\frac { No.\quad of\quad moles\quad of\quad CO2 }{ Total\quad moles } \\ p_{ CO_{ 2 } }=\frac { 2\times 20.5 }{ 2.496 } =16.4\quad atm\\$

Total partial pressure=(16.4+4.07)atn=20.47 atm=20.5 atm

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
Volume percent of

${H}_{2}$ is:

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

Explanation

In a mixture of gases the percent of each gas in the volume is same as the percent of each partial pressure in the total pressure.

Total partial pressure=

$150+300+200+100=750\\$ torr

$volume\quad percent H2=\frac { 150 }{ 750 } \times 100\\ =20$

$1.0\ g$ sample of air consists of approximately

$0.76\ g$ of nitrogen and

$0.24\ g$ of oxygen. This sample occupies a

$1.0\ L$ vessel at

$20^{o}C$ . Then:

0%
The partial pressure of $N_{2}$ is $0.65\ atm$
0%
The partial pressure of $O_{2}$ is $0.36\ atm$
0%
The total pressure is $0.83\ atm$
0%
The total pressure is $1.05\ atm$

Explanation

conclusion: hence the option (C) is correct.

The data needed for use are

${ P }_{ { N }_{ 2 }= }?\quad \quad \quad \quad { n }_{ N_{ 2 } }=?\quad \quad \quad V=1.00L\quad \quad \quad T=293K\quad ({ 20 }^{ 0 }C)$

${ P }_{ O_{ 2 } }=?\quad \quad \quad { n }_{ { O }_{ 2 } }=?\quad \quad \quad V=1.00L\quad \quad \quad \quad T=293K\quad ({ 20 }^{ 0 }C)$

The number of

${ n }_{ N_{ 2 } }$ and

${ n }_{ { O }_{ 2 } }$ can be obtained from the masses and molar masses (28.02g/mol for

${ N }_{ 2 }$ 32.00g/mol of

${ O }_{ 2 }$ 0

Moles of

${ N }_{ 2 }$ = 0.76G

${ N }_{ 2 }$ x =

$\dfrac { 1\quad mol\quad { N }_{ 2 } }{ 28.02g\quad { N }_{ 2 } } =0.0271\quad mol\quad { N }_{ 2 }$

Moles of

${ O }_{ 2 }$ =0.24g

${ O }_{ 2 }$ x=

$\dfrac { 1\quad mol\quad { O }_{ 2 } }{ 32.00{ gO }_{ 2 } } =0.00750mol\quad { O }_{ 2 }$

using

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

${ P }_{ { N }_{ 2 }= }0.0271mol{ N }_{ 2 }\times \dfrac { 0.08206Latm }{ 1Kmol } \times \dfrac { 293 }{ 1.00L } =0.65\quad atm{ N }_{ 2 }$

${ P }_{ O_{ 2 } }=0.00750mol{ O }_{ 2 }\times \dfrac { 0.08206Latm }{ 1Kmol } \times \dfrac { 293 }{ 1.00L } =0.18\quad atm{ O }_{ 2 }$

The total pressure is the sum of these partial pressure

$P=0.65 atm+0.18atm=0.83atm$

The value of

${K}_{p}$ for the reaction

${CO}_{(2(g)}+{C}_{(g)}\rightleftharpoons 2{CO}_{(g)}$ is

$3.0$ at

$1000K$ . If initially

${P}_{{CO}_{2}}=0.48$ bar and

${P}_{CO}=0$ bar and pure graphite is present. The equilibrium partial pressure of

$CO$ and

${CO}_{2}$ are:

0%
${P}_{CO}=0.33,{P}_{{CO}_{2}}=0.15$
0%
${P}_{CO}=0.66,{P}_{{CO}_{2}}=0.33$
0%
${P}_{CO}=1.44,{P}_{{CO}_{2}}=0.66$
0%
${P}_{CO}=0.66,{P}_{{CO}_{2}}=0.15$

Explanation

Given

${ K }_{ p }$ =3,

${ P }_{ C{ O }_{ 2 } }$ =0.48 bar,

${ P }_{ CO }$ =0 bar

${ CO }_{ 2 }(g)+C(s)\Longleftrightarrow 2CO$

$t=0\quad { CO }_{ 2 }=0.48\quad { CO }=0\\ \quad t={ t }_{ eq }\quad { CO }_{ 2 }=0.48-p\quad { CO }=2p$

${ K }_{ P }=\frac { (p_{ co })^{ 2 } }{ { p }_{ co } }$

$3=\frac { (2p)^{ 2 } }{ 0.48-p }$

p=0.3325 bar

${ p }_{ co }=2p=2\times 0.3325\quad bar=0.665\quad bar$

${ p }_{ co_{ 2 } }=0.48-p=0.48-0.3352\quad bar=0.1475\quad bar=0.15bar$

$N_{2}$ is found in a litre flask under

$100\ kPa$ pressure and

$O_{2}$ is found in another

$3$ litre flask under

$320\ kPa$ pressure. If the two flask are connected, the resultant pressure is:

0%
$265\ kPa$
0%
$210\ kPa$
0%
$420\ kPa$
0%
$365\ kPa$

Explanation

At initial condition

${ P }_{ 1 }=100kPa$

$=1atm{ V }_{ 1 }$

$=1L{ P }_{ 2 }$

$=320k{ Pa=3atm{ V }_{ 2 } }$ =

$3L$

Applying ideal gas law,

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

total mole at initial conditions,

$(1+{ n }_{ 2 })$ =

${ n }_{ i }=\dfrac { 2 }{ RT } +\dfrac{ 9.3 }{ RT } =\dfrac { 11.3 }{ RT }$

After mixing the gases the number of moles remains constant gases do not undergo reaction.

${ n }_{ i }={ n }_{ f }={ P }_{ f }\times \dfrac { { V }_{ f } }{ RT } \\ Where,{ P }_{ f }=final\quad pressure\\ \quad \quad \quad \quad \quad { V }_{ f }=final\quad volume\\ { V }_{ f }={ V }_{ 1 }+{ V }_{ 2 }=4L\\ thus,{ P }_{ f }\times \dfrac { 4 }{ RT } =\dfrac { 11.3 }{ RT } \\ { P }_{ f }=2.85atm$

${ P }_{ f }=285kPa\approx 265kPa$

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%
$3L$

$0.5\ \text{mole}$ of each

$H_{2},SO_{2}$ and

$CH_{4}$ are kept in a container. A hole was made in the container. After 3 hours, decreasing order of partial pressures of gases in the container will be:

0%
$P_{SO_{2}} > P_{CH_{4}} > P_{H_{2}}$
0%
$P_{H_{2}} > P_{SO_{2}} > P_{CH_{4}}$
0%
$P_{CH_{4}} > P_{SO_{4}} > P_{H_{2}}$
0%
$P_{CH_{4}} > P_{H_{2}} > P_{SO_{2}}$

Explanation

Answer : D.

$P_{SO_{2}} > P_{CH_{4}} > P_{H_{2}}$

The partial pressure of a gas is directly proportional to mole fraction, therefore directly proportional to the number of moles.

Initially, there are equal number of moles (0.5 moles) of each gases. So their partial pressures are equal.

After 3 hours, some amount of each gases will be effused out through the hole. Gas with larger effusion rate will be effused more through the hole and will have less number of moles remaining in the container, resulting in lower partial pressure compared to others.

We know that,

Molar mass of

$H_{2}$ ,

$M_{H_{2}}$ =

$1\times2$ =

$2$ g

Molar mass of

$SO_{2}$ ,

$M_{SO_{2}}$ =

$32 + (16\times2)$ =

$64$ g

Molar mass of

$CH_{4}$ ,

$M_{CH_{4}}$ =

$12 + (1\times4)$ =

$16$ g

Therefore, the order of molar masses,

$M_{SO_{2}} > M_{CH_{4}} > M_{H_{2}}$

$\textrm{Rate of effusion of gas}$

$\alpha$

$\dfrac{1}{\sqrt{\textrm{molar mass}}}$

$\Rightarrow$ Order of rates of effusion,

$r_{SO_{2}} < r_{CH_{4}} < r_{H_{2}}$

$\textrm{Number of moles remaining in the container}$

$\alpha$

$\dfrac{1}{\textrm{Rate of effusion}}$

$\Rightarrow$ Order of number of moles remaining in the container,

$n_{SO_{2}} > n_{CH_{4}} > n_{H_{2}}$

$\textrm{Partial pressure}$

$\alpha$

$\textrm{Number of moles remaining in the container}$

$\Rightarrow$ Order of partial pressures,

$P_{SO_{2}} > P_{CH_{4}} > P_{H_{2}}$

The density of a gas

$A$ is twice that of

$B$ . Molar mass of

$A$ is half that of

$B$ . The ratio of partial pressure of

$A$ to

$B$ is:

0%
$1/4$
0%
$1/2$
0%
$4/1$
0%
$2/1$

Explanation

According to Ideal gas equation

$PV=nRT$

$n=\dfrac{m}{M}$

$PV=\dfrac { m }{ M } RT\\ P=\dfrac { m }{ V } \times \dfrac { RT }{ M } \\ P=d \times \ \dfrac { RT }{ M }$

$R$ and

$T$ are constant

$P_{ A }=\dfrac { { d }_{ A } }{ { M }_{ A } } \\ P_{ B }=\dfrac { { d }_{ B } }{ { M }_{ B } }$

Given

${ M }_{ A }=\dfrac { 1 }{ 2 } { M }_{ B }\\ d_{ A }=2d_{ B }$

$\dfrac { { P }_{ A } }{ { P }_{ B } } =\dfrac { { d }_{ A }{ M }_{ B } }{ { M }_{ A }{ d }_{ B } }$

Putting the value of

${ d }_{ A }$ and

${ M }_{ A }$ in the above equation

$\dfrac { { P }_{ A } }{ { P }_{ B } } =4$

A container is divided into two compartment. One compartment contains

$2$ mole of

$N_{2}$ gas at

$1$ atm and

$300\ K$ & other compartment contains

$H_{2}$ gas at the same temperature and pressure. Volume of

$H_{2}$ compartment is four times the volume of

$N_{2}$ compartment. [assume no reaction under these condition]
If the container containing

$N_{2}$ and

$H_{2}$ are further heated to

$1000\ K$ forming

$NH_{3}$ with

$100\%$ yield. Calculate the final total pressure.

0%
$2.22\ atm$
0%
$3\ atm$
0%
$2\ atm$
0%
$3.33\ atm$

Explanation

Compartment containing

$N_2$

$n=2$

$P=1$ atm

$T=300$ K

Finding for V

Using Ideal gas equation

$PV=nRT$

$V=\dfrac{2\times0.0821\times300}{1}=49.26$ L

Compartment containing

$H_2$

Givn volume is 4 times than

$N_2$

$V_2 =4\times49.26=197$ L

Finding moles for

$H_2$ Compartment

given pressure and temperature is equal to

$N_2$ compartment

$n=\dfrac{197}{0.0821\times300}=8$

Calculating final pressure

$T=1000$ K

$P=\dfrac{8\times0.0821\times1000}{197}=3.33$ atm

A glass tube of volume

$112ml$ . containing a gas is partially evacuated till the pressure in it drops to

$3.8 \times 10^{-5}\ torr$ at

$0^{o}C$ . The number of molecules of the gas remaining in the tube is?

0%
$3\times 10^{17}$
0%
$1.5\times 10^{14}$
0%
$4.5\times 10^{18}$
0%
$6\times10^{18}$

Explanation

Pressure

$=3.8\times 10^{-5} torr$ and temperature

$=0^oc+273k$

$R=62.36367\ L\ Torr |c^{-1} mo|^{-1}$

Volume

$=112\ ml =112 \times 10^{-3}c$

number of mole

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

$=\dfrac{3.8\times 10^{-5}\times 112\times10^{-3}}{62.36367\times 273}$

$=2.5\times10^{-10}$

number of molecules of gas

$=6.022\times 10^{23} \times 2.5\times 10^{-10}$

$=1.5\times10^{14}$

So, the correct option is

$B$

Vapour pressure of a saturated solution of a sparingly soluble salt

$A_2B_3$ is 31.8 mm of Hg at

$40^{\circ}C$ . If vapour pressure of pure water is 31.8 mm of Hg at

$40^{\circ}C$ the solubility product of

$A_2B_3$ at

$40^{\circ}C$ is:

0%
$5.67\times 10^{-6}$
0%
$1.42\times 10^{-6}$
0%
$6.30\times 10^{-5}$
0%
1

An LPG cylinder can withstand pressure difference of 15 atm across its boundaries. If at room temp $({ 27 }^{ o })$ it is filled with 2 atm pressure. Determine the temperature at which it will explode?

0%
${ 216 }^{ o }C$
0%
${ 2127 }^{ o }C$
0%
${ 2400 }^{ o }C$
0%
${ 27 }^{ o }C$

Explanation

Given,

Pressure difference across LPG cylinder boundaries= 15 atm

We know atmospheric pressure = 1 atm

It is given that pressure difference across wall =15 atm

therefore,Pressure inside cylinder's wall=15+1=16 atm

${P}_{1}=16atm$

${P}_{2}=2atm$

${T}_{2}=300K$

Now according to gay lusaac's law at constant

$V$

$P\propto T$

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

${T}_{1}=16 \times 300/2$

${T}_{1}=2400K$

$18.0\ g$ of glucose $\left( {{C_6}{H_{12}}{O_6}} \right)$ is added to $178.2\ g$ of water. The vapour pressure of water for this aqueous solution at ${100^o}\ C$ is :

0%
$759..00$ Torr
0%
$7.60$ Torr
0%
$76.00$ Torr
0%
$752.40$ Torr

Explanation

Molecular mass of water

$=2\times 1+1\times 16=18$ g

For

$178.2g$ water

$n_A$ =9.9

Molecular mass of glucose

$=6\times 12+12\times 1+6\times 16=180$ g

For 18g glucose

$n_B$ =0.1

$X_B=\dfrac{0.10}{(0.1+9.9)}=0.01$

$X_A=0.99$

For lowering of vapour pressure ,

$p=p^0_AX^A=P^0_A(1−X_B)$

$P=760(1−0.01)$

$=760−7.6$

$=752.4torr$

100 g of liquid

$A$ (molar mass 140 g

$mol^{-1}$ ) was dissolved in 1000 g of liquid

$B$ (molar mass 180 g

$mol^{-1}$ ) The vapour pressure of pure liquid

$B$ was found to be 500 torr.

Calculate the vapour pressure of pure liquid

$A$ and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr:

0%
$3.19, 500$
0%
$283.18, 31.999$
0%
$500, 300$
0%
$190, 31.965$

Explanation

Given:

$P_T= 475, W_A= 100g, M_A=140, {P^o_B}=500$

$W_B=1000g, M_B= 180, P^o_A ?, P_A=?$

$P_T=P^o_AX_A+P^o_BX_B\longrightarrow (1)$ &

$X_A+X_B=1 \longrightarrow (2)$

$\Rightarrow \cfrac {X_B}{X_A}=\cfrac {W_BM_A}{W_AM_B}=\cfrac {1000\times 140}{100\times 180}=7.78$

$\therefore X_B= 7.78 X_A$

Substituting

$X_B$ in

$(2)$

$X_A+7.78X_A=1$

$\therefore X_A=0.113$ & hence

$X_B=1-0.113=0.886$

Substituting

$X_A$ &

$X_B$ in equation

$(1)$

$\Rightarrow 475= P^o_A(0.113)+500(0.886)$

$\Rightarrow 0.113P^o_A=32 \Rightarrow P^o_A=\cfrac {32}{0.113}=283.18 torr$

Now,

$P_A=P^o_A(1-X_B)=P^o_AX_A$

$= 283.18 \times 0.113$

$= 31.999 torr$

Moles of

$Na_2SO_4$ to be dissolved in 12 mole of water to lower its vapour pressure by 10 millimeter Mercury at a temperature at which vapour pressure of pure water is 50 millimeters?

0%
1.5 moles
0%
2 moles
0%
1 mole
0%
3 moles

The vapour pressure of pure

$A$ is 10 torr and at the same temperature when 1 g of

$B$ is dissolved in 20 gm of

$A$ , its vapour pressure is reduced to 9.0 torn If the molecular mass of

$A$ is 200 amu, then the molecular mass of

$B$ is:

0%
100 amu
0%
90 amu
0%
75 amu
0%
120 amu

Explanation

Given data

Wt of

$A=20$ gm

${ P }_{ A }^{ 0 }=10$ torr

molecular mass of

$A=200$ amu

Wt of

$B=1$ gm

${ P }_{ solution }=9$ torr

molecular mass of

$B=?$

This problem can be solved by using the relative lowering of vapour pressure.

$\dfrac { { P }^{ 0 }-P }{ { P }^{ 0 } } =\dfrac { { n }_{ 2 } }{ { n }_{ 1 }+{ n }_{ 2 } }$

${ P }^{ 0 }=$ Vapour pressure of pure solvent

$P=$ Vapour pressure of solution

${ n }_{ 2 }=$ moles of solute

${ n }_{ 1 }=$ moles of solvent

$\dfrac { 10-9 }{ 10 } =\dfrac { \dfrac { 1 }{ x } }{ \dfrac { 20 }{ 200 } +\dfrac { 1 }{ x } }$

$\left[ \dfrac { 1 }{ 10 } +\dfrac { 1 }{ x } \right] \dfrac { 1 }{ 10 } =\dfrac { 1 }{ x }$

$\dfrac { 1 }{ 100 } +\dfrac { 1 }{ 10x } =\dfrac { 1 }{ x }$

$\dfrac { 1 }{ 100 } =\dfrac { 1 }{ x } -\dfrac { 1 }{ 10x } =\dfrac { 9 }{ 10x }$

$\therefore$ The molecular mass of compound (solute) B is

$90$ amu

$10x=900$

$\boxed { x=90amu }$

Hence, the correct option is

$B$

${ \Delta }_{ f }{ G }^{ }$ at 500 K for substance

$S$ in liquid state and gaseous state are +100.7 kcal

${ }^{ -1 }$ and +103 kcal

${ }^{ -1 }$ , respectively. Vapour pressure of liquid

$S$ at 500 K is approximately equal to:
(

$R=2\ cal$

${ K }^{ -1 }{ }^{ -1 }$ )

0%
0.1 atm
0%
10 atm
0%
100 atm
0%
1 atm

Explanation

Solution:- (B)

$10 \; atm$

As we know that, at equilibrium,

${\Delta{G}}^{°} = -2.303 \; RT \; \log{{K}_{p}} ..... \left( 1 \right)$

${S}_{\left( l \right)} \rightleftharpoons {S}_{\left( g \right)}$

From the above reaction,

${\Delta{G}}^{°} = {{\Delta{G}}_{f}}_{\left( \text{product} \right)} - {{\Delta{G}}^{°}}_{\left( \text{reactant} \right)}$

$\Rightarrow {\Delta{G}}^{°} = 103 - 100.7 = 2.3 \; {kcal}/{mol} = 2.3 \times {10}^{3} \; {kcal}/{mol}$

Given:-

$T = 500 \; K$

$R = 2 {cal}/{mol-K}$

From

${eq}^{n} \left( 1 \right)$ , we have

$2.3 \times {10}^{3} = -2.303 \times 2 \times 500 \times \log{{K}_{p}}$

$\Rightarrow \log{{K}_{p}} = -1$

$\Rightarrow {K}_{p} = {10}^{-1} \; atm$

Now, from the above reaction,

${K}_{p} = \cfrac{1}{{P}_{{S}_{\left( l \right)}}}$

$\Rightarrow {P}_{{S}_{\left( l \right)}} = \cfrac{1}{{10}^{-1}} = 10 \; atm$

Hence the vapou pressure of liquid

$S$ is approximately equal to

$10 \; atm$ .

Which of the following change is observed occurs when a substance X is converted from liquid to vapour phase at the standard boiling point?
I.Potential energy of the system decreases
II. The distance between molecules increases
III. The average kinetic energy of the molecules in both phases are equal.

0%
I only
0%
II only
0%
III only
0%
II and III only

In the given equilibrium,

${ H }_{ 2 }O(l)\rightleftharpoons { H }_{ 2 }O(g)$ at

$100^oC$ the vapour pressure is

$1 \ atm$ . If the volume of the container is halved, after sometime the vapour pressure becomes:

(assuming constant temperature)

0%
$1.5 atm$
0%
$2.5 atm$
0%
$2 atm$
0%
$1 atm$

The vapour pressure of a pure liquid

$A$ is

$70$ torr at

$27^oC$ . It forms an ideal solution with another liquid

$B$ . The mole fraction of

$B$ is

$0.2$ and total vapour pressure of the solution is

$84$ torr at

$27^oC$ . The vapour pressure of pure liquid

$B$ at

$27^o$ C is:

0%
$14$
0%
$56$
0%
$140$
0%
$70$

Explanation

We know Raoult's law,

For an ideal solution,

Partial pressure in solution

$=$ mole fraction

$\times$ Vapour pressure in pure state

Given,

${ P }_{ A }^{ 0 }=70$ torr at

${ 27 }^{ 0 }C$

${ X }_{ B }=0.2$

then

${ X }_{ A }=1-{ X }_{ B }=1-0.2=0.8$

${ P }_{ B }^{ 0 }=?$

${ P }_{ total }$ or

${ P }_{ solution }$

$=84$ torr

By, Raoult's law

$\boxed { { P }_{ solution }={ X }_{ A }{ P }_{ A }^{ 0 }+{ X }_{ B }{ P }_{ B }^{ 0 } }$

$84=0.8\times 70+0.2{ P }_{ B }^{ 0 }$

${ P }_{ B }^{ 0 }=\dfrac { 84-56 }{ 0.2 }$

${ P }_{ B }^{ 0 }=140$ torr

$\therefore$ Vapour pressure of pure liquid

$B$ at

${ 25 }^{ 0 }C$ is

$140$ torr.

Hence, the correct option is

$\text{C}$

Which solution has the highest vapour pressure?

0%
0.02 M $NaCl$ at $50^o C$
0%
0.03 M sucrose at $15^oC$
0%
0.005 M $CaCl_2$ at $50^oC$
0%
0.005 M $CaCl_2$ at $25^oC$

Ammonium carbamate dissociates as:

$N{ H }_{ 2 }COON{ H }_{ 4 }(s)\quad \rightleftharpoons \quad 2N{ H }_{ 3 }(g)\quad +\quad C{ O }_{ 2 }(g)$
In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that partial pressure of

$N{ H }_{ 3 }$ now equals to the original total pressure. Calculate the ratio of partial pressure of

$C{ O }_{ 2 }$ now to the original partial pressure of

$C{ O }_{ 2 }$ :

0%
4
0%
9
0%
$\dfrac { 4 }{ 9 }$
0%
$\dfrac { 2 }{ 9 }$

Explanation

$\underset {Initial}{{NH_2COONH_4}_{(s)}} \rightleftharpoons \underset {2P}{{2NH_3}_{(g)}}+\underset {P}{{CO_2}_{(g)}}$

$K_P=P_{NH_3}^2 P_{CO_2}$

$\Rightarrow K_P=2P.P \longrightarrow (1)$

Now, in second case

$P_t=3P$

$\underset {Initial}{{NH_2COONH_4}_{(s)}}\rightleftharpoons \underset {3P}{{2NH_3}_{(g)}}+\underset {P'}{{CO_2}_{(g)}}$

$K_P=(3P)^2(P')\longrightarrow (2)$

From

$(1)$ &

$(2)$ ,

$(2P)^2P=(3P)^2(P')$

$\Rightarrow P'=\cfrac {4P^3}{9P^2}=\cfrac {4P}{9}$

$\therefore \cfrac {P'}{P}=\cfrac {4}{9}$

A 3L container at 3 atm pressure consist a mixture of

$N_2$ gas and

$H_2O$ vapours. Solid sphere of volume 0.5L is also present in same container. What will be the pressure of gas, if volume of container is reduced to 2L, at same temperature (AQ. tension = 0.2 atm)?

0%
5.4 atm
0%
4.2 atm
0%
4.87 atm
0%
5 atm

At same temperature

$N_2O_4$ is dissolved to 40% and 50% at total presure

$p_1$ and

$p_2$ atm respevtively in

$NO_2$ .Then the ratio of

$P_1$ &

$P_2$ is?

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

$20^{o}C$ , the vapour pressure of

$0.1\ M$ solution of urea is

$0.0311\ mm$ less than that of water and the vapour pressure of

$0.1\ M$ solution of

$KCl$ is

$0.0574\ mm$ less than that of water. The apparent degree of dissociation of

$KCl$ at this dilution is :

0%
$92.1\%$
0%
$84.6\%$
0%
$68.4\%$
0%
$54.1\%$

Explanation

The Vant Hoff factor is the property of a solute. So it is directly proportional to any colligative property of solute (like vapour pressure).

So, vapour pressure

$\propto$ Vant Hoff factor (i)

$\implies \dfrac{vapour\:pressure_{(urea)}}{vapour\:pressure_{(KCl)}}=\dfrac{i_{urea}}{i_{KCl}}$

But

$i_{(urea)}=1$

$\implies \dfrac{vapour\:pressure_{(KCl)}}{vapour\:pressure_{(urea)}}={i_{KCl}}=\dfrac{0.0574}{0.0311}$ ...(i)

$KCl\leftrightharpoons K^++Cl^-$

$1$

$0$

$1-\alpha$

$\alpha$

For dissociation, we have:

$\alpha=\dfrac{i-1}{n-1}$

Here,

$n=2$

$\implies \alpha=\dfrac{i-1}{2-1}$

$\implies i={1+\alpha}$

From (i), we have:

$\dfrac{0.0574}{0.0311}={1+\alpha}$

$\implies \alpha=0.846$

So,

$\%\:\alpha=\alpha\times 100=84.6\%$

Hence, the correct option is (B).

${ CuSO }_{ 4 }\cdot 5H_{ 2 }O\left( s \right) \rightleftharpoons { CuSO }_{ 4 }\cdot { 3H }_{ 2 }O\ (s)+{ 2H }_{ 2 }O\ (l){ K }_{ p }=1.086\times { 10 }^{ -4 }{ atm }^{ 2 }\ at\ { 25 }^{ o }C$ The efflorescent nature of

${CuSO}_{4}\cdot \ {5H}_{2}O$ can be noticed when vapour pressure of

${H}_{2}O$ in atmosphere is:

0%
$>7.92\ mm$
0%
$<7.92\ mm$
0%
$\gtrless7.92\ mm$
0%
none of these

Explanation

Efflorescent nature of

$CuSO_4.5H_2O$ means the loss of some water molecules so the compound can float on the surface rather than at bottom.

The pressure constant contains only the partial pressure of gas, the partial solid and liquid is taken unity.

So,

$CuSO_4.5H_2O(s)\leftrightharpoons CuSO_4.3H_2O(S)+2H_2O(g)$

$K_p=(P_{H_2O})^2$

$(P_{H_2O})^2=1.086\times 10^{-4}atm^2$

$P_{H_2O}=1.04\times 10^{-2}atm$

Now,

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

$P_{H_2O}=7.92\:mm\:of\:Hg$

So, for the reaction to proceed forward, the

$P_{H_2O}<7.92\:mm\:of\:Hg$

Hence, the correct option is (B).

A solution of a non-volatile solute in water has a boiling point of 375.3K. Calculate the vapour pressure of water above this solution at 338K. Given, Po (water) = 0.2467 atm at 338K and Kb for water = 0.52.

0%
$0.18\ atm$
0%
$0.23\ atm$
0%
$0.34\ atm$
0%
$0.925\ atm$

Which of the following has least vapour presure ?

0%
$0.1\ M\ NaCl$
0%
$0.1\ M\ MgCl_{2}$
0%
$0.1\ M\ AlCl_{3}$
0%
$0.1\ M\ K_4[Fe(CN)_{6}]$

The empirical formula of a monobasic acid is

${CH}_{2}O$ . The vapour density of its ethyl ester is

$44$ . What is the Molecular formula of the acid?

0%
${C}_{2}{H}_{4}{O}_{2}$
0%
${CH}_{2}{O}_{2}$
0%
${C}_{2}{H}_{2}{O}_{2}$
0%
${C}_{3}{H}_{6}{O}_{3}$

Explanation

Empirical formula

$CH_{2}O\,\,\rightarrow weight=12+2+16=30\,gm$

Molecular mass of acid=(Molecular mass of ethyl ester)-(Molecular weight of

$C_{2}H_{5}$ )+(Atomic mass of

$H$ atom)

$=(2\times V\cdot D)-(2\times 12\,+5\times 1)+1$

$=88-29+1=60$

$\therefore$ Molecular formula

$=\cfrac{Molecular \,Weight}{Empirical \,Weight}\times Empirical \,Formula$

$=\cfrac{60}{30}\times C(H_{2}O)=C_{2}H_{4}O_{2}$

The colourless gas X in the mixture is:

0%
Bromine
0%
Pure nitrogen dioxide
0%
Dinitrogen tetraoxide
0%
Nitric oxide

Explanation

Given,

Colorless gas - $X$

Brown gas - $Y$

heat dinitrogen tetraoxide

${N_2}{O_4}$ $\xrightarrow{{{{160}^{\rm O}}C}}$ $2N{O_2}$ $\xrightarrow{{{{600}^{\rm O}}C}}$ $2NO$ $+$ ${O_2}$

in this reaction,

${N_2}{O_4}$ - dinitrogen tetraoxide (colorless)

$2N{O_2}$ - nitrogen dioxide (brown)

$2NO$ - Nitric oxide (colorless)

Therefore, the name of $X$ gas is nitric oxide that is colorless.

Hence, option(d) is the correct answer.

Plots of

$\cfrac { PV }{ RT } vsP$ for 1 mole of

${ H }_{ 2 }$ ,

${ NH }_{ 3 }$ and

${ CH }_{ 4 }$ gases are given. Match the curve with corresponding gases
Curve Gas
(i) Curve 'a' p-

${ H }_{ 2 }$
(ii) Curve 'b' q-

${ NH }_{ 3 }$
(iii) Curve 'c' r-

${ CH }_{ 4 }$

0%
(i)-p, (ii)-q, (iii)-r
0%
(i)-q, (ii)-r, (iii)-p
0%
(i)-p, (ii)-r, (iii)-q
0%
(i)-r, (ii)-q, (iii)-p

3.6 gm of

$O_2$ is adsorbed on 1.2 gr of metal powder. What volume of

$O_2$ adsorbed per gram of the absorbant at 1 atm and 273K?

0%
2.1
0%
0.19
0%
1
0%
None

Explanation

$1.2 g \rightarrow 3.6 g$ oxygen

$1 g \rightarrow \dfrac{3.6}{1.2} = 3 g$

$PV = \dfrac{x}{M} \times RT$

$1 \times V = \dfrac{3}{32} \times 0.0821 \times 273$

$V=2.1\ L$

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 12 - 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 12 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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