MCQ Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 12

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

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

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

$A$

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)

$pA^0=550mm$

$pB^0=250\ mm$

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.

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.

$Fc$ and

$Fa$ denote cohesive and adhesive force on a liquid molecule near the surface of a solid. then the surface of request is concave , when;

0%
$F_{A}<\frac{F_{0}}{\sqrt{2}}$
0%
$F_{A}=\frac{F_{c}}{\sqrt{2}}$
0%
$F_{n}>\frac{F_{c}}{\sqrt{2}}$
0%
$F_{A}>F_{C}$

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

$4.40\ g$ 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

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

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

${N}_{2}+3{H}_{2}\rightleftharpoons 2{NH}_{3}$ . 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=

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

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

$\frac { P }{ 2 }$

Select correct option(s).

0%
Pressure in container $-I$ is $3\ atm$ before opening the valve
0%
Pressure after opening the valves is $3.57\ atm$
0%
Moles in each compartment are same after opening the valve
0%
Pressure in each compartment are same after opening the valve

Explanation

Container 1

$T=300$ K

$V=16.4$ L

$n=2$

$R=0.0821$

Applying Ideal gas equation

$PV=nRT$

Before opening the valve

$P=\dfrac{2\times0.0821\times300}{16.4}=3$ atm

The total pressure for volatile components

$A$ and

$B$ is 0.02 bar at equilbibrium. If the mole fractions of component

$A$ is

$0.2$ , then what will be partial pressure of compount

$B$ ?

0%
$0.02\ bar$
0%
$0.04\ bar$
0%
$0.016\ bar$
0%
$0.2\ bar$

Explanation

Partial pressure of A

$=P_{A}=x_{A}\times P_{total}=0.2\times 0.02=0.004$

Partial pressure of B

$=P_{B}$

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

$P_{B}=0.020-0.004=0.016\, bar$

$2.24 L$ cylinder of oxygen at

$1atm$ and

$273 K$ is found to develop a leakage. When the leakage was plugged the pressure dropped to

$570 mm$ of Hg. The number of moles of gas that escaped will be:

0%
$0.025$
0%
$0.050$
0%
$0.075$
0%
$0.09$

Explanation

Original pressure,

$\left( {P}_{i}\right) = 1 \text{ atm} = 760 \text{ mmHg}$

Final pressure,

$\left( {P}_{f} \right) = 570 \text{ mmHg}$

$\therefore$ Drop in pressure,

$\left( P \right) = {P}_{i} - {P}_{f} = \left( 760 - 570 \right) \text{ mmHg} = 190 \text{ mmHg} = \cfrac{190}{760} \text{ atm}$

Volume,

$\left( V \right) = 2.24 L$

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

Temperature,

$\left( T \right) = 273 K$

According to ideal gas equation,

$PV = nRT$

$\Rightarrow \; n = \cfrac{PV}{RT}$

$\Rightarrow \; n = \cfrac{\left( \cfrac{190}{760} \right) \times 2.24}{0.0821 \times 273} = 0.02498 = 0.025 \text{ moles}$

Hence the number of moles of gas that escaped will be

$0.025 \text{ moles}$ .

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

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

${100}^{o}C$ 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.

$100\space g$ 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

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

$0.5\ \text{mole}$ 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 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.

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

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

$18.0\ g$ 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 ,

$p=p^0_AX^A=P^0_A(1−X_B)$

$P=760(1−0.01)$

$=760−7.6$

$=752.4torr$

Factors affecting the rate of diffusion are:

I. Gradient of concentration
II. Permeability of the membrane
III. Temperature
IV. Pressure
V. Size of diffusing material

0%
I, III and V are correct
0%
I and V are correct
0%
I, II, III, IV, V are correct
0%
Only V is correct

Explanation

$\textbf{Correct Option A}$

$\textbf{Solution}$

Diffusion process is explained as the movement of substance from a region of higher concentration to a region of lower concentration.Diffusion occurs in three physical states namely solids, liquids and gas

Various external factors are there which affect the rate of the diffusion such as gradient concentration, temperature and size of the diffusing material.

$\bullet$ Option I. In case of concentration gradient, if the difference in the concentration is greater then the rate of diffusion is also higher. When the medium have a lower concentration of solute then refusing molecule have higher possibility of moving away from the central area.

$\bullet$ Option II. The permeability of the membrane is the rate of passive diffusion through the membrane and it does not affect the rate of diffusion.

$\bullet$ Option III. When we increase the temperature, the kinetic energy of all the particles in the system increases due to which the particles starts moving at an increased rate which in turn increases the rate of diffusion.
At lower temperature the energy of the molecule decreases which also decreases the rate of diffusion.

$\bullet$ Option IV. Pressure does not affect the rate of diffusion.

$\bullet$ Option V. The diffusion of the smaller particles is more rapid than the large side molecules. Large molecules require larger surface area due to which the diffuse slowly but and lighter particles can diffuse more quickly because they consume smaller surface area.

So, the correct option is A.

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

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

$10 \; atm$ .

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

$\text{C}$

At what temperature, the sample of neon gas would be heated to double its pressure, if the initial volume of gas is reduced by

$15$ % at

${75}^{o}C$ ?

0%
${319}^{o}C$
0%
${592}^{o}C$
0%
${128}^{o}C$
0%
${60}^{o}C$

Explanation

By the equation,

$PV=nRT$

$n$ and

$R$ are constants.

Hence

${ P }_{ 1 }\times \dfrac { V_{ 1 } }{ { T }_{ 1 } } ={ P }_{ 2 }\times \dfrac { { V }_{ 2 } }{ { T }_{ 2 } }$

${ T }_{ 1 }$ =273+75=348 k ,

${ P }_{ 2 }$ =2

${ P }_{ 1 }$

${ V }_{ 2 }$ =

${ V }_{ 1 }$

$= -15$

$\dfrac { { V }_{ 1 } }{ 100 }$

$=0.85\dfrac { { V }_{ 1 } }{ { T }_{ 2 } }$

substituting the values in the derived equation:

${ P }_{ 1 }\times \dfrac { { V }_{ 1 } }{ 348 } =2{ P }_{ 1 }\times \dfrac { 0.85{ V }_{ 1 } }{ { T }_{ 1 } }$

${ T }_{ 2 }=348\times 1.7=591.6 k$

If in °C then,

$591.6 - 273$ =

$318.6°C$

Hence,318.6

$\approx 319$ is the answer.

$20^{o}C$ , 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).

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,

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

$= 283.18 \times 0.113$

$= 31.999 torr$

${ CuSO }_{ 4 }\cdot 5H_{ 2 }O\left( s \right) \rightleftharpoons { CuSO }_{ 4 }\cdot { 3H }_{ 2 }O\ (s)+{ 2H }_{ 2 }O\ (l){ K }_{ p }=1.086\times { 10 }^{ -4 }{ atm }^{ 2 }\ at\ { 25 }^{ o }C$ 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%
$0.1\ M\ K_4[Fe(CN)_{6}]$

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%
$1 atm$

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 $25^oC$

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

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

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

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

$V=2.1\ L$

The solubility of a specific non-volatile salt is 4g in 100 g of water at

${25}^{0}$ C. If 2.0g, 4.0g and 6.0g of the salt added of 100g of water at

${25}^{0}$ C, in system X, Y and Z. The vapour pressure would be in the order:

0%
$Z>Y>X$
0%
$X>Y>Z$
0%
$Z>X=Y$
0%
$X>Y=Z$

The pressure exerted by a mass of x mg resting on the area of 1.00

${ cm }^{ 2 }$ is 1.00 Pa, then x is

0%
10.6
0%
10.3
0%
103
0%
10.2

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.

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.

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

According to kinetic theory of gases :

0%
collisions are always elastic
0%
between collisions, the molecules move in straight lines with constant velocities
0%
only a small number of molecules have a very high velocity
0%
at of the above

A spherical balloon of

$21\ cm$ diameter is to be filled up with

$H_2$ at

$NTP$ from a cylinder containing the gas at

$20\ atm$ at

$27^{\circ}C$ . The cylinder can hold

$2.82$ litre of water. The number of balloons that can be filled up

0%
11
0%
10
0%
8
0%
1

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

A beaker is filled with a liquid of density

$\rho$ upto a height h if the beaker is at rest, the mean pressure at the walls is :

0%
0
0%
$h \rho g$
0%
$h \rho g/2$
0%
$2 h \rho g$

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%
$52.5\ mm\ Hg$

The reduction potential of hydrogen electrode will be positive if

0%
${P_{{H_2}}} = 2\;atm;\left[ {H^ + } \right] = 1\;M$
0%
${P_{{H_2}}} = 2.5\;atm;\left[ {H^ + } \right] = 1.5\;M$
0%
${P_{{H_2}}} = 2.5\;atm;\left[ {H^ + } \right] = 1\;M$
0%
${P_{{H_2}}} = 1\;atm;\left[ {H^ + } \right] = 2\;M$

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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