MCQ Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 7

A Solution of non-volatile solute in water freezes at

$-0.{ 30 }^{ }C.$ The vapour-pressure of pure water at

$298K$ is

$23.51mm Hg$ and

${ K }_{ 1 }$ for water is

$1.86 K. kg { mol }^{ -1 }$ Calculate the vapour-pressure of this solution at

$298K.$

0%
$23.4mm$
0%
$24.8mm$
0%
$34.8mm$
0%
$40mm$

Explanation

We know that,

$\triangle { T }_{ f }={ K }_{ f }m$

$\therefore \quad m=\cfrac { \triangle { T }_{ f } }{ { K }_{ f } } =\cfrac { 0.30 }{ 1.86 } =0.161$

According to Raooh's law,

${ P }^{ \circ }-P/{ P }^{ \circ }=$ No. of moles of Solute/No of moles of solvent

$\therefore \quad 23.51-P/23.51=\cfrac { 0.161 }{ 1000/18 } =0.0020898$

$\therefore \quad P=23.51-23.51\times 0.0020898$

$\therefore \quad P=23.44mmHg$

Calculate the volume occupied by 4 mole of an ideal gas at

$2.5\times { 10 }^{ 5 }{ Nm }^{ -2 }$ pressure and 300K temperature.

0%
$0.4\ dm^3$
0%
$0.04\ dm^3$
0%
$0.3\ dm^3$
0%
None of these

Explanation

$T=300K$

$P=2.5\times 10^{5}Nm^{-2}$

$n=4 \, moles$

$V=\dfrac{n\times R\times T}{p}$

$=\dfrac{4\times \dfrac{25}{3}\times 300}{2.5\times 10^{5}}$

$=\dfrac{10^{4}}{2.5\times 10^{5}}\Rightarrow 4\times 10^{-2}$

$=0.04 m^{3}$

What would be the vapour pressure of 0.5 molal solution of non volatile solute in benzene at

$30^oC$ ? The vapour pressure of pure benzene at

$30^oC$ is 119.6 torr.

0%
119.6 torr
0%
121.23 torr
0%
118.52 torr
0%
None of these

Explanation

$\dfrac{\triangle p}{p^0}=x_2$

$\dfrac{\triangle p}{p^0}=\dfrac{n_2}{n_1}$

$=\dfrac{m_2\times M_1}{M_2\times m_1}$

$=\dfrac{m\times m_1}{1000}$

$\dfrac{\triangle p}{p^0}=\dfrac{0.5\times 18}{1000}$

$=0.009$

$\dfrac{p^0-p}{p^0}=0.009$

$[m=molarity=\dfrac{m_2}{M_2}\times \dfrac{1000}{m_1}$

$\dfrac{119.6-p}{119.6}=009$

$19.6-p=1.0764$

$\boxed{p=118.52}torr$

A mixture of

$He$ and

$SO_2$ at one bar pressure contains

$20\%$ by weight of

$He$ . Partial pressure of

$He$ be:

0%
0.2 bar
0%
0.4 bar
0%
0.6 bar
0%
0.8 bar

Explanation

Mixture contains 20g $He$ and 80g $S{O}_{4}$

No of moles of $He$ = 20/4 = 5 mol

No of moles of $S{O}_{4}$ = 80/64 = 1.25 mol

Mole Fraction of $He$ = 5/6.25 = 0.8

And Partial Pressure of $He$ = 0.8 x 1 bar

= 0.8 bar

The vapour pressure of pure water is 760 mm at

$25^{ 0 }C$ .The vapour pressure of solution containing 1(m) solution of glucose will be:

0%
761.0 mm
0%
746.5 mm
0%
734.5 atm
0%
746.5 Pa

Explanation

Vapor pressure

$\dfrac{P^{o}- P}{P^{o}}= \dfrac{W_{2}}{M_{2}} \times \dfrac{M_{1}}{W_{1}}$

$P^{o}= 760\ mm$

$M_{1} = 18$

Molality

$(m)= 1m= \dfrac{mole\ of\ solute}{(weight)\ kg\ of\ solvent }= \dfrac{W_{2}/M_{2}}{W_{i} m \ gm}$

$\dfrac{760-P}{760} = 1 \times 10^{-3} \times 18$

$P= 746.5\ mm$

Hence, the correct option is

$B$

For a binary ideal solution, the variation in total vapour pressure versus composition of solution is given by which of the curves?

A balloon weighting

$50 Kg$ is filled with

$685 kg$ of helium at

$1 atm$ pressure and

${ 25 }^{ o }C$ . What will be its pay load if it displaced

$5108 kg$ of air?

0%
$4373 kg$
0%
$4423 kg$
0%
$5793 kg$
0%
None of these

Explanation

The payload can be explained as the difference between the mass of the air displaced and that of the mass of balloon whereas mass of balloon is equal to the sum of the mass of balloon skin and mass of Helium filled in it.

Given that:-

Mass of air displaced

$= 5108 \; Kg$

mass of balloon skin

$= 50 \; Kg$

Mass of helium filled in it

$= 685 \; Kg$

$\therefore \text{ Payload} = 5108 - \left( 50 + 685 \right) = 5108 - 735 = 4373 \; Kg$

Air contains

$79\% \ N_2$ and

$21\% O_2$ by volume. If the pressure is

$750 \ mm$ of

$Hg$ , the partial pressure of oxygen

$(O_2)$ is:

0%
$157.5\ mmHg$
0%
$175.5\ mmHg$
0%
$315.0\ mmHg$
0%
$257.5\ mmHg$

Explanation

Mole Fraction of Oxygen = 0.21

Barometric Pressure = 750 mg

Partial Pressure of oxygen = Mole Fraction x Barometric Pressure

Partial Presuure of Oxygen

$= 0.21 \times 750$

$= 157.5\ mmHg$

In the following graphical representation for reaction

$A \to B$ there are two types of regions:

0%
$I$ and $II$ both represent kinetic region at different interval
0%
$I$ and $II$ both represent equilibrium regions at different time interval
0%
$I$ represents kinetic while $II$ represents equilibrium region
0%
$I$ represents equilibrium while $II$ represents kinetic region

$2.8 g$ of a gas at

$1atm$ and

$273K$ occupies a volume of

$2.24$ litres. The gas can not be:

0%
${ O }_{ 2 }$
0%
$CO$
0%
${ N }_{ 2 }$
0%
$C_{ 2 }{ H }_{ 4 }$

Explanation

Let the molecular weight of gas be M.

Weight of gas

$= 2.8 g$

no. of moles of gas,

$\left( n \right) = \cfrac{\text{wt.}}{\text{mol. wt.}} = \cfrac{2.8}{M}$

Pressure,

$\left( P \right) = 1 atm$

Volume,

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

Temperature,

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

$R = 0.082 \; L \; atm \; {mol}^{-1} \; {K}^{-1}$

From ideal gas equation,

$PV = nRT$

$\Rightarrow \; 1 \times 2.24 = \cfrac{2.8}{M} \times 0.082 \times 273$

$\Rightarrow \; M = \cfrac{62.68}{2.24}$

$\Rightarrow \; M = 27.98 g \approx 28 g$

Hence the molecular mass of gas must be 28 g.

Since molecular mass of

${O}_{2}$ is 32 g, hence the gas cannot be

${O}_{2}$ .

The volume of

$3.7 g$ of a gas at

${ 25 }^{ o }C$ is the same as the volume of

$0.184 g$ of

${ H }_{ 2 }$ at

${ 17 }^{ o }C$ and same pressure. What is the molecular weight of the gas?

0%
$31.33 g/mol$
0%
$41.33 g/mol$
0%
$21.45 g/mol$
0%
$11.45g/mol$

Explanation

From ideal gas equation-

$PV = nRT$

whereas,

$P =$ pressure of the gas

$V =$ volume it occupies

$n =$ number of moles of gas present in the sample

$R =$ universal gas constant

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

$T =$ absolute temperature of the gas

At constant pressure and volume,

$n \propto \cfrac{1}{T}$

$\therefore \; \cfrac{{n}_{1}}{{n}_{2}} = \cfrac{{T}_{2}}{{T}_{1}}$

$\Rightarrow \; {n}_{2} = \cfrac{{n}_{1} \times {T}_{1}}{{T}_{2}}$

Let

${n}_{1}$ and

${n}_{2}$ be the no. of moles of

${H}_{2}$ and the other gas respectively.

Let

${T}_{1}$ and

${T}_{2}$ be the temperature for

${H}_{2}$ and the other gas respectively.

Let the molecular weight of other gas be 'M' g.

Given weight of

${H}_{2} = 0.194 g$

Molecular wt. of

${H}_{2} = 2 g$

Given wt. of other gas

$= 3.7 g$

Molecular weight of other gas

$= M = ?$

As we know that,

$\text{no. of moles} = \cfrac{\text{given wt.}}{\text{mol. wt.}}$

Therefore,

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

${H}_{2} = \cfrac{0.184}{2} = 0.092 \text{ mole}$

${n}_{2} =$ moles of other gas

$= \cfrac{3.7}{M}$

${T}_{1} = 17 ℃ = \left( 17 + 273 \right) K = 290 K$

${T}_{2} = 25 ℃ = \left( 25 + 273 \right) K = 298 K$

$\therefore \; \cfrac{3.7}{M} = \cfrac{0.092 \times 290}{298}$

$\Rightarrow \; M = \cfrac{3.7 \times 298}{0.092 \times 290} = 41.3268 g \approx 41.33 g$

Hence the molecular weight of other gas is 41.33 g.

$CuSO_5H_2O_{(s)} \rightleftharpoons {CuSO_4}_{(aq)}$ ;

$K_p = 4 \times 10^{-4}\ atm^2$

If the vapor pressure of water is 38 torr then the percentage of relative humidity is:

0%
4
0%
10
0%
40
0%
1

The vapour pressure of water at

$20^{o}C$ is

$17.54mm Hg$ . Then the vapour pressure of the water in the apparatus is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (temp. constant) is:

0%
$5.77\ mmHg$
0%
$16\ mmHg$
0%
$35.08\ mmHg$
0%
between $8.77$ and $15.54\ mmHg$

Explanation

We know that,

\dfrac {\text{ Moles of sucrose} }{\text{ Moles of sucrose}+\text{ Moles of water} }

So,

$X_{ sucrose }=\dfrac { \dfrac { 200g }{ 342.13gmol^{ -1 } } }{ \dfrac { 200g }{ 342.13gmol^{ -1 } } +\dfrac { 112.3g }{ 18.01gmol^{ -1 } } } =0.0857$

Mole fraction of water:

$X_{ water }=\dfrac { \dfrac { 112.3g }{ 18.01gmol^{ -1 } } }{ \dfrac { 112.3g }{ 18.01gmol^{ -1 } } +\dfrac { 200g }{ 342.13gmol^{ -1 } } } =0.914$

Note that by definition,

$x_{ sucrose }+x_{ water }=1$

Because sucrose is involatile, the vapour pressure of the solution is proportional to the mole fraction of water:

$P_{ solution }=X_{ water }+17.54\ mm$ of

$Hg$

$=0.914\times 17.54\ mm$ of

$Hg=16\ mm$ of

$Hg$
Conclusion: hence the correct answer is not mentioned in the above options.

A liquid A and B form an ideal solution. If vapour pressure of pure A and B are

$500N{ m }^{ -2 }$ and

$200N{ m }^{ -2 }$ respectively, the vapour pressure of a solution of A and B containing 0.2 mole fraction of A would be:

0%
$700N{ m }^{ -2 }$
0%
$300N{ m }^{ -2 }$
0%
$260N{ m }^{ -2 }$
0%
$140N{ m }^{ -2 }$

Explanation

Vapour pressure of a solution containing A & B liquid is

$P_T=P_A+P_B$

$P_A= X_A-P^o_A$

$X_A$ = mole fraction of A;

$P^o_A$ = Pure vapour pressure of A.

$\Rightarrow P_A= 0.2 \times 500 Nm^{-2}$ &

$P_B= (1-0.2)\times 200 Nm^{-2}$

Thus

$P_T= 0.2 \times 500 Nm^{-2} + 0.8 \times 200 Nm^{-2}$

$\Rightarrow P_T= 260 Nm^{-2}$

Thus total vapour pressure of solution is

$260 Nm^{-2}$ .

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of

$290\ mm$ at

$300\ K$ . The vapour pressure of propyl alcohol is

$200\ mm$ . If the mole fraction of ethyl alcohol is

$0.6$ , its vapour pressure (in

$mm$ ) at the same temperature will be:

0%
$300$
0%
$700$
0%
$360$
0%
$350$

Explanation

By Roult's law,

${ P }_{ T }={ P }_{ A }^{ \circ }{ x }_{ A }+{ P }_{ B }^{ \circ }{ x }_{ B }$

$\therefore \quad 290mm={ P }_{ A }^{ \circ }\times 0.6+200\times \left( 1-0.6 \right)$

$={ P }_{ A }^{ \circ }\times 0.6+80$

$\therefore \quad { P }_{ A }^{ \circ }=350\ mm$

Hence, the correct option is

$D$

Which of the following is not a matter ?

0%
Book
0%
Air
0%
Idea
0%
Cold drink

For the equilibrium ,

${ 2SO }_{ 3 }(g)\rightleftharpoons 2{ SO }_{ 2 }(g)+{ O }_{ 2 }(g)$ ; the partial pressure of

${ SO }_{ 3 }$ ,

${ SO }_{ 2 }$ and

${ O }_{ 2 }$ gases at

$650 K$ are respectively

$0.2 bar,0.6 bar$ and

$0.4 bar$ . If the moles of both oxides of sulphur are so adjusted as equal, what will be the partial pressure of

${ O }_{ 2 }$ ?

0%
$4.4$
0%
$0.44$
0%
$3.6$
0%
$1.12$

Explanation

$2 SO_3 \rightleftharpoons 2SO_2 + O_2$

$K_p = \dfrac{[SO_2]^2 [O_2]}{[SO_3]^2} = \dfrac{(0.6)^2 (0.4)}{(0.2)^2} = 3.6$

$SO_2$ and

$SO_3$ are made equal,

$3.6 = \dfrac{[SO_2]^2}{[SO_3]^2} \times [O_2] \Rightarrow [O_2] = 3.6$

Because moles of

$SO_2 \, \& \, SO_3$ are equal,

$P_{SO_2}, \, \& \, P_{SO_3}$ came each other

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

${ 25 }^{ 0 }C$ . The fraction of total pressure exerted by hydrogen is?

0%
1/2
0%
8/9
0%
16/19
0%
1/9

Explanation

Molar ratio of

$CH_4$ :

$H_2$ = 16 : 2 = 8 : 1 (Mixed in equal quantities)

Pressure is directly proportional to number of moles.

Methane is 8/9 of total number of moles.

So,

Fraction of pressure by

$CH_4$ = 8/9

Therefore B is the correct option.

$3$ moles of

$P$ and

$2$ moles of

$Q$ are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are

$80$ and

$60$ respectively?

0%
$80$ torr
0%
$140$ torr
0%
$720$ torr
0%
$70$ torr

Explanation

Mole fraction of

$P=\dfrac { 3 }{ 3+2 } =\dfrac { 3 }{ 5 }$

Mole fraction of

$Q=\dfrac { 2 }{ 3+2 } =\dfrac { 2 }{ 5 }$

Hence,

Total vapour pressure =(Mole fraction of

$P$

$\times$ Vapour pressure of

$P$ ) + (mole fraction of

$Q$

$\times$ Vapour pressure of

$Q$ )

$=\left( \dfrac { 3 }{ 5 } ×80 \right) +\left( \dfrac { 2 }{ 5 } ×60 \right) =48+24=72\ torr$

A quantity of gas is collected in a graduated tube over the mercury. The volume of gas at

${18}^{o}C$ is

$50.0mL$ and the level of mercury in the tube is

$100mm$ above the outside mercury level The barometer reads

$750$ torr. Hence, volume at STP is approximately:

0%
$22mL$
0%
$40mL$
0%
$20mL$
0%
$44mL$

Explanation

$P_{ 1 }=650torr$

$V_{ 1 }=50ml$

$T_{ 1 }=18C+273=291K$

$K_{ P_{ 2 } }=760Torr$

$V_{ 2 }=?$

$T_{ 2 }=273K$

Ideal gas law

$\dfrac { P_{ 1 }V_{ 1 } }{ T_{ 1 } } =\dfrac { P_{ 2 }V_{ 2 } }{ T_{ 2 } }$

Rearranges to

$\therefore V_{ 2 }=\dfrac { (P_{ 1 }V_{ 1 }T_{ 2 }) }{ (P_{ 2 }T_{ 1 }) }$

$V_{ 2 }=\left( \dfrac { 650)(50)(273) }{ (760)(291) } \right)$

$V_{ 2 }=40.11ml$

Which of the following expression is true regarding gas laws?

(

$w=$ weight,

$m=$ molar mass)

0%
$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 2 } }{ { M }_{ 2 }{ w }_{ 1 } }$
0%
$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 2 }{ w }_{ 1 } }{ { M }_{ 1 }{ w }_{ 2 } }$
0%
$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 1 } }{ { M }_{ 2 }{ w }_{ 2 } }$
0%
$\dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { { M }_{ 1 }{ w }_{ 1 } }{ { M }_{ 2 }{ w }_{ 2 } }$

Explanation

According to ideal gas equation

$PV=nRT$

Where,

$P=$ Pressure

$V=$ Volume

$n=$ No. of mole

$R=$ Gas constant

$T=$ Temperature

$n=\dfrac {w_1}{M_1}$

Where

${w}_{1}=$ weight of the gas

${m}_{1}=$ Molar mass

Taking

$P,\ V$ and

$R$ constant for two gas 1 and 2

${ T }_{ 1 }=\dfrac { { P }{ \times V\times }{ M }_{ 1 } }{ { w }_{ 1 }\times { R }_{ } }$

${ T }_{ 2 }=\dfrac { { P }{ \times V\times }{ M }_{ 2 } }{ { w }_{2 }\times { R }_{ } }$

$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 2 } }{ { M }_{ 2 }{ w }_{ 1 } }$

How many grams of sucrose must be added to

$320g$ of water to lower the vapour pressure by

$1.5mm$

$Hg$ at

${25}^{o}C$ ?
(Given: The vapour pressure of water at

${25}^{o}C$ is

$23.8mm$

$Hg$ and molar mass of sucrose is

$324.3g/mol$ )

0%
$21.5g$
0%
$140g$
0%
$363.36g$
0%
$160.12g$

Explanation

$w_{1}=$ Wsucrose = ?

$M_{1}=$ Molecular weight

$+32,4.3\ g/mol$

$w_{2}=320\ g$

$M_{2}=H_{2}O=18\ g/mol$

$p-ps=1.5\ mm$

$p=23.8\ mm$

$\dfrac{p-ps}{p}=\dfrac{W_{1}}{M_{1}}\times \dfrac{M_{2}}{W_{2}}$

$\dfrac{1.5}{23.8}=\dfrac{W_{1}}{324.3}\times \dfrac{18}{320}$

$W_{1}=363.36\ g$

The vapour pressure of a liquid decreases by

$10 torr$ when a non-volatile solute is dissolved. The mole fraction of the solute in solution is

$0.1$ . What would be the mole fraction of the liquid if the decrease in vapour pressure is

$20torr$ , the same solute being dissolved?

0%
$0.2$
0%
$0.9$
0%
$0.8$
0%
$0.6$

Explanation

Answer is option

$C.$

According to Rault's law,

$\cfrac{P^0-P}{P}=X_a\longrightarrow (1)$ Where

$X_a=\text{Mole fraction of solute.}$

$\underline {\text{Case 1}:}$

$P^0-P=10$

$torr,$

$X_a=0.1$

$\therefore$

$\cfrac{10}{P}=0.1$ or

$P=\cfrac{10}{0.1}=100$

$torr$

$\underline {\text{Case 2}:}$

$P^0-P=20$

$torr,$

$\therefore (1)\longrightarrow \cfrac{20}{100}=X_a\Longrightarrow X_a=0.2$

We have

$X_a+X_b=1,$ where

$X_b=\text{Mole fraction of solvent.}$

$\therefore X_b=1-0.2=0.8$

A gas at a pressure of

$5.0$ bar is heated from

$0^{o}$ to

$546^{o}C$ and is simultaneously compressed to one-third of its original volume. Hence, final pressure is:

0%
$\dfrac {5}{9}$ bar
0%
$15.0$ bar
0%
$30.0$ bar
0%
$45.0$ bar

Explanation

Initial volume of the gas

$={ V }_{ 1 }L$

Initial Temperature of the gas

$={ T }_{ 1 }=(0+273)=273K$

Final pressure of the gas

$={ P }_{ 2 }=?$

Final volume of the gas

$={ V }_{ 2 }L$

Final Temperature of the gas

$={ T }_{ 2 }=(546+273)K=819K$

Given condition is compressed to one third of its original volume.

${ V }_{ 1 }=3{ V }_{ 2 }$

We know from gas laws,

$\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ 819 } =\dfrac { 5\times 3{ V }_{ 2 } }{ 273 }$

${ P }_{ 2 }=\dfrac { 5\times 3\times 819 }{ 273 } bar=45bar$

Which gas does have density

$1,58$ by air?

0%
$NO_2$
0%
$N_2O$
0%
$N_2O_3$
0%
$NO$
0%
$CO_2$

Explanation

$N_2O$ molecular weight is 44

Density of N

$_2$ O at STP =

$\dfrac{44}{22.4}$

Air has 78% of

$N_2$ . Its density =

$\dfrac{28}{22.4}$

Relative density of

$N_2$ O with air is =

$\dfrac{Density \ of\ N_2O}{Density\ of\ N_2\ in \(air)}$

$\dfrac{44/22.4}{28/22.4}=\dfrac{44}{28}$ =1.58

$N_2O$ has density 1.58 that of air.

Two liquids A and B form an ideal solution. The vapour pressure of pure A and pure B are 66mm of Hg and 88mm of Hg, respectively. Calculate the composition of vapour A in the solution which is equilibrium and whose molar volume is 36%.

0%
0.43
0%
0.70
0%
0.30
0%
0.50

Explanation

$P_t= P_A+P_B =66mm+88mm=154mm$ of

$Hg$

Now,

$P_A$

$=X_A \times P_T$

Therefore,

$X_A$ =

$\cfrac{P_A}{P_T}=\cfrac{66}{154}$

$=0.43$

Therefore, composition of Vapour A is

$0.43$ .

Which of the assumptions of the kinetic-molecular theory best explains the observations that a gas can be compressed?

0%
Gas molecules move at random with no attractive forces between them
0%
The velocity of gas molecules is proportional to their kelvin temperature
0%
The amount of space occupied by a gas is much greater than the space occupied by the actual gas molecules
0%
In collisions with the walls of the container or with other molecules, energy is conserved

Which of the following statement is not a postulate of kinetic molecular theory of gases?

0%
$K.E.$ is dependent on temperature
0%
$K.E.$ is dependent on pressure
0%
The molecular collisions are perfectly elastic collisions
0%
The pressure of the gas is due to collisions of gas molecules on the walls of the vessel

Which of the following relationships is valid?

0%
$\dfrac {pV}{T}=\dfrac {2}{3}K$
0%
$\dfrac {pV}{T}=\dfrac {3}{2}K$
0%
$\dfrac {pV}{T}=2K$
0%
$\dfrac {pV}{T}=3K$

Explanation

$pV=\dfrac{1}{3}mN{C}_{2}$ (

$mN=$ molar mass)

For 1 mole of gas pV=RT, n=N(Av. constant)

$RT =\dfrac { 1 }{ 3 }mN{C}_{2}$

$RT =\dfrac{2}{3}\times\dfrac{1}{2}mN{c}_{2}$

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

So,

$pV=\dfrac{2}{3}$ K

$C=$ Speed of molecule

$p=$ Pressure

$V=$ Volume

$R=$ Gas constant

$T=$ Temperature

$n=$ No. of moles

$m=$ Molar mass

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is

$1:4$ . The ratio of the number of their molecule is _______.

0%
$1:4$
0%
$7:32$
0%
$1:8$
0%
$3:16$

Explanation

We know

Moles ratio

$=$ number of molecules ratio

$\dfrac { { n }_{ { O }_{ 2 } } }{ { n }_{ { N }_{ 2 } } } =\dfrac { \dfrac { m_{O_2} }{ { M }_{ { O }_{ 2 } } } }{ \dfrac { { m }_{ { N }_{ 2 } } }{ { M }_{ { N }_{ 2 } } } }$
Where,

${ m }_{ O_{ 2 } }=$ given mass of

${ O }_{ 2 }$

${m }_{ { N }_{ 2 } }=$ given mass of

${N}_{2}$

${M}_{O_{2}}=$ Molecular mass of

${ O }_{ 2 }$

${M}_{N_{2}}=$ molecular mass of

${N}_{2}$

${n }_{ { O }_{ 2 } }=$ Number of moles of

${ O }_{ 2 }$

${n }_{ { N }_{ 2 } }=$ Number of moles of

${N}_{2}$

$=\left[ \dfrac { { m }_{ { O }_{ 2 } } }{ { m }_{ { N }_{ 2 } } } \right] \dfrac { 28 }{ 32 } =\dfrac { 1 }{ 4 } \times \dfrac { 28 }{ 32 } =\dfrac { 7 }{ 32 }$

Therefore, the answer is 7:32

The mass of glucose that should be dissolved in

$100g$ water in order to produce same lowering of vapour pressure as produced by dissolving

$1g$ urea in

$50g$ of water is?

0%
$1g$
0%
$2g$
0%
$6g$
0%
$12g$

Explanation

$\left( { H }_{ 2 }NCON{ H }_{ 2 } \right) \rightarrow M=60g/mol$

$m=$ molar mass

${ M }_{ { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 } }=12\times 6+12\times 1+6\times 6=72+12+96+180 g/mol$

Relative lowering in Vapour pressure

$=no.\quad of\quad moles\times \cfrac { molar\quad mass\quad of\quad solvent }{ mass\quad of\quad water } \\ =0.33\times \cfrac { 18 }{ 50 }$

$\cfrac { W }{ 180 } \times \cfrac { 18 }{ 100 } =\cfrac { 1 }{ 60 } \times \cfrac { 18 }{ 50 } \\ W=6g$

For which of the following binary solution

${P}_{Total}={P}_{A}^{o}\times {X}_{A}+ {P}_{B}^{o}\times {X}_{B}$ is followed?

0%
Phenol+ Aniline
0%
Benzene + Toluene
0%
Acetone + Chloroform
0%
water + $HCl$

In a gaseous mixture at 4 atm pressure, 25% of molecules are Nitrogen, 40% of molecules are carbon dioxide and the rest are oxygen. The partial pressure of oxygen in the mixture is:

0%
1.40 atm
0%
1.6 atm
0%
1 atm
0%
0.9 atm

Explanation

total pressure

$= 4atm$

nitrogen

$= 25\%$

$Co_{2}=40\%$

oxygen

$=35%$

(portal pressure )

$O_{2}=$ total pressure

$\times$ (mole tractra)

$O_{2}$

$4\times \dfrac{35}{100}$

$\dfrac{7}{5}$

$1.40atm$

The vapour pressure is least for?

0%
pure water
0%
0.1m aqueous urea
0%
0.2m aqueous urea
0%
0.3m aqueous urea

Explanation

Vapour pressure decreases as content of any impurity increases in solution.

$\therefore V.P$ will be least for

$0.3m$ aq.urea

$0.157$ g of a certain gas collected over water occupies a volume of

$135ml$ at

${ 27 }^{ 0 }C$ and

$750 mm$ of Hg. Assuming ideal behaviour , the molecular weight of the gas is:

(aqueous tension at

${ 27 }^{ 0 }C$ is

$26.7 mm$ of

$Hg$ )

0%
$30$
0%
$32$
0%
$28$
0%
$16$

Explanation

Pressure of the gas= Total pressure- Aqueous tension

$\Rightarrow 750-26.7=723.3 mm$ of

$Hg$

$\because PV=\cfrac {m}{M}RT$

$\therefore \left(\cfrac {723.3}{760}\right)\times 135 \times 10^{-3}= \cfrac {0.157}{M}\times 0.0821\times 300$

$\therefore M= 30$

${ 20 }^{ 0 }C$ , the vapour pressure of diethyl either is 44mm. When 6.4 g. of a non-volatile solute is dissolved in 50g. of either, the vapour pressure falls to 410mm. The Molecular weight of the solute is:

0%
150
0%
130
0%
160
0%
180

Explanation

$\Delta _{P}=440-410=30mm$

$\therefore \dfrac{\Delta _{P}}{P_{0}}=x_{solute}$

$\Rightarrow \displaystyle \frac{30}{440}=\frac{\frac{6.4}{M}}{\frac{6.4}{M}+\frac{50}{74}}$

$\Rightarrow \displaystyle \frac{19.2}{M}+\frac{150}{74}=\frac{44\times 6.4}{M}$

$\Rightarrow \displaystyle \frac{150}{74}=\frac{6.4(44-3)}{M}$

$\Rightarrow \displaystyle M=\frac{6.4\times 41\times 74}{150}$

$\approx 130$

Equal masses of

${ SO }_{ 2, }{ CH }_{ 4 }$ and

${ CH }_{ 4 }$ are mixed in empty container at 298 K, when total pressure is 2.1 atm. The partial pressure of

$O_2$ in the mixture is:

0%
$0.5$ atm
0%
$0.75$ atm
0%
$1.2$ atm
0%
$0.6$ atm

When a student was given aviscometer, the liquid was sucked with difficulty, the liquid may be:

0%
benzene
0%
toluene
0%
water
0%
glycerine

The gas Z in the above fraction is ?

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

18 g glucose is added to 178.2 g of water the vapour pressure of water for this aqueous solution at

${ 100 }^{ 0 }C$ is:

0%
704 torr
0%
759 torr
0%
7.6 torr
0%
None of these

Explanation

Molar mass of water $=18$ $g$

For $178.2$ $g$ of $H_2$ , $n_A = 9.9$ $X_B= \cfrac {0.1}{0.1 + 9.9} =0.01$

Molar mass of Glucuse $=180$ $g$ $X_A = 0.99$

For $18$ $g$ , $n_B = 0.1$

For lowering of vapour pressure ,

$P = P^o \times A= P^oA(1-X_B) = 760(1-0.01)$

$P = 760-7.6$

$P= 752.4$ $torr$

The density of

$CO_2$ gas at

$27^0 C$ and

$1 \,atm$ pressure is _____ (gram/litre).

0%
$1.78$
0%
$1.52$
0%
$1.96$
0%
$1.20$

Explanation

$PV=nRT$

$\Rightarrow 1 \times V= \cfrac {m}{44} \times 0.0821 \times 300$

$\Rightarrow d= \cfrac {m}{V}$

$\Rightarrow d=\cfrac {44}{0.0821 \times 300}$

density=

$1.78g/Litre$

Which of the following curve does not represent Gay-lussac's law?

A vessel has nitrogen gas and water vapours at a total pressure of 1 atm. The partial pressure of water vapour is 0.4 atm. The content of this vessel are transferred completely to another vessel having one-third of the capacity of the original volume at the same temperature, then the total pressure of this system in the new vessel is:

0%
2.5 atm
0%
3 atm
0%
3.4 atm
0%
4.2 atm

Explanation

It is given that, total pressure of water vapor and nitrogen is

${{P}_{{{N}_{2}}}}+{{P}_{{{H}_{2}}O}}=1\,atm$

$\because {{P}_{{{H}_{2}}O}}=0.4\,atm$

$so,$

${{P}_{{{N}_{2}}}}=0.6\,atm$

$\because {{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$

$0.6\times {{V}_{1}}={{P}_{2}}\dfrac{{{V}_{1}}}{3}$

${{P}_{{{N}_{2}}}}=2.1\,atm$

${{P}_{{{H}_{2}}O}}=0.4\,atm$

${{P}_{T}}=2.5\,atm$

Vapour pressure of

${ CCI }_{ 4 }$ at

${ 24 }^{ o }C$ is

$143 mmHg 0.05$ g of the non-volatile solute (mol.wt.=65) is dissolved in

$100 ml { CCI }_{ 4 }$ . Find the vapour pressure of the solution (Density of

${ CCI }_{ 4 }$ =

$1.58 g/{ cm }^{ 2 }$ )

0%
$143.99 mm$
0%
$94.39 mm$
0%
$199.34 mm$
0%
$14.197 mm$

Explanation

The equation of relative lowering vapour pressure is given as-

$\cfrac{{p}^{0} - p}{{p}^{0}} = {X}_{2}$

Whereas,

${p}^{0} =$ Vapour prssure of solvent

$p =$ Vapour pressure of solution

${X}_{2} =$ Mole fraction of solute

Given that density of

$C{Cl}_{4}$ is

$1.58 \; {g}/{{cm}^{3}}$

As we know that,

$\text{density} = \cfrac{\text{mass}}{\text{volume}}$

$\Rightarrow 1.58 = \cfrac{\text{mass}}{100}$

$\Rightarrow \text{mass of } C{Cl}_{4} = 158 \; g$

Molecular weight of

$C{Cl}_{4} = 154 \; g$

$\therefore$ No. of moles of

$C{Cl}_{4} = \cfrac{158}{154} = 1.026 \text{ mol}$

Given weight of solute

$= 0.05 \; g$

Molecular weight of solute

$= 65 \; g$

$\therefore$ No. of moles of solute

$= \cfrac{0.05}{65} = 0.00077 \text{ mol}$

$\therefore$ Total no. of moles

$= 1.026 + 0.00077 = 1.02677 \text{ mol}$

$\therefore$ Mole fraction of solute

$= \cfrac{0.00077}{1.02677} = 0.00075$

Vapour pressure of solvent

$= 143 \text{ mm of Hg}$

Noe from equation of relative lowering vapour pressure, we have

$\cfrac{143 - p}{143} = 0.00075$

$\Rightarrow 143 - p = 0.107$

$\Rightarrow p = 142.893 \text{ mm of Hg}$

Hence the vapour pressure of solution is

$142.893 \text{ mm of Hg}$ .

The vapour pressure of water at

$20$ is 17.5 mm Hg. If 18 g of glucose

${ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }$ is added to 178.2 g water

$20$ , the vapour pressure of the resulting solution will be:

0%
17.675 mm Hg
0%
15.750 mm Hg
0%
16.500 mm Hg
0%
17.325 mm Hg

Explanation

No. of moles of glucose=

$\cfrac{18}{180}=0.1$

No. of moles of water=

$\cfrac{178.2}{18}=9.9$

Mole fraction of glucose=

$\cfrac{0.1}{4.9+0.1}=\cfrac{0.1}{10}=0.01$

Using Raoult's law, as solute is non-volatile,

${ x }_{ \beta }=\cfrac { { P }_{ A }^{ o }-{ P }_{ A } }{ { P }_{ A }^{ o } }$

$P_{A}^{o}=$ vapour pressure of pure water

$P_{A}=$ vapour pressure of the solution

$x_{\beta}=$ mole fraction of glucose

$\Longrightarrow 0.01=\cfrac { 17.5-{ P }_{ A } }{ 17.5 } \\ \Longrightarrow { P }_{ A }=17.5-0.175=17.325$

The vapor pressure of the mixture=17.325 mm Hg.

Hence, the correct option is

$D$

The vapour pressure of o-nitrophenol at any given temperature is predicted to be:

0%
Higher than that of p-nitrophenol
0%
Lower than that of p-nitrophenol
0%
Same as that of p-nitrophenol
0%
Higher or lower depending upon the size of the vessel

The surface tension of the liquid at its boiling point

0%
becomes zero
0%
becomes infinity
0%
is equal to the value at room temperature
0%
is half to the value at room temperature

Four one litre flasks are separately filled with gases

$O_2, F_2, CH_4$ and

$CO_2$ under same conditions.

The ratio of the number of molecules in these gases are:

0%
$2 : 2 : 4 : 3$
0%
$1 : 1 : 1 : 1$
0%
$1 : 2 : 3 : 4$
0%
$2 : 2 : 3 : 4$

Explanation

Avogadro's law: It states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

$\\ n_{1} : n_{2} : n_{3} : n_{4} = \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} = 1 : 1 : 1 : 1$

So, the ratio of no. of molecules

$= 1: 1: 1: 1$

The correct option is

$B.$

The vapour pressure of pure liquid is

$70$ Torr at

$27^o C$ . The vapour pressure of a solution of this liquid and another liquid (mole fraction 0.2) is

$84$ Torr at

$27^o C$ . The vapour pressure of pure liquid

$B$ at

$27^o C$ is?

0%
$140$ Torr
0%
$280$ Torr
0%
$160$ Torr
0%
$200$ Torr

Explanation

Vapour Pressure of pure liquid

$A=70$ torr (

$P_{A}^{o}$ )

Mole fraction of liquid

$B=0.2$ (

$x_{B}$ )

Thus mole fraction of liquid

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

Total vapour pressure

$=84$ torr

$P=x_{A} P_{A}^{o} + x_{B} P_{B}^{o}$

$84=(0.8) \times 70 + 0.2 \times P_{B}^{o}$

$P_{B}^{o}$

$=140$ torr

$=$ vapour pressure of the pure liquid

$B$

Hence, the correct option is

$\text{A}$

Which of the liquids in each of the following pairs has a higher vapour pressure?

0%
Alcohol, glycerine
0%
Mercury, water
0%
petrol, kerosene
0%
None of these

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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