MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 6

Select correct statement(s) :

0%
Kinetic energy is zero at ${ O }^{ 0 }$ C
0%
RMS velocity of ${ O }_{ 2 }$ at ${ 27 }^{ 0 }$ C is = $\sqrt { \frac { 3\times 8.314\times 300 }{ 32 } } { ms }^{ -1 }$
0%
Distribution of molecules is very small when $u\rightarrow O\quad or\quad u\rightarrow \infty$
0%
All the statement are correct

Explanation

We know that RMS velocity of gas is given by,

$V_{RMS}=\sqrt{\dfrac{3RT}{M}}$

where, R-Universal gas constant

T-temperature of gas in Kelvin

M-molecular weight of gas

Here, we know

$M_{O_2}=32$

and

$T=273+27=300K$

So,

$V_{RMS}=\sqrt{\dfrac{3\times8.314\times 300}{32}}$ at

$T=27^{0}C$

Read the Paragraph carefully and answer the following questions:

100 gof an ideal $\left( {mol.wt. = 40} \right)$ gas is present in a cylinder at ${27^0}c\;and\,2atm$ pressure. During transportation, cylinder fell and a dent was developed in the cylinder. The valve attached to cylinder cannot keep the pressure greater than $2atm$ and therefore $'10g'$ of gas leaked-out through cylinder.

The volume of the cylinder before a dent was:

0%
$10.79litre$
0%
$20.79litre$
0%
$30.79litre$
0%
$40.79litre$

Explanation

Initial volume can be calculated:

$PV=nRT$

$V_1=(\dfrac{100}{40})(\dfrac{0.0821 \times 300}{2}) = 30.79 dm^3$

The vapour pressure of two pure liquids

$(A)$ and

$(B)$ are

$100$ and

$80\ torr$ respectively. The total vapour pressure of solution obtained by mixing

$2$ moles of

$(A)$ and

$3$ moles of

$(B)$ would be :-

0%
$120\ torr$
0%
$36\ torr$
0%
$88\ torr$
0%
$180\ torr$

Explanation

Total vapour pressure

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

${ P }_{ A }=100\times \dfrac { 2 }{ 2+3 } =100\times \dfrac { 2 }{ 5 } =40$ torr

${ P }_{ B }=80\times \dfrac { 3 }{ 2+3 } =80\times \dfrac { 3 }{ 5 } =48$ torr

${ P }_{ Total }={ P }_{ A }+{ P }_{ B }=40+48=88$ torr.

A closed bulb capacity 200ml containing

${ CH }_{ 4 }$ ,

${ H }_{ 2 }$ and He at 300K. The ratio of partial pressure of

${ CH }_{ 4 }$ ,

${ H }_{ 2 }$ and He, respectively is 2:3:Calculate the ratio of their weights present in the container.

0%
2:4:6
0%
16:8:100
0%
16:3:10
0%
4:9:25

Explanation

Partial pressure

$=$ mole fraction

$\times$ total pressure

$\therefore$ Mass are also in

$2:3:5$

$\therefore$ Ratio of mass

$=2\times 16:3\times 2:5\times 4$

$=16:3:10$

Ideal gas equation is also called equation of states because:

0%
it depends on states of matter
0%
it is relation between four variables and describes the state of any gas
0%
it is combination of various gas laws and any variable can be calculated
0%
none of these

Explanation

Assuming an equilibrium state, the three properties needed to completely define the state of a system are pressure (

$P$ ), volume (

$V$ ), and temperature (

$T$ ). Hence, the definition: An Equation of State (EOS) is a semi-empirical functional relationship between pressure, volume and temperature of a pure substance. Ideal gas equation is also called equation of states because it is relation between four variables and describes the state of any gas. Hence option

$B$ is correct answer.

The vapour pressure of two liquids

$P$ and

$Q$ are

$80$ torr and

$60$ torr respectively. The total vapour pressure obtained by mixing

$3$ mole of P and

$2$ mole of

$Q$ would be:

0%
$68$ torr
0%
$20$ torr
0%
$140$ torr
0%
$72$ torr

Explanation

Mole fraction of

$P(X_P)=\cfrac {3}{3+2}={3}{5}$

Mole fraction of

$Q(X_Q)=\cfrac {2}{3+2}=\cfrac {2}{5}$

Now, Total vapour pressure

$(P_T)=X_PP_P+X_QP_Q$

$=\cfrac {3}{5}\times 80+\cfrac {2}{5}\times 60$

$\therefore$ Total vapour pressure

$(P_T)=72 torr$

Volume of a cylinder containing

$10$ marbles and gas is

$1$ litre at

$2$ atm pressure. Now pressure on the cylinder is increased to

$4$ atm, at which volume becomes

$725$ ml. Calculate volume of each marble.

0%
$60 \ ml$
0%
$45 \ ml$
0%
$57\ ml$
0%
$39 \ ml$

Explanation

Let us assume that the volume of each marble be

$v$ ml. Let the gas behave as an ideal gas. Assume that the compression is done as an isothermal process.

$P_1 = 2 atm$

Initial volume of the gas

$= V_1 = 1000 - 10v$ ml

$P_2 = 4 atm$

$V_2 = 725 - 10 v$ ml

Using Gas equation:

$P_1 V_1 = P_2 V_2$

$\implies 2 (100 - 10v) = 4 (725 - 10v)$

$\implies 20 v = 900$

$\implies v = 45$ ml

The vapour pressure of a pure liquid A is

$70$ torr at

$27^o$ C. 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

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

$27^o$ C.

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

Explanation

Here we will use the simple total vapour pressure equation:

$P_T=x_AP_A^o+x_BP_B^o$

$\Rightarrow 84\ torr= 0.8 \times 70\ torr+ 0.2 \times P_B^o$

$\Rightarrow P_B^o=140\ torr$ .

One mole of

${ N }_{ 2 }{ O }_{ 4 }$ at 300 K is kept in a closed container under 1 atm pressure. It is heated to 600 K when 20% by mass of

${ N }_{ 2 }{ O }_{ 4 }$ decomposes to

$'{ NO }_{ 2 }'\left( g \right)$ . The resultant pressure is:

0%
1.2 atm
0%
2.4 atm
0%
2 atm
0%
1 atm

Explanation

$20\%$ of

$1=\dfrac{20}{100}\times 1=0.2$

$N_2O_4\rightleftharpoons 2NO_2$

$1$

$0$

$-0.2$

$0.2\times 2$

$\overline { \quad 0.8 \quad \quad \quad \quad0.4 \,\,\,\,\,\,\,\,\,\,}$

$300K$ ideal gas

$eq^n$

$PV=nRT$

$1\times V=1.0\times R\times 300$ .....(1)

$600K=PV=nRT$

$P_1\times V=(0.8+0.4)\times R\times 600$ .......(2)

Divide

$eq^n$ by

$1$

$P_1\dfrac{1.2\times 600}{1\times 300}$

$2.5atm$

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

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

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

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

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

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

$=0.009$

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

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

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

$19.6-p=1.0764$

$\boxed{p=118.52}torr$

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

$T=300K$

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

$n=4 \, moles$

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

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

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

$=0.04 m^{3}$

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

$B$

Equal masses of

$SO_2, CH_4$ and

$O_2$ are mixed in empty container at

$298$ K, where total pressure is

$2.1$ atm. The partial pressures of

$CH_4$ in the mixture is:

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

Explanation

Partial pressure is directly proportional to the mole fraction of the gas

Partial pressure

$=$ Total pressure

$\times$ Mole fraction

Let the mass of each constituent

$= x$

Moles of

$SO_2$

$=$

$\frac{x}{64}$

Moles of

$CH_4$

$=$

$\frac{x}{16}$

Moles of

$O_2$

$=$

$\frac{x}{32}$

Total Moles

$=$

$\frac{7x}{64}$

Partial pressure

$=mole\ fraction \times total\ pressure =$

$\frac{\frac{x}{16}}{\frac{7x}{64}}$ x21

$= 1.2\ atm$

Hence partial pressure of

$CH_4= 1.2\ atm$

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

$= 685 \; Kg$

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

The Total pressure in a mixture of

$8g$ of dioxygen and

$4g$ of dihydrogen confined in a vessel of

$1dm^34$ at

$27^oC$ is ?

0%
$R = 83bar\ dm^3k^{-1}mol^{-1}$
0%
$R = 93bar\ dm^3k^{-1}mol^{-1}$
0%
$R = 56bar\ dm^3k^{-1}mol^{-1}$
0%
$R = 63bar\ dm^3k^{-1}mol^{-1}$

Explanation

Mass of oxygen=

$8g= 8/32=0.25 mol$

Mass of hydrogen=

$4g=4/2=2 mol$

From

$PV=nRT$

$\Rightarrow P \times 1=(0.25+2)0.083 \times 300= 56.02$ bar

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

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

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

$\therefore \quad P=23.44mmHg$

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

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.

Which of the following gases has the highest density under standard conditions?

0%
$CO$
0%
$N_2O$
0%
$C_3H_8$
0%
$SO_2$

Explanation

Under Standard conditions The Compound with Highest Molecular has Highest Density (PM =

$\rho$ RT)

$S{O}_{2}$ has high molecular weight so it has the highest density

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

$= 0.21 \times 750$

$= 157.5\ mmHg$

The measured density at

$NTP$ of a gaseous sample of a compound was found to be

$1.78\ g/L$ . What is the weight of

$1\ mole$ of the gaseous sample?

0%
$45.6\ g$
0%
$39.9\ g$
0%
$28.25\ g$
0%
None of these

Explanation

$1.78g$ of gas

$\rightarrow$ 1lt at

$NTP\\$

$xgram\rightarrow 39.9g\\$

$x=1.789\times 39.9\\$

$x=71g$

so it is iron sulphate,which is having

$70g$ of molecular weight.
conclusion: hence the answer is option(D) None of these.

$2.8 g$ 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

${O}_{2}$ .

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

$D$

If P is the pressure of gas, then the kinetic energy per unit volume of the gas is:

0%
$P/2$
0%
$P$
0%
$3P/2$
0%
$2P$

Explanation

$K.E = \dfrac{3}{2}nRT$

From the Ideal gas law equation,

$PV = nRT$

On substituting the values, we get-

${K.E} =\dfrac{3PV}{2}$

Kinetic energy per unit volume,

$\dfrac{K.E.}{V} = \dfrac {3P}{2}$

Hence, option

$C$ is correct.

$3.00\ L$ oxygen gas is collected over water at

${27}^{o}C$ when the barometric pressure is

$787\ Torr$ . If vapour pressure of water is

$27\ Torr$ at

${27}^{o}C$ , moles of

$O_{2}$ gas will be?

0%
$0.122$
0%
$0.126$
0%
$0.004$
0%
$0.152$

Explanation

Given Volume (V)

$=3.00$ L

Temperature(T)

$=27^{0}C=300 K$

Barometric pressure

$=787$ torr

Vapour pressure of water

$=27$ torr

total pressure

$=787-27=760\quad{torr}=1\quad{atm}$

Applying in ideal gas equation

PV=nRT

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

$n=\dfrac{{1}\times{3}}{{0.0821}\times{300}}=0.122$

A baloon blown up with

$1$ mole of gas has a volume of

$480 mL$ at

$5^oC.$ the balloon is field to

$(7/8)th$ of its maximum capacity suggest. The minimum temperature at which it will brust?

0%
$31.4^oC$
0%
$44.7^o C$
0%
$49.1^o C$
0%
$37.6^o C$

Explanation

Given,

$n_1=1$ mole

$R= 0.0821$

$T_1=5^oC= 278K$

$V_1=480ml= 0.48L$

$P_1= \cfrac {n_1RT_1}{V_1}=\cfrac {1\times 0.0821 \times 278}{0.48}$

$\Rightarrow P_1= 47.55 atm$

Now, as the given volume i.e.

$480ml$ is

$7/8^{th}$ of the total capacity.

Thus total volume will be

$\Rightarrow 480ml \times \cfrac {8}{7}= 548.57 ml$

The new temperature at which the balloon can burst will be at highest volume of balloon,

$P_1V_2=n_1RT_2$

$\Rightarrow 47.55 atm \times 548.57 ml= 1\times 0.0821 \times T_2$

$\Rightarrow T_2= 317.7 K$

$\Rightarrow T_2= 44.7^oC$

This is the minimum temperature at which balloon burst.

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.

The surface of a lake is at

$2^{o}C$ and depth of the lake is

$20\ m$ . Find the temperature of the bottom of the lake?

0%
$552^{o}C$
0%
$42^{o}C$
0%
$4^{o}C$
0%
$825^{o}C$

$4.40\ g$ piece of solid

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

$1.00\ L$ at

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

0%
$0.122$
0%
$2.46$
0%
$1.22$
0%
$24.6$

Explanation

$moles\quad of\quad { CO }_{ 2 }=\frac { 4.40 }{ 44 } =0.1$

$P=\frac { 0.1\times 0.0821\times 300 }{ 1 }$

$=2.46V$

conclusion: hence the answer option (B) is correct.

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.

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

$P, V, M, T$ and

$R$ are pressure, volume, molar mass, temperature and gas constant respectively, then for an ideal gas, the density is given by:

0%
$\dfrac{RT}{PM}$
0%
$\dfrac{P}{RT}$
0%
$\dfrac{M}{V}$
0%
$\dfrac{PM}{RT}$

Explanation

For ideal gas

PV=RT

V=RT/P

Density is given by

D=M/V

Where M is mass and V is volume

Putting the value of V

$\frac { M }{ RT/P }$

$\frac { MP }{ RT }$

The volume of

$2.8\ g$ of carbon monoxide at

$27^{o}C$ and

$0.821 atm$ pressure is:

0%
$0.03\ L$
0%
$3\ L$
0%
$0.3\ L$
0%
$1.5\ L$

Explanation

The volume of carbon monoxide=

$\dfrac { nRT }{ P }$
Mass of carbon\quad monoxide

$=2.8g$

$P=0.821\ atm,$

$T=27+273=300K$

$R=0.0821$
because gram molecular weight of

$CO=12+16=28$
because number of moles in 2.8g of

$CO=\dfrac { 2.8 }{ 28 } =0.1$
therefore volume

$=\dfrac { 0.1\times 0.0821\times 300 }{ 0.821 } =3\ litre$

$3$ 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

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

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

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

$4$ moles

$CO, 5$ mole

$N_{2}$ and

$2$ mole

$C_{2}H_{4}$ are placed in a

$5$ litre at

$27^{o}C$ . The ratio of kinetic energy per molecular of

$CO_{2}$ ,

$He$ and

$NH_{3}$ is:

0%
$4:5:2$
0%
$5:5:4$
0%
$2:4:5$
0%
$1:1:1$

Explanation

Kinetic energy

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

Where n, R and T is number of moles, gas constant and temperature respectively

For 4 moles

$CO$

$=\dfrac{3}{2}\times{4}RT$

For 5 Mole of

${N}_{2}$

$=\dfrac{3}{2}\times{5}RT$

For 2 mole of

${C}_{2}{H}_{4}$

$=\dfrac{3}{2}\times{2}RT$

$CO:{N}_{2}:{C}_{2}{H}_{4}=4:5:2$

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 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

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

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 + $HCl$

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

$+32,4.3\ g/mol$

$w_{2}=320\ g$

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

$p-ps=1.5\ mm$

$p=23.8\ mm$

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

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

$W_{1}=363.36\ g$

$0.157$ 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

$Hg$

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

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

$\therefore M= 30$

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

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

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

$\therefore X_b=1-0.2=0.8$

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

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 6 - MCQExams.com

0:0:2

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 6 - MCQExams.com

0:0:2

Practice Class 11 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.