MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 8

The boiling point of

$C_{6}H_{6},CH_{3}OH, C_{6}H_{5}NH_{2}$ and

$C_{6}H_{5}NO_{2}$ are

$80^{\circ},\ 65 ^{\circ},\ 184^{\circ}$ and

$212^{\circ}$ respectively. Which will show highest vapourr pressure at room temperature?

0%
$C_{6}H_{6}$
0%
$CH_{3}OH$
0%
$C_{6}H_{5}NH_{2}$
0%
$C_{6}H_{5}NO_{2}$

Explanation

Boiling point is the temperature at which the vapor pressure of a liquid become equal to atmospheric pressure.

$\rightarrow$ So, the compounds which has low boiling point will have high vapor pressure.

$\rightarrow$ Given

${ CH }_{ 3 }OH$ has boiling point of

${ 65 }^{ 0 }$ .

$\rightarrow$ It has highest vapor among given with

$0.166$ atm.

Ans :- Option B.

Kinetic energy of gas depends on:

0%
temperature
0%
pressure
0%
surface area
0%
both $A$ and $B$

Explanation

We know that

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

Here

$T$ denotes temperature and this energy doesn't depend on any other parameter.

Ans :- Option A.

$127^0C$ and 1 atm pressure, a mixture of a gas contains 0.3 mole of

$N_2$ and 0.2 mole of

$O_2$ . The volume of the mixture is?

0%
15 L
0%
18.2 L
0%
16.4 L
0%
22.4 L

Explanation

Let

$V$ be volume of mixture.

Partial pressure of

$O_2=X_{O_2}\times P_{total}=\cfrac{n_{O_2}RT}{V}$

$\therefore V=\cfrac{n_{O_2}RT}{X_{O_2}\times P_{total}}=\cfrac{0.2\times 0.0821\times 400}{\left(\cfrac{0.2}{0.5}\right)\times 1}=16.4L$

Dry ice is solid carbon dioxide. A

$0.05\ g$ sample of dry ice is placed in an evacuated

$4.6$ L vessel at

$30^0$ C. Calculate the pressure inside the vessel after all the dry ice has been converted to

$CO_2$ gas.

0%
$6.14$ atm
0%
$0.614$ atm
0%
$0.0614$ atm
0%
$6.14 \times 10^{-3}$ atm

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

$290mm$ at

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

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

$0.6$ its vapour presssure at same temperature will be:-

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

Explanation

$P_{sol} = X_{sol} . P^o_{sol}$

$X \rightarrow$ mole fraction

$P^o \rightarrow$ pure vapour pressure of solution

$290 mm =X_{ethanal} . P^o_{ethanal} + X_{propanal} . P^o_{propanal}$

$290 mm =0.6 \times P^o_{ethanal} + (1-0.6) \times 200 mm$
on solving

$P^o_{ethanal} =350$

$nm$

The value of the molar gas constant is:

0%
$8.3145\times 10^3 \ J\ mol^{-1}K^{-1}$
0%
$1.987 \ cal\ mol^{-1} K^{-1}$
0%
$0.083145\times 10^3dm^3 bar\ mol^{-1}K^{-1}$
0%
$0.983145\ dm^3\ bar\ mol^{-1}K^{-1}$

Explanation

The gas constant is also known as the molar, universal, or ideal gas constant, denoted by the symbol R and is equivalent to the Boltzmann constant, but expressed in units of energy per temperature increment per mole.

Its value is

$8.31446J.K^{−1}.mol^{−1}$

We know 1 joule = 4.182 cal

so its value becomes

$8.31446 \times 4.182 cal.K^{−1}.mol^{−1}$ =

$1.987 cal.mol^{−1}.K^{−1}$

The amount of solute (mol mass=

$60$ ) that must be added to

$180g$ of water so that the vapour pressure of water is lowered by

$10 \%$ is?

0%
$30$ g
0%
$60$ g
0%
$120$ g
0%
$12$ g

The vapour pressure of a pure liquid A is 70 torr at 300 K. It forms an ideal solution with another liquid B. The mole fraction of B in the solution is 0.2 and total pressure of solution is 84 torr at 300 K. The vapour pressure of pure liquid B at 27C is ?

0%
0.14 torr
0%
560 torr
0%
140 torr
0%
70 torr

Explanation

Solution:- (C)

$140 \text{ torr.}$

Mole fraction of

$B = 0.2$

As we know that,

${x}_{A} + {x}_{B} = 1$

$\Rightarrow {x}_{A} = 1 - 0.2 = 0.8$

${{p}^{°}}_{A} = 70 \text{ torr.}$

${{p}^{°}}_{B} = ?$

${p}_{total} = 84 \text{ torr.}$

From Dalton's law of partial pressure,

${p}_{total} = {p}_{A} + {p}_{B} ..... \left( 1 \right)$

From Raoult's law of vapour pressure,

${p}_{A} = {{p}^{°}}_{A} \cdot {x}_{A}$

$\Rightarrow {p}_{A} = 70 \times 0.8 = 56$

Again, from Raoult's law of vapour pressure,

${p}_{B} = {{p}^{°}}_{B} \cdot {x}_{B}$

${p}_{B} = {{p}^{°}}_{B} \times 0.2$

Substituting these values in equation

$\left( 1 \right)$ , we have

$84 = 56 + 0.2 \times {{p}^{°}}_{B}$

$\Rightarrow {{p}^{°}}_{B} = \cfrac{84 - 56}{0.2} = 140 \text{ torr.}$

Hence the vapour pressure of pure liquid

$B$ is

$140 \text{ torr.}$ .

How much pressure will be felt by a gas ballon which at 100 m depth in a sea form sea level?

0%
$1081320$ Pa
0%
$98$ Pa
0%
$9.8 \times 10^5$ Pa
0%
$1.58 \times 10^6$ Pa

Explanation

$P=P_O+\rho_gh$

$P_O=$ atmospheric pressure

$=1.013 \times 10^5 Pa$

$\rho=$ density of

$H_2O= 1000 kg/m^3, g=gravity= 9.8 m/s^2$

$P=$ pressure experienced by gas balloon

$h=$ depth of the balloon from sea level=

$100m$

$P=(1.0132 \times 10^5)+(1000)(9.8)(100)$

$P=1081320 Pa$

So, the correct option is

$A$

One mole of non volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is:

0%
$\dfrac{2}{3}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{1}$

Explanation

Relative lowering in vapor pressure

$=\cfrac{P^{\circ}-P_s}{P^{\circ}}=x_A$

Where

$x_A=$ mole fraction of non-volatile solute.

$\therefore \cfrac{P^{\circ}-P_s}{P^{\circ}}=\cfrac{1}{3}\\ \therefore \cfrac{P_s}{P^{\circ}}=1-\cfrac{1}{3}=\cfrac{2}{3}$

Hence, the correct option is

$A$

Which is not true in case of an ideal gas?

0%
It cannot be converted into a liquid
0%
There is no interaction between the molecules
0%
All molecules of the gas move with same speed
0%
At a given temperature, $PV$ is proportional to the amount of the gas

The no. of moles per liter in the equation

$PV=nRT$ is expressed by:

0%
$\dfrac { P }{ RT }$
0%
$\dfrac { PV }{ RT }$
0%
$\dfrac { RT }{ PV }$
0%
none of these

Explanation

We know,

$PV=nRT$

$\Rightarrow P= \cfrac {n}{V}RT$

$\Rightarrow n/V=P/RT$

$\Rightarrow \cfrac {n}{V}=\cfrac {\text{Number of moles}}{\text{Volume}}=\cfrac {P}{RT}$

Two liquids, A and B, form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mm Hg while that of pure B is 75 mm Hg. If the vapour over the mixture consists of 50 mol percent A, what is the mole percent A in the liquid?

0%
20
0%
76
0%
27
0%
65

The relative decrease in vapour pressure of an aqua solution containing 2 moles of

$4NaCl$ in 3 moles of

${H}2_{O}$ is 0.On reaction with

$AgNO_3$ this solution will form:

0%
1 mol of $AgCl$
0%
0.25 mol of $AgCl$
0%
2 mol of $AgCl$
0%
0.4 mol of $AgCl$

A gas volume 100 cc. is kept in a vessel at pressure

$10^4$ atm maintained at temperature

$24^0C$ . If now the pressure is increased to

$10^5$ atm, keeping the temperature constant, then the volume of the gas becomes:

0%
10 cc
0%
100 cc
0%
1 cc
0%
1000 cc

Explanation

$P_1V_1=P_2V_2$

$\Rightarrow 10^4\times 100= 10^5 \times V_2$

$\Rightarrow V_2=\cfrac {10^4 \times 100}{10^5}$

$\Rightarrow V_2= 10 cc$

The vapour pressure of water is

$12.3k \, Pa$ at

$300K$ . Calculate the vapour pressure

$1$ molal solution of a non-volatile solute in it.

0%
$12.08 \ kPa$
0%
$1.208 \ Ppa$
0%
$2.4 \ kPa$
0%
$0.4 kPa$

Explanation

$1$ molal solution means 1 mol of the solute is present in

$1000 g$ of the solvent (water).

The molar mass of water

$= 18 g$

$mol^{- 1}$

Number of moles present in

$1000$ g of water

$= 1000/18$

$= 55.56 \ mol$

Therefore, the mole fraction of the solute in the solution is

$x_2 = 1 / (1+55.56)$

$= 0.0177$

It is given that,

Vapour pressure of water,

$P^0_1$

$= 12.3 kPa$

Applying the relation,

$(P^0_1 - P_1) / P^0_1 = X_2$

$= (12.3 - P_1) / 12.3$

$=0.0177$

$= 12.3 - P_1$

$= 0.2177$

$= P_1$

$= 12.0823$

$= 12.08 kPa$

A glass tube of volume 112ml containing a gas is partially evacuated till the pressure in it drops to

$3.8\times { 10 }^{ -5 }$ torr at 0 degree C.The number of molecules of the gas remaining in the tube?

0%
$3\times { 10 }^{ 17 }$
0%
$1.5\times { 10 }^{ 14 }$
0%
$112.5\times { 10 }^{ 20}$
0%
Name of the gas is required

Explanation

volume = 0.112 L

pressure = 3.75 atm

temp = 273K

gas law is PV=nRT

$n=\dfrac{3.75\times 0.112}{0.0821\times 273}$

n=0.01874 moles

A mole has

$6.022\times 10^{23}$ atoms or molecules of the pure substance being measured

no. of molecules

$= 0.01874 \times 6.022\times 10^{23} = 112.5\times 10^{20}$

Hence, the correct option is

$\text{C}$

A container contains equal mass of

$C{ O }_{ 2 },{ O }_{ 2 },C{ H }_{ 4 },S{ O }_{ 2 }$ gases at a particular temperature.Which gas will exert the maximum amount of partial pressure?

0%
$C{ O }_{ 2 }$
0%
${ O }_{ 2 }$
0%
$C{ H }_{ 4 }$
0%
$S{ O }_{ 2 }$

Explanation

Molecular weight of $C{{O}_{2}}=12+16+16=44$

Molecular weight of ${{O}_{2}}=16+16=32$

Molecular weight of $C{{H}_{4}}=12+4=16$

Molecular weight of $S{{O}_{2}}=32+16+16=64$

So;

$S{{O}_{2}}$ has maximum Molecular weight.

Hence $S{{O}_{2}}$ exert maximum amount of partial pressure.

The kinetic energy for 14 grams of nitrogen gas at

$127^0C$ is nearly (mol. mass of nitrogen = 28 and gas constant = 8.31

$JK^{-1} mol^{-1}$ )

0%
$1230\ J$
0%
$4015\ J$
0%
$2493\ J$
0%
$990\ J$

Explanation

$KE=\cfrac {3}{2}nRT$

$n=\cfrac {14}{28}=0.5$

$=\cfrac {3}{2}\times 0.5 \times 8.314 \times (127+273)$

$=2494.2 J$

If number of molecules of

$H_{2}$ are double than that of

$O_{2}$ , then ratio of mean kinetic energy per molecule of hydrogen to that of oxygen at 300 K is:

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

Explanation

Mean Kinetic energy is given as,

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

$n=$ no. of molecules.

$R=$ Constant.

$T=$ Temperature.

For

$H_2$ ,

$K.E_{H_2}=\cfrac{3}{2}n_{H_2}RT\longrightarrow (1)$

For

$O_2$ ,

$K.E_{O_2}=\cfrac{3}{2}n_{O_2}RT\longrightarrow (1)$

On dividing equation

$(1)$ by equation

$(2)$ we get,

$\cfrac{K.E_{H_2}}{K.E_{O_2}}=\cfrac{n_{H_2}}{n_{O_2}}$

given

$n_{H_2}=2n_{O_2}$ ,

$\cfrac{K.E_{H_2}}{K.E_{O_2}}=\cfrac{n_{2O_2}}{n_{O_2}}=\cfrac{2}{1}$

Thus the ration is

$2:1.$

Two liquids A and B have

${P}_{A}^{o}$ and

${P}_{B}^{o}$ in the ratio

$1:3$ and the ratio of number of moles of

$A$ and

$B$ in liquid phase are

$1:3$ . Then mole fraction of A in vapour phase in equilibrium with the solution is equal to:

0%
$0.1$
0%
$0.2$
0%
$0.5$
0%
$1.0$

Explanation

Partial pressure of

$A=K\times\cfrac{1}{4},$ where

$K$ is ratio constant.)

Partial pressure of

$B=3K\times \cfrac{3}{4}$

Total pressure is

$\cfrac{10K}{4}$ .

Vapor phase mole fraction of

$A$ is,

$\cfrac{\text{Partial pressure of }A}{\text{Total pressure}}=\cfrac{K\times \cfrac{1}{4}}{\cfrac{10K}{4}}=\cfrac{1}{10}=0.1$

An amount of 1.00 g of a gaseous compound of borbon and hydrogen occupies 0.820 liter at 1.00 atm and at

$3^{0}C$ . The compound is (R=0.0820 liter atm

$mole^{-10}K^{-1};$ at. wt:H=1.0,B=10.8)

0%
$BH_{3}$
0%
$B_{4}H_{10}$
0%
$B_{2}H_{6}$
0%
$B_{3}H_{12}$

Explanation

Given,

volume - $0.820\ l$

mass - $1.00\ g$

pressure - $1.00\ atm$

temperature - ${3^{\rm O}}C$

Using ideal gas equation - $PV = nRT$

$n$ - number of mole = $\dfrac{m}{M}$

$m$ - given mass

$M$ - molar mass

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

here, find the molar mass of the compound

$M = \dfrac{{mRT}}{{PV}}$

Firstly, change the temperature in Kelvin.

${3^{\rm O}}C + 273 = 276K$

Now, put the value in the formula.

$M = \dfrac{{1 \times 0.0820 \times 276}}{{1 \times 0.820}}$

= $27.6\ g/mole$

write each compound molar mass according to the given value and match it with this value.

(a) $B{H_3}$ - $10.8 + 1 \times 3 = 13.8$

(b) ${B_4}{H_{10}}$ - $4 \times 10.8 + 1 \times 10 = 53.2$

(d) ${B_3}{H_{12}}$ - $3 \times 10.8 + 12 \times 1 = 44.4$

So, option (c) is the correct answer.

The gas Z in the above fraction is ?

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

A gas occupies

$600ml$ at

${27}^{o}C$ and

$730mm$ pressure. What would be its volume at STP?

0%
$0.5244$ lit.
0%
$1.5244$ lit.
0%
$2.5244$ lit
0%
$3.5244$ lit

Explanation

$V_1=600mL\\T_1=27°C=273+27=300K\\P_1=730mm\\ \cfrac{P_1V_1}{T_1}=\cfrac{P_2V_2}{273}\\\cfrac{730\times 600}{300}=\cfrac{760\times V_2}{273}\\V_2=524.44mL=0.5244L$ At STP,

$T_2=273K\\P_2=760mm\\V_2=?$

Two flask

$A$ and

$B$ of equal volumes maintained temperature 300 K and 700 K contain equal mass of

$He(g)$ and

$N_{2}(g)$ respectively. What is the ratio of total translation kinetic energy of gas in flask

$A$ to that of flask

$B$ ?

0%
$1:3$
0%
$3:1$
0%
$3:49$
0%
None of these

Explanation

$K.E = \dfrac{1}{2}m{v}^{2}$

$v = \sqrt{\dfrac{3RT}{M}}$ , Its Given Masses of

${H}_{2}$ and

${N}_{2}$ are Equal .

Ratio of K.E would become

$\dfrac{{T}_{1}}{{M}_{1}} : \dfrac{{T}_{2}}{{M}_{2}}$

Ratio of Total translation kinetic energy of Flask A to that of flask B =

$\dfrac{300}{4} : \dfrac{700}{28}$ =

$3 :1$

Equal mass which of the following substance of occupy maximum volume at room temperature?

0%
Water
0%
Sugar
0%
Oxygen
0%
Graphite

Explanation

Lesser the molecular weight , the greater the substance occupies the volume

So here graphite has lower molecular weight. So It occupies more volume.

Option

$D$ is correct.

A container contains

$O_{2}$ and

$N_{2}$ in equal molar concentration at same temperature, what is the

$CORRECT$ statement about the average molar kinetic energy of the two gasses?

0%
Depends upon volume
0%
$KE_{N2}=KE_{O2}$
0%
$KE_{N2}> KE_{O2}$
0%
$KE_{N2}< KE_{O2}$

Explanation

Molar kinetic energy

$=KE= \dfrac 32 RT$

It is dependent on temperature.

Average kinetic is directly proportional to temperature.

Hence average KE is same as temperature is same.

So the correct answer is(B)

$KE_{N_2}=KE_{O_2}$

Normal boiling point of a liquid is that temperature at which vapour of the liquid pressure of the liquid is equal to:

0%
zero
0%
380mm of Hg
0%
760mm of Hg
0%
100mm of Hg

A container of volume

$40$ liters consist of some gas at a pressure of 2 atm and temperature of

$300 K$ . It is heated to

$400 K$ such that half of the gas escape and volume changes to 60 liters. The new pressure of the gas at the final condition will be:

0%
$2 atm$
0%
$1 atm$
0%
$\dfrac{8}{9}$ atm
0%
$\dfrac{9}{8}$ atm

Correct gas equation is :

0%
$\dfrac{V_1T_2}{P_!}=\dfrac{V_2T_1}{P_2}$
0%
$\dfrac{P_1V_1}{P_2V_2}=\dfrac{T_1}{T_2}$
0%
$\dfrac{P_1T_2}{V_1}=\dfrac{P_2V_2}{T_2}$
0%
$\dfrac{V_1V_2}{T_1T_2}=P_1P_2$

A relation between vapour pressure and the temperature is known as

0%
Ideal gas equation
0%
Boltzman equation
0%
Clausious equation
0%
Clausius-Clapeyron equation

Lowering in vapour pressure is highest for:

0%
$0.2m$ urea
0%
$0.1m$ glucose
0%
$0.1m \, MgSO_4$
0%
$0.1m \, BaCl_2$

Explanation

Lowering in vapour pressure is directly proportional

to molality which in turn is directly proportional

to n factor (no. of ions it gives)

The concentration of urea, i.e,

$0.2 \ M$ is much

more than glucose,

$MgSO_{4}$ &

$BaCl_{2}$

Therefore urea will have greater molality &

therefore it will have highest vapour pressure.

Ans is A)

$0.2 M$ urea

Consider the followimg Vapour Pressure - Mole fraction graph. SP (Total vapour pressure of the resultant solution) is equal to

0%
PQ+RS
0%
PQ+QR+RS
0%
SR+SQ
0%
PQ+QR

The total kinetic energy (in joules) of the molecules in

$8 \ g$ of methane at

$27^oC$ is:

0%
$374100 \ J$
0%
$935.3 \ J$
0%
$1870.65 \ J$
0%
$700 \ J$

Explanation

$\textbf{Step 1: Analyzing the given data}$

$\text{Mass of methane=8 g}$

$\text{Molar mass of methane=16 g}$

$\text{Number of moles}=\cfrac{Mass\ of\ substance}{Molar\ mass\ of\ substance}$

$\text{Number of moles of methane}=\dfrac{8}{16}=0.5$

$\text{Temperature}=27^oC=273+27=300\ K$

$\text{Gas constant(R)}=8.314\ JK^{-1}mol^{-1}$

$\textbf{Step 2: Calculation kinetic energy}$

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

$K.E=\dfrac{3}{2}\times 0.5mol\times 300K \times 8.314\ JK^{-1}mol^{-1}$

$K.E=1870.65\ J$

So, total kinetic energy is 1870.65J.

A mixture contains 16 g of oxygen ,28g of nitrogen and 8g of methane.Total pressure of the mixture is 740mm. What is the partial pressure of nitrogen in mm?

0%
185
0%
370
0%
555
0%
740

Explanation

As no of moles of $O_2=16/32=0.5$ moles

no of moles of $N_2=28/28=1$ moles

no of moles of $CH_4=8/16=0.5$ moles

Total pressure=740mm

Partial pressure=total pressure $\times$ mole fraction of $N_2$

Mole fraction of $N_2$ = No of moles of $N_2$ /total no of moles

$=1/0.5+0.5+1=1/2$

then partial pressure $=1/2\times740$

$=370\ mm$

if the mass ratio of hydrogen gas is to oxygen gas in a mixture is 1:2 then the ratio of their number of molecules is:

0%
4:1
0%
1:2
0%
8:1
0%
1:4

Explanation

No of molecules =

$\dfrac{Weight}{Gram\space Molecular\space Weight} \times {N}_{A}$

Molecular wt of hydrogen

${H}_{2}$ = 2

Molecular wt of oxygen

${O}_{2}$ = 32

So Ratio of number of molecules =

$\dfrac{1}{2} : \dfrac{2}{32}$ =

$8 : 1$

Option

$C$ is correct.

Mass of a given gas occupies two litre at STP. By keeping the pressure constant, at what temperature the gas will occupy 4 litre?

0%
$546\ K$
0%
$273\ K$
0%
$100^oC$
0%
$50^oC$

Explanation

At constant pressure,

$V \propto T$

$V= KT$

So if volume becomes double then temperature also becomes double.

Initial temperature at STP is 273 K which becomes 546 K.

Hence, option A is correct.

When the pressure of 5L of

$N_2$ is double and its temperature is raised from 300K to 600K, the final volume of the gas would be:

0%
10 L
0%
5 L
0%
15 L
0%
20 L

Explanation

Since

$\dfrac{PV}{T}=nR=constant$

$\Rightarrow \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$\Rightarrow \dfrac{P_1\times5}{300}=\dfrac{2P_1\times V_2}{600}$

$\Rightarrow V_2=5\ L$

In an ideal solution of non-volatile solute B in solvent A in 2: 5 molar ratio has vapour pressure 250 mm. If another solution in ratio 3: 4 prepared then vapour pressure above this solution is:

0%
200 mm
0%
250 mm
0%
350 mm
0%
400 mm

Explanation

Given molar ratio

$=2:5$ v.p

$=250\ mm\left(\text{one solute}\right)$

Moral ratio

$=3:4\quad v.p=? \quad\left(\text{another solute}\right)$

Mole fraction of solven

$=\dfrac{5}{2+5}=\dfrac{5}{4}=0.71\quad \left({1}^{st}{\text{solute}}\right)$

According to Raoult's law;

$p$ (solution)

$\times$ mole fraction of solven

$\Rightarrow250=p\left(\text{pure solvent}\right)0.71$

$p\left(\text{pure solvent}\right)= \dfrac{250}{0.71}=352.11\ mm$

mole fraction of solvent

$=\dfrac{4}{3+4}=\dfrac{4}{7}=0.57$

$p\left(\text{solution}\right)=352.11\times 0.57=200.7\approx 200\ mm$

Hence, the correct option is

$A$

The product formed in the reaction of

$SOCl_2$ with white phosphorus is

0%
$PCl_3$
0%
$SO_2Cl_2$
0%
$SCl_2$
0%
$POCl_3$

A sample of 16g charcoal was brought into contact with

$C{ H}_{ 4}$ gas contained in a vessel of 1 litre at

$2{ 7 }^{0 }C$ . The pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is 1.6g/

$c{ m}^{ 3 }$ . What is the volume of the

$C{ H}_{ 4}$ gas adsorbed per gram of the adsorbent at 608 torr and

$2{ 7 }^{0 }C$ ?

0%
125 mL/g
0%
16.25 mL/g
0%
26 mL/g
0%
None of these

Explanation

Final volume of gas, at

$608$ torr pressure

$V_{2} = \dfrac{P_1V_1}{P_2}$

$= \dfrac{760 \times 1}{608}$

$\Rightarrow$

$1.25$ or

$V_2 = 1250\,mL$

Volume occupied by gas = Volume of vessel - Volume occupied by charcoal

$= 1000 - \dfrac{16}{1.6} = 990 \,mL$

Difference of volume is due to adsorption of gas by charcoal.

$\therefore$ Volume of gas adsorbed by charcoal

$= 1250 - 990$

$\Rightarrow 260\,mL$
Volume of the gas adsorbed per gram of charcoal

$= \dfrac{260}{16} = 16.25\,mL / g$ at

$608$ torr and

$27^{\circ}C$ .

The vapour pressure of a solution containing 5 g of a non electrolyte in 100 g of water at a particular temperature is

$2985 N{m}^{-2}$ . If the vapour pressure of pure water is

$3000 N{m}^{-2}$ , the molecular weight of solute is:

0%
$60$
0%
$120$
0%
$180$
0%
$380$

Explanation

we have formula of

$\text{Relative lowering of vapour pressure}$ which is

$\dfrac{P_o-P}{P_o}=X_{solute}=\dfrac{n_{solute}}{n_{solute}+n_{solvent}}\approx\dfrac{n_{solute}}{n_{solvent}}$

Now, pute the values

$\dfrac{3000-2985}{3000}=\dfrac{\dfrac{5}{M}}{\dfrac{100}{18}}$

$\dfrac{15}{3000}=\dfrac{5\times 18}{100\times M}$

$M=180$

correct answer is C.

The temperature and pressure at which ice, liquid. water and water vapour can exist together are ?

0%
${ 0 }^{ o }C,\quad 1 atm$
0%
${ 2 }^{ o }C,\quad 4.7atm$
0%
${ 0 }^{ o }C,\quad 4.7mm$
0%
${ -2 }^{ o }C,\quad 4.7mm$

Explanation

The temperature and pressure at which ice, liquid, water and water vapour exist together at triple point of water.

Triple point of water is

$0^\circ C \ and \ 4.7mm$

Hence, Option "C" is the correct answer.

Kinetic theory of gases presumes the collision between the molecules to be perfectly elastic because:

0%
the molecules are rigid
0%
the temperature remains constant irrespective of collisions
0%
collisions will not split the molecules
0%
the molecules are tiny

In the gas equation

$PV = nRT$ , the value of

$R$ depends upon:

0%
Nature of gas
0%
The pressure of gas
0%
Unit of measurement
0%
Temperature of gas

Explanation

$R$ is the universal gas constant. It has a fixed value which depends upon the units in which

$P, V, n$ and

$T$ are expressed in the ideal gas equation,

$PV = nRT$ .

It's values are

$0.0821\ L\cdot atm\cdot mol^{-1}\cdot K^{-1}$ ,

$8.314\ J\cdot mol^{-1}\cdot K^{-1}$ or

$2 \ cal\cdot mol^{-1}\cdot K^{-1}$

Thus, option C is correct.

According to the equipartition principle, each degree of freedom contribute following amount to the translational kinetic energy of molecule: (hence k = Boltzmann constant)

0%
$\dfrac{1}{2} kT$
0%
$kT$
0%
$\dfrac{3}{2} kT$
0%
$\dfrac{1}{3} kT$

Explanation

According to the equipartition principle, each degree of freedom contributes an amount of

$\dfrac{1}{2}kT$ to the kinetic energy of the molecule.

The solubility of a specific non-volatile salt is

$4\mathrm { g }$ in

$100\mathrm { g }$ of water at

$25 ^ { \circ } \mathrm { C } .$ If

$2.0 g,$

$4.0\mathrm g$ and

$6.0 g$ of the salt added of

$100 g$ of water at

$25 ^ { \circ } \mathrm { C } ,$ in system

$X , Y$ and

$Z .$ The vapour pressure would be in the order:

0%
$X < Y < Z$
0%
$X > Y > Z$
0%
$Z > X = Y$
0%
$x > Y = z$

If volume of

$2$ moles of a gas at

$27^\circ C$ is

$8.21 L$ then pressure of this gas will be.

0%
$0.6$ atm
0%
$6$ atm
0%
$60$ atm
0%
$600$ atm

$500ml$ of a gas

$A$ at

$1000$ torr, and

$1000ml$ of gas

$B$ at

$800$ torr are placed in

$2L$ container the final pressure will be -

0%
$100$ torr
0%
$650$ torr
0%
$1800$ torr
0%
$2400$ torr

Explanation

According to Boyle's Law (Modified)

$P_1V_1+P_2V_2+........=P_fV_f$

Given,

$P_1=1000;\quad V_1=500$

$mL$

$P_2=800;\quad V_2=1000$

$mL$

$P_f=?;\quad V_f=2000$

$mL$

$\implies P_1V_1+P_2V_2=P_fV_f$

$\implies (1000)(500)+(800)(1000)=(P_f)(2000)$

$\implies \cfrac {1300}{2}=P_f$

$\implies P_f=650$

$torr$

The number of

$H_{3}O^{+}$ ions present in 10ml of water at

$25^{\circ}C$ is:

0%
$6.023 \times 10^{-14}$
0%
$6.023 \times 10^{14}$
0%
$6.023 \times 10^{-19}$
0%
$6.023 \times 10^{19}$

Explanation

No. of hydronium ions present in

$10$ ml water at

${ 25 }^{ 0 }C$ .

$pH=7$ at

${ 25 }^{ 0 }C$

$\left[ { H }^{ + } \right] ={ 10 }^{ -7 }M$

No. of moles

$=$ Concentration

$\times$ Volume

$=\left( { 10 }^{ -7 } \right) \left( 10 \right) \times { 10 }^{ -3 }$

$={ 10 }^{ -9 }$ moles

1 mole has

$6.023\times { 10 }^{ 23 }$ ions of hydronium.

$\Rightarrow \quad 6.023\times { 10 }^{ 23 }\times { 10 }^{ -9 }$ ions in

$10$ ml water

$\Rightarrow 6.023\times { 10 }^{ 14 }$ ions are there in

$10$ ml water at

${ 25 }^{ 0 }C$ .

Hence, option B is correct.

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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