MCQ Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 8

${ \Delta }_{ f }{ G }^{ o }$ at

$500\ K$ for substance

$S$ in liquid state and gaseous state are

$+100.7\ kcal\ { mol }^{ -1 }$ and

$+103\ kcal\ { mol }^{ -1 }$ respectively. Vapour pressure of liquid

$S$ at

$500\ K$ is approximately equal to:

$(R= 2\ { cal }\ K^{ -1 }\ { mol }^{ -1 })$

0%
$0.1\ atm$
0%
$1\ atm$
0%
$10\ atm$
0%
$100\ atm$

Explanation

Substance (s)

$\rightarrow$ Substance (g)

$\triangle G^{o}=103-100.7$

$=2.3\ K\ cal\ mol^{-1}$

$K_{p}=$ Vapour pressure of the substance

$K_{p}=P$

$\triangle G=-RT\ In\ K_{p}$

$(\because K_{p}=P)$

$2.3=-2\times 10^{-3}\times 500\ In\ P$

$2.3=-2\times 500\times10^{-3}\times 2.3\ \log P$

$\log P=-1$

$P=0. 1\ atm$

The density of a gas is

$2.5$ g/L at

$127^0$ C and

$1$ atm. The molecular weight of the gas is:

0%
$82.1$
0%
$41.05$
0%
$56$
0%
$28$

Explanation

$PV=nRT$

$\Rightarrow P=\dfrac { n }{ V } RT=\dfrac { W }{ MV } RT=\dfrac { d }{ M } RT$

$\therefore$

$P=\dfrac { d }{ M } RT\Rightarrow M=\dfrac { dRT }{ P }$

$\Rightarrow M=\dfrac { 2.5gm/lit\times 0.0821Lit-atm/K-mol\times 400K }{ 1atm }$

$\Rightarrow M=82.1gm/mol$ .

A solution has a

$1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at

$20^{\circ}C$ are

$440$ mm of

$Hg$ for pentane and

$120$ mm of

$Hg$ for hexane. The mole fraction of pentane in the vapour phase would be?

0%
$0.549$
0%
$0.200$
0%
$0.786$
0%
$0.478$

Explanation

Mole fraction of pentane solution=

$\cfrac {1}{1+4}=\cfrac {1}{5}$

Mole fraction of hexane solution=

$\cfrac {4}{1+4}=\cfrac {4}{5}$

Total pressure of solution,

$P_S=X_{P}P_P^O+X_HP_H^O$

$=\cfrac {1}{5}\times 440\ mm\ Hg+\cfrac {4}{5}\times 120\ mm\ Hg$

$=184\ mm\ Hg$

$\therefore$ The fraction of pentane in the vapour phase

$=\cfrac {X_PP_P^O}{P_S}=\cfrac {88}{184}=0.478$ .

Hence, the correct option is

$D$

An ideal gas obeying kinetic gas equation :

0%
can be liquefied if its temperature is more than critical temperature
0%
can be liquefied at any value of any value T and P
0%
can not be liquefied under any value of T and P
0%
can be liquefied if its pressure is more than critical pressure

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

$9.0$ torr. 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

According to Dalton's law of partial pressure,

$p_{total}=\chi_Ap_A+\chi_Bp_B$

Since B is a non-volatile substance,

$p_B=0$

$\Rightarrow p_{total}=p_A\times \dfrac{\dfrac{m_A}{M_A}}{\dfrac{m_A}M_A+\dfrac{m_B}{M_B}}$

$\Rightarrow 9=10\times \dfrac{\dfrac{20}{200}}{\dfrac{20}{200}+\dfrac{1}{M_B}}$

$\Rightarrow M_B=90\ amu$

Hence, option

$B$ is the correct answer.

The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?

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

Explanation

$\cfrac {\Delta P}{P_o}=\cfrac {n}{n+N}$

$\Rightarrow \cfrac {10}{P_o}=0.2$

$\Rightarrow P_o=50mm$

For other solutions of same solvent,

$\Rightarrow \cfrac {20}{P_o}=\cfrac {n}{n+N}$

$\Rightarrow \cfrac {20}{50}=$ mole fraction of solute=

$0.4$

$\therefore$ Mole fraction of solvent=

$1-0.4=0.6$

Hence, the correct option is

$\text{B}$

A balloon filled with methane

$CH_4$ is pricked with a sharp point and quickly plunged into a tank of hydrogen at the same pressure. After sometime the balloon will have?

0%
Enlarged
0%
Collapsed
0%
Remained unchanged in size
0%
Ethylene $(C_2H_4)$ inside it

Explanation

Given: initially balloon has methane.

Balloon having methane is led to diffuse methane by pricking.

Again the same balloon suddenly plunged to a tank to effuse hydrogen to it.

According to graham's law of effusion/diffusion.

$\dfrac { { r }_{ 1 } }{ { r }_{ 2 } } =\sqrt { \dfrac { { M }_{ 1 } }{ { M }_{ 2 } } }$

$\dfrac { { r }_{ { CH }_{ 4 } } }{ { r }_{ { H }_{ 2 } } } =\sqrt { \dfrac { 2 }{ 16 } }$

We know,

$M =$ Molecular weight

$r =$ Rate of effusion/diffusion

$\dfrac { { r }_{ { CH }_{ 4 } } }{ { r }_{ { H }_{ 2 } } } = { \dfrac { 1 }{ 2\sqrt { 2 } } }$

${ r }_{ { H }_{ 2 } }=2\sqrt { 2 } \times { r }_{ { CH }_{ 4 } }$

Rate of hydrogen effusion is

$2\sqrt { 2 }$ times the rate of methane diffusion.

Therefore, balloon enlarges as hydrogen moves in faster.

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,

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

With increase in temperature, the fluidity of liquids

0%
increases
0%
decreases
0%
remains constant
0%
may increase or decrease

The volume of helium gas is 44.8 L at:

0%
100 $^oC$ and 1 atm
0%
0 $^o C$ and 1 atm
0%
0 $^o \ C$ and 0.5 atm
0%
100 $^o \ C$ and 0.5 atm

Explanation

The conditions at STP are:

$P=1\ atm$

$T=273.15\ K$

We know that the volume occupied by any gas at STP is

$22.4\ L$

Hence using the relation

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

$\dfrac{1\times22.4}{273.15}=\dfrac{P_2\times44.8}{T_2}$

$\dfrac{T_2}{P_2}=546.3$

Since this condition is satisfied by a pressure of 0.5 atm and a temperature of 273 K.

Hence option

$C$ is the correct answer

Equal weights of

$HF$ ,

$HCl$ ,

$HBr$ , and

$HI$

${at\ 87}^{0}C$ and

$750\ mm$

$Hg$ pressure are taken. The correct sequence of these gases in the increasing order of their volume is ?

0%
HI<HBr<HCl<HF
0%
HBr<HI<HCl<HF
0%
HF<HBr<HCl<HI
0%
HI<HBr<HF<HCl

If the pressure and absolute temperature of

$2\ liters$ of carbon dioxide are doubled the volume of the gas would be:

0%
$2\ litres$
0%
$4\ litres$
0%
$5\ litres$
0%
$7\ litres$

Explanation

$\text{From ideal gas equation} \hspace{40mm} \text{Given: P$_{2}$ = 2P$_{1}$ ; T$_{2}$ = 2T$_{1}$} \\ \cfrac{P_{1} V_{1}}{T_{1}} = \cfrac{P_{2} V_{2}}{T_{2}} \Rightarrow \cfrac{P_{1}(2)}{T_{1}} = \cfrac{(2P_{1})(V_{2})}{2 T_{1}} \\ \Rightarrow V_{2} = 2 litres$

According to Avogadro's law the volume of a gas will ____ as _____ if ____ are held constant.

0%
increases, number of moles; P & T
0%
decreases, number of moles; P & T
0%
increases; T & P; number of moles
0%
decreases; P & T; number of moles

Explanation

According to Avogadro's law: Equal volume of all gases at same temperature and pressure will have same no. of molecules.

For a given mass of ideal gas,

Volume

$\propto$ Number of moles of the gas (if temperature and pressure are constant)

So, volume of the gas will increase as the number of moles if pressure (P) and temperature (T) are held constant.

Vapour pressure of pure benzene is 119 torr and of toluene is 37.0 torr at the same temperature mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be:

0%
$0.137$
0%
$0.205$
0%
$0.237$
0%
$0.435$

In atmosphere which gas have high partial pressure?

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

Explanation

Since mole fraction of nitrogen is the highest in atmosphere & partial pressure is directly proportional to mole fraction, Nitrogen will have the highest partial pressure.

Partial pressure= P

$\times$ mole fraction.

0.5 mole of each of

$H, \ SO_2$ and

$CH_4$ , are kept in a container. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be:

0%
$P(CH_{4}) > P(SO_{2}) > P(H_{2})$
0%
$P(H_{2}) > P(CH_{4}) > P(SO_{2})$
0%
$P(SO_{2}) > P(CH_{4}) > P(H_{2})$
0%
$P(H_{2}) > P(SO_{2}) > P(CH_{4})$

Explanation

$SO_2$ Molar mass=

$32+32=64g/mol$

$CH_4$ Molar mass=

$12+4=16 g/mol$

$H_2$ Molar mass=

$2 \times 1= 2 g/mol$

Rate of diffusion is inversely related to molecular weight. Lighter the gas, more is the diffusion. The rate of diffusion is given by,

$r_{H_2} >r_{CH_4} >r_{SO_2}$

Hence the order of partial pressure will be

$P_{SO_2}>P_{CH_4}>P_{H_2}$

A mixture of ethanol and propanol has vapor pressure 279 mm of Hg at 300 K. The mol fraction of ethanol in the solution is 0.6 and vapour pressure of pure propanol is 210 mm of Hg. Vapour pressure of pure ethanol will be:

0%
$336.6$ mm
0%
$325$ mm
0%
$375$ mm
0%
$415$ mm

Explanation

$p=p_1+p_2$

$\Rightarrow p={p_1}^0x_1+{p_2}^0x_2$ ( Raoult's law )

$p={p_1}^0\left( 1-x_2\right) +{p_2}^0x_2$

$p={p_1}^0+\left( {p_2}^0-{p_1}^0\right) x_2$

$279mm={p_1}^0+\left( {p_2}^0-{p_1}^0\right) x_2$

Let

$2,$ ethanol

$1,$ propanol

$\therefore 279mm=210mm+\left( {p_2}^0-210\right) 0.6$

$69=\left( {p_2}^0-210\right)0.6$

${p_2}^0-210=115$

$\Rightarrow {p_2}^0=325mm$

Hence, the answer is

$325mm.$

Vapours of a liquid can exist only at ?

0%
below boiling point
0%
below critical temperature
0%
below inversion temperature
0%
above critical temperature

The molecular weight of a gas isAt 400K, if 120g of this gas has a volume of 20 litres, the pressure of the gas is?

0%
5.02 atm
0%
0.546 atm
0%
4.92 atm
0%
49.6 atm

Explanation

Moles of gas

$=\cfrac{120}{40}=3=n$

$\because PV=nRT$

$\therefore P\times 20=3\times 0.0821\times 400$

$P=4.926$ 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

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

0%
Sugar
0%
Water
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.

The type of molecular force of attraction present in the following compound is :

0%
Intermolecular H-bonding
0%
Intramolecular H-bonding
0%
van der WaaLs' force
0%
All of these

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

$140 \text{ torr.}$ .

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.

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

$\text{C}$

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 of the following exists as a liquid at room temperature due to the formation of associated molecules only?

0%
Benzene
0%
Ammonia
0%
Bromine
0%
Carbon disulphide

Explanation

Benzene is a non-polar aromatic molecule. Benzene has a melting point of

${ 5.5 }^{ 0 }C$ and a boiling point of

${ 80 }^{ 0 }C$ . It is therefore liquid at room temperature.

${ NH }_{ 3 }$ liquid boils at

${ -33.3 }^{ 0 }C$ and freezes at

${ -77.3 }^{ 0 }C$ to white crystals. So at room temperature is gas.

Bromine evaporates readily to form coloured gas at room temperature.

Carbon disulphide evaporates at room temperature.

Hence, the correct answer is option

$\text{A}$ .

$P^o$ and

$P^s$ are the vapour pressure of the solvent and solution respectively.

$n_ 1$ and

$n_2$ are the mole fraction of the solvent and solute respectively, then?

0%
$P^S=P^o n_1$
0%
$P^S=P^o n_2$
0%
$P^o=P^S n_2$
0%
$P^S=P^o \dfrac { { n }_{ 1 } }{ { n }_{ 2 } }$

Vapour pressure diagram of some liquids plotted against temperature are shown below most volatile liquid is

0%
$A$
0%
$B$
0%
$C$
0%
$D$

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 $AgCl$

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,

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

$= (12.3 - P_1) / 12.3$

$=0.0177$

$= 12.3 - P_1$

$= 0.2177$

$= P_1$

$= 12.0823$

$= 12.08 kPa$

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,

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

$4.4gm$ of a gas at STP occupies a volume of

$2.24$ lit the gas can be:

0%
${O}_{2}$
0%
$CO$
0%
${NO}_{2}$
0%
${CO}_{2}$

Explanation

At STP,

$2.24 litre$ is ocupied by

$4.4gm$ of gas

Then,

$22.4L$ is occupied by

$= \cfrac {22.4 \times 4.4}{2.24}=44g$

Molar mass of

$CO_2$

$C= 12, O=16$

$\Rightarrow 12+2(16) \Rightarrow 12+32= 44gm$

So, the gas will be

$44g$ .

In an experiment during the analysis of a carbon compound,

$145\ cc$ of

${H}_{2}$ was collected at

$760\ mm$ of Hg pressure and

${27}^{o}\ C$ temperature. The mass of

${H}_{2}$ is nearly:

0%
$10\ mg$
0%
$12\ mg$
0%
$24\ mg$
0%
$6\ mg$

Explanation

Applying Ideal gas law,

$PV=nRT$

$P = \dfrac{760}{760} = 1\ atm$

$n=\dfrac{PV}{RT}=\dfrac{1\times 145}{0.082\times 300}=5.8$ or

$6$ mole.

Hence, option D is correct.

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

For gaseous state, if most probable speed is denoted by

${C}^{\ast}$ , average speed by

$\overline { C }$ and mean square speed by

$C$ , then for a large number of molecules the ratios of these speeds are:

0%
${C}^{\ast}:\overline { C } :C=1.225:1.128:1$
0%
${C}^{\ast}:\overline { C } :C=1.128:1.225:1$
0%
${C}^{\ast}:\overline { C } :C=1:1.128:1.225$
0%
${C}^{\ast}:\overline { C } :C=1:1.225:1.128$

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%
$\dfrac{V_1V_2}{T_1T_2}=P_1P_2$

Pure hydrogen sulphide is stored in a tank of

$100$ litre capacity at

${20^0}C$ and

$2$ atm pressure . The mass of the gas will be:

0%
$34$
0%
$340$ g
0%
$282.68$ g
0%
$28.24$ g

Explanation

Given that:

Pressure = 2 atm

Temperature

$= 20^oC$

T = 293 K

Volume = 100 L

Using ideal gas equation as:

PV=nRT

The gas is hydrogen sulfide.

Molar mass = 34 g/mol

$2\times 100 = \dfrac{\text{Mass of the gas}}{34} \times 0.0821\times 293$

Mass of the gas

$= 282.68\ gm$

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

Equal weight of

${CH}_{4}$ and

${H}_{2}$ are mixed in a container at

$25^o C$ , the fraction of total pressure exerted by hydrogen is :

0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { 1 }{ 3 }$
0%
$\cfrac { 1 }{ 9 }$
0%
$\cfrac { 8 }{ 9 }$

Explanation

Let the weight of methane and hydrogen added be x grams.

Then, moles of methane $= \dfrac{x}{{16}}$

And, moles of hydrogen $= \dfrac{x}{2}$

Now , we know that $PV = nRT$ .(where P is pressure, n is the number of moles, V is the volume, T is the temperature, and R is the gas constant.)

In this case, V, R and T are constant. Hence, $P\propto n$ $\Rightarrow {P_{Total}}= \dfrac{x}{{16}} + \dfrac{x}{2} = \dfrac{9}{{16}}x$ $\to (1)$

Also, ${P_{Hydrogen}}\propto \dfrac{x}{2}$ $\to (2)$

Now, dividing equation (2) by (1), we get fraction of total pressure exerted by hydrogen $= \dfrac{{\dfrac{x}{2}}}{{\dfrac{{9x}}{{16}}}} = \dfrac{8}{9}$ , which is the required answer.

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

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.

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.

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 relative lowering of vapour pressure of a solution containing

$6\ g$ of urea dissloved in

$90\ g$ of water is

0%
$0.0196$
0%
$0.05$
0%
$1.50$
0%
$0.01$

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

$27^{\circ}C$ .

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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