MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 7

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.

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 ,

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

$P = 760-7.6$

$P= 752.4$ $torr$

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

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

${ 20 }^{ 0 }C$ , 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

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

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

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

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

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

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

$\approx 130$

One litre of a gas weighs

$2$ g at

$300$ K and

$1$ atm pressure. If the pressure is made

$0.75$ atm, at which temperature will one litre of the same gas weighs

$1$ g?

0%
$600$ K
0%
$800$ K
0%
$900$ K
0%
$450$ K

Explanation

$\text{As } PV = \cfrac{m}{M}RT \Rightarrow M=\cfrac{mRT}{PV} \\ \cfrac{m_{1}RT_{1}}{P_{1} V_{1}} = \cfrac{m_{2}RT_{2}}{P_{2} V_{2}} \text{ [for the same gas]} \\ \Rightarrow \cfrac{(2)(R)(300)}{1} = \cfrac{(1)(R)(T)}{0.75} \\ T= (600) (0.75) =450 K$

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

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 given laws of chemical combination is satisfied by the figure?

0%
Law of multiple proportion
0%
Gay Lussac's law
0%
Avogadro law
0%
Law of definite proportion

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 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

$PV=nRT$

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

$\therefore$

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

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

$\Rightarrow M=82.1gm/mol$ .

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 molecular weight of a gas is

$40$ . At

$400$ K if

$120$ g of this gas has a volume of

$20$ litres, the pressure of the gas in atm is:

0%
$4.92$
0%
$5.02$
0%
$49.6$
0%
$0.546$

Explanation

$PV = \cfrac{m}{M}RT \\ \Rightarrow P = \cfrac{m}{MV}RT = \cfrac{120}{(40)(20)}(0.0821)(400) = (60)(0.0821) = 4.926 atm$

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

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

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

$B.$

A vessel contains equal masses of three gases $A,\ B,\ C$ and recorder a pressure of 3.5 bar at ${ 25 }^{ o }$ C. The molecular mass of $C$ is twice that of $B$ and molecular mass of $A$ is half of B. Find the partial pressure of $B$ (in bar the vessel).

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

Explanation

$m_{A} = m_{B} = m_{C} = m \hspace{20mm} P_{total}=3.5 bar \hspace{20mm} M_{C} = 2M_{B} \hspace{20mm} 2M_{A}=M_{B} \\ \text{Partial Pressure of B = } \cfrac{\cfrac{m_{B}}{M_{B}}}{\cfrac{m_{A}}{M_{A}}+\cfrac{m_{B}}{M_{B}} +\cfrac{m_{C}}{M_{C}}} \cdot P_{total} = \cfrac{\cfrac{m}{M_{B}}}{\cfrac{m}{M_{B}/2}+\cfrac{m}{M_{B}} +\cfrac{m}{2M_{B}}} \cdot P_{total} = \cfrac{1}{2+1+\cfrac{1}{2}} \cdot P_{total} \\ \hspace{35mm} = \cfrac{2}{7} \cdot P_{total} = \cfrac{2}{7} \times 3.5 = 1bar$

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

$0.43$ .

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

Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as :
Assertion: At equilibrium, vapour phase will be always rich in component which is more volatile.
Reason: The composition of vapour phase in equilibrium with the solution is determined by the partial pressures of the components.

0%
If both assertion and reason are true and reason is the correct explanation of assertion.
0%
If both assertion and reason are true but reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both assertion and reason are false.

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

$K_{p}=P$

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

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

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

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

$\log P=-1$

$P=0. 1\ atm$

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

$D$

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

$\text{B}$

$20$ litre of an ideal gas, present in a piston fitted cylinder at

$10$ atm is allowed to expand in a process in which

$P/V$ is a constant. If final volume of gas is

$60$ litre, then work done by gas is:

0%
$81.12\ kJ$
0%
$-81.12\ kJ$
0%
$40.56\ kJ$
0%
$-40.56\ kJ$

Explanation

$W=-P(V_2-V_1)$

$=10(60-20)$

$=10(40)=-400 atmL$

$1 atm= 101325 Pa$

$\Rightarrow 1 atm= 101325J/m^3$

$\Rightarrow 1L= 10^{-3}m^3$

$\therefore W= 400 \times 101325 \cfrac {J}{M^3}\times 10^{-3}M^{3}$

$=-40.56 kJ$

A bulb of unknown volume contained an ideal gas at 650mm pressure. A certain amount of the gas was withdrawn and found to occupy 1.5 cc at 700 mm pressure. The pressure of the gas remaining in the bulb was found to be 600 mm. If all measurements are made at the same temperature, the volume of the bulb is ?

0%
21 cc
0%
42 cc
0%
105 cc
0%
63 cc

Explanation

The number of moles of gas withdrawn,

$n_1=\cfrac {RT}{PV}=\cfrac {RT}{1.5 cc\times 700 mm}=\cfrac {RT}{1050}$

If the total volume of bulb is

$x$ cc

The number of moles of gas remaining in the bulb,

$n_2=\cfrac {RT}{x\times 600}$

Again, before withdrawal of gas

$n_1+n_2=\cfrac {RT}{x\times 650}$

$\Rightarrow RT\left(\cfrac {1}{1050}+\cfrac {1}{600 \times x}\right)=\cfrac {RT}{x\times 650}$

$\Rightarrow \cfrac {1}{1050}=\cfrac {1}{x}\left(\cfrac {1}{650}-\cfrac {1}{600}\right) \Rightarrow x= 21$ cc

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$

An open bulb containing air at

$19^{\circ}C$ was cooled to a certain temperature at which the number of moles of the gaseous molecules increased by

$25\%$ . The final temperature is:

0%
$-39.4^{\circ}C$
0%
$233.6^{\circ}C$
0%
$39.4^{\circ}C$
0%
$240^{\circ}C$

Explanation

From the ideal gas relation,

$PV=nRT$

Since the pressure and volume in the given problem is constant, we get

$nT=constant$

$\Rightarrow n_1T_1=n_2T_2$

$\Rightarrow T_2=\dfrac{T_1}{1.25}=\dfrac{292}{1.25}=233.6\ K=-39.4^o\ C$

A vessel contains 0.5 mole each of

${ SO }_{ 2 }{ ,H }_{ 2 }$ and

$CH_{ 4 }$ . Its outlet was made open and closed after sometime. Thus, order of partial pressure will be:

0%
${ PSO }_{ 2 }>{ PCH }_{ 4 }>{ PH }_{ 2 }$
0%
${PH}_{2}>{PCH}_{4}>{PSO}_{2}$
0%
${PH}_{2}>{PSO}_{2}>{PCH}_{4}$
0%
${PH}_{2} ={PSO}_{2} = {PCH}_{4}$

Explanation

Given gases are

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

Molecular weight of given gases are as follows.

${ SO }_{ 2 }>{ CH }_{ 4 }>{ H }_{ 2 }$

As the molecular weight is high, Partial pressure is also high.

So,

${ PSO }_{ 2 }>{ PCH }_{ 4 }>{ PH }_{ 2 }$

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%
$0.435$

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.

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.

At a pressure of 760 torr and temperature of 273.15K, the indicated volume of which system is not consistent with the observation?

0%
14 g of $N_2$ + 16 g of $O_2$ ; Volume = 22.4 L
0%
4 g of He + 44 g of $CO_2$ ; Volume = 44.8 L
0%
7 g of $N_2$ + 36 g of $O_3$ ; Volume = 22.4 L
0%
17 g of $NH_3$ + 36.5 g of $HCl$ , Volume = 44.8 L

Explanation

Given,

Pressure=1atm

temperature=

$73.15K$

Now,For volume calculation,we know

1 mole of any gaseous mixture=

$22.4L$

n mole of any gaseous mixture

$=n\times 22.4L$

Volume of mixture

$=n\times 22.4$

Now,

a)14g of

${N}_{2}$ +16g of

${O}_{2}$

$=\cfrac{14}{28}+\cfrac{16}{32}=\cfrac{1}{2}+\cfrac{1}{2}=1$ mole

Volume =

$22.4L$

b) 4g of

$He+44g$ of

${CO}_{2}$

$=\cfrac{4}{4}+\cfrac{44}{44}=1+1=2$ mole

Volume

$=2\times 22.4=44.8L$

c)7g of

${N}_{2}$ +36g of

${O}_{3}$

$=\cfrac{7}{28}+\cfrac{36}{48}=\cfrac{1}{4}+\cfrac{3}{4}=1$ mole

Volume

$=22.4L$

d)17g of

${NH}_{3}$ +36.5g of HCl

$=\cfrac{17}{17}+\cfrac{36.5}{36.5}=2$ mole

Volume of mixture

$=44.8L$

Because

${ NH }_{ 3 }(g)+HCl(g)\longrightarrow { NH }_{ 4 }Cl(s)$

Volume of solid<<volume of gas

Therefore, Volume of mixture is

$<<44.8L$

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

How much should the pressure be increased in order to decrease the volume of a gas by

$15\%$ at a constant temperature?

0%
$15.6\%$
0%
$17.6\%$
0%
$6.7\%$
0%
$8.9\%$

Explanation

Given,volume is decreased by 15%

${V}_{1}=V$

${V}_{2}=V-15V/100$

$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =1.176$ ---------1

Now,we know

$PV=nRT$

at T constant,R constant and n constant

PV=K

$P\propto \dfrac { 1 }{ V }$

$\dfrac { { P }_{ 2 } }{ { P }_{ 1 } } =\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =1.176$

Now % of pressure increased:

$(\dfrac { { P }_{ 2 } }{ { P }_{ 1 } } -1)\times 100$

$=(1.176-1)\times100$

$=17.6\%$

A gaseous mixture contain 1 gm of ${ H }_{ 2},$ 4 gm of $He$ 7gm of ${ N }_{ 2 }$ and 8gm of ${ O }_{ 2 }$ . The gas having the highest partial pressure is:

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

Explanation

Given,

1gm of hydrogen, 4gm of helium,7gm of nitrogen,8gm of oxygen

Now,

Given mole of hydrogen

$\cfrac{1}{2}$ mole

mole of helium

$=\cfrac{4}{4}$ =1mole

mole of nitrogen

$=\cfrac{7}{28}=\cfrac{1}{4}$ mole

mole of oxygen

$=\cfrac{8}{32}=\cfrac{1}{4}$ mole

We know according to ideal gas equation

$P\propto n$ , P=pressure, n=no of moles

Therefore,the gas having highest no of moles will have height partial pressure.

hence helium will have highest partial pressure.

An aquatic species need at least 4mg/L of

$O_{2}$ for their survival.

$O_{2}$ in water at 273K and 1 atm pressure is

$2.21 \times 10^{-3}$ mol/L. The partial pressure of

$O_{2}$ above water(in an atmosphere at 273K) needed for the survival of species is:

0%
0.56 atm
0%
0.056 atm
0%
5.525 atm
0%
5.525 $\times$ 10 $^{-3}$ atm

Explanation

Given, required solubility of oxygen gas for the survival of marine species=4mg/L

We know 1 mole of oxygen

$= 32g$ of oxygen

0.004g of

${O}_{2}=125 \times {10}^{-6}mol/L$

Now we know initial pressure above water = 1 atm

Also given,

Initially solubility of

${O}_{2}$ gas in water at 1 atm

$=2.21 \times {10}^{-3}mol/L$

Now,we know according to Henry's law of partial pressure.

Solubility of gas is directly proportional to partial pressure of that gas over liquid.

$S\propto P$

$\dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { { P }_{ 1 } }{ { P }_{ 2 } }$

$\dfrac { 2.21\times { 10 }^{ -3 } }{ 125\times { 10 }^{ -6 } } =\dfrac { 1 }{ { P }_{ 2 } }$

${P}_{2}=0.056atm$

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.

A gaseous mixture containing

$0.35_{g}$ of

$N_{2}$ and 5600 ml of

$O_{2}$ at STP has kept in a 5 liters at 300K. The total temperature of the gaseous mixture is?

0%
1.293 atm
0%
1.2315 atm
0%
12.315 atm
0%
0.616 atm

Explanation

Moles of

$O_2=\cfrac{5600}{22400}=0.25=n_{O_2}$

Moles of

$N_2=\cfrac{0.35}{28}=0.0125=n_{N_2}$

Initial pressure of

$N_2=\cfrac{n_{N_2}RT}{V}=\cfrac{0.0125\times 0.0821\times 300}{5}=0.061\text{ }atm$

Potential pressure of

$O_2=\cfrac{n_{O_2}RT}{V}=\cfrac{0.25\times 0.0821\times 300}{5}=1.231\text{ }atm$

$\therefore$ Total pressure

$=1.231+0.061=1.292\text{ }atm$

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

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

The molecular weight of

$O_2$ and

$SO_2$ are

$32$ and

$64$ respectively. If one litre of

$O_2$ at

$15^0$ C and

$759$ mm pressure contains N molecules, the number of molecules in two litre of

$SO_2$ under the same conditions of temperature and pressure will be:

0%
$N/2$
0%
$N$
0%
$2N$
0%
$4N$

Explanation

Solution:- (C)

$2 N$

From ideal gas equation-

$PV = nRT$

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

For

${O}_{2}$ -

$V = 1 \; L \; \left( \text{Given} \right)$

No. of moles of

${O}_{2} = \cfrac{P}{RT}$

Therefore,

No. of molecules of

${O}_{2} = n \times {N}_{a} = N \; \left( \text{Given} \right)$

$\Rightarrow \cfrac{P}{RT} \times {N}_{a} = N ..... \left( 1 \right)$

For

$S{O}_{2}$ -

$V = 2 \; L \; \left( \text{Given} \right)$

No. of moles of

$S{O}_{2} = \cfrac{2P}{RT}$

Therefore,

No. of molecules of

$S{O}_{2} = n \times {N}_{a}$

$\Rightarrow$ No. of moles of

$S{O}_{2} = \cfrac{2P}{RT} \times {N}_{a}$

$\Rightarrow$ No. of molecules of

$S{O}_{2} = 2 N \; \left( \text{From } \left( 1 \right) \right)$

Hence the no. of molecules of

$S{O}_{2}$ is

$2 N$ .

Air contains 79%

${ N }_{ 2 }$ and 21%

${ O }_{ 2 }$ by volume. If the pressure is

$750$ of

$Hg$ , the partial pressure of

${ O }_{ 2 }$ is:

0%
$157.5$ of $Hg$
0%
$175.5$ of $Hg$
0%
$315.0$ of $Hg$
0%
$257.5$ of $Hg$

Explanation

Solution:- (A)

$157.5 \text{ mm of Hg}$

Given:-

Total pressure

$\left( P \right) = 750 \text{ mm of Hg}$

Percentage of

${O}_{2}$ in air

$= 21 \%$

Therefore,

${P}_{{O}_{2}} = \cfrac{\% \text{ of } {O}_{2} \text{ in the air}}{100} \times P$

$\Rightarrow {P}_{{O}_{2}} = \cfrac{21}{100} \times 750 = 157.5 \text{ mm of Hg}$

Hence the partial pressure of

${O}_{2}$ is

$157.5 \text{ mm of Hg}$ .

Vapours of a liquid can exist only at ?

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

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.

Suppose that we change

$U_{RMS}$ of gas in a closed container from

$5 \times 10^{2}$ cm/sec to

$10 \times 10^{2}$ cm/sec, which one of the following might correctly explain how this change was accomplished?

0%
By heating the gas w double the temperature
0%
By removing 75% of the gas at constant volume we decrease the pressure to one quarter of its original value
0%
By heating the gas we quadruple the pressure
0%
By pumping in more gas at constant temperature we quadruple the pressure.

Explanation

We know that

$U_{RMS}=\sqrt{\cfrac{SRT}{M}}\Longrightarrow \cfrac{U_2}{U_1}=\sqrt{\cfrac{T_2}{T_1}}$

$\Longrightarrow \cfrac{10\times 10^2\text{ }cm/sec}{5\times 10^2\text{ }cm/sec}=\sqrt{\cfrac{T_2}{T_1}}$

$\Longrightarrow \cfrac{T_2}{T_1}=\cfrac{4}{1}$

again,

$C_{RMS}=\sqrt{\cfrac{3P}{P}}$

$\Longrightarrow \cfrac{P_2}{P_1}=\cfrac{4}{1}$

By heating the gas we quadruple the pressure .

So, the correct answer is option

$C.$

For a fixed amount of perfect gas, which of these statements must be true?

0%
$E$ and $H$ each depend only on $T$
0%
$C_p$ is constant
0%
$PdV = nRdT$ for every infinitesimal process
0%
Both $(A)$ and $(C)$ are true

Explanation

For an ideal gas

$=, PV=nRT$

$E$ and

$H$ are dependent only on temperature where

$H$ is enthalpy and

$E$ is energy.

$\cfrac{C_p}{C_v}$ remains constant for a real gas.

A weather balloon is inflated with helium. The balloon has a volume of

$100 m^3$ and it must be inflated to a pressure of

$0.10 atm$ . If

$50 L$ gas cylinders of helium at a pressure of

$100 atm$ are used, how many cylinders are needed? Assume that temperature is constant.

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

Explanation

If the cylinders were at 1.00 atm, they would contain 50.00 L each, so at 100.0 atm they contain 5000. L each.

$1m^3=1000L$ ,

$100.0 m^3= 1.000 \times 10^5 L$ .

Since the pressure in the balloon is only 0.10 atm, though, the gas volume equals only

$1.000 \times 10^4 L$ at 1 atm.

(1.000 × 10⁴ L) / (5000. L/cylinder) = 2.00 cylinders.

$\dfrac{1.000\times 10^4 L}{5000L}$ /cylinder

$=2$ cylinders

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

$325mm.$

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.

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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