MCQ Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 6

$I$ is surface molecule whereas

$II$ is interior molecule. Choose the correct one.

0%
$I$ results net attraction into the liquid
0%
$II$ are attracted in all directions
0%
Both $A$ and $B$
0%
None of the above

At high altitudes, water boils at a lower temperature because

0%
the atmospheric pressure is high at high altitudes
0%
the viscosity of water is reduced at high altitudes
0%
the atmospheric pressure is low at high altitudes
0%
the surface tension of water is reduced at high altitudes

Explanation

The boiling of water depends on the atmospheric pressure at that altitude. At high altitudes, the pressure is lower, so water can not boil at

$100°$ . As the altitude increases, the boiling point of water reduces.

The correct expression of partial pressure in terms of mole fraction is :

0%
$p_{1} = x_{1}P_{total}, p_{2} = x_{2}P_{total}$
0%
$p = x_{1}x_{2}P_{total}$
0%
$P_{total} = p_{1}x_{1}, P_{total} = p_{2}x_{2}$
0%
$p_{1} + p_{2} = x_{1} + x_{2}$

Explanation

$x_{1}$ and

$x_{2}$ are mole fractions of two gases.
Their partial pressures are

$p_{1} = x_{1} P_{total}, p_{2} = x_{2} P_{total}$ .

The pressure of a

$1 : 4$ mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?

0%
$0.8\times 10^{4} atm$
0%
$0.008\ N\ m^{-2}$
0%
$8\times 10^{4} N\ m^{-2}$
0%
$0.25\ atm$

Explanation

Correct Option: Option C

Hint: $P_{partial} \propto n_{gas}$

Explanation

Step 1: Preparing data

We know that, $P_{partial} = \chi_{gas} \times P_{total}$ where, … $(1)$
$P_{partial}=$ Partial pressure of gas
$\chi_{gas}=$ mole fraction of gas
$P_{total}=$ Total pressure of mixture of gases
Also, $\chi_{n_{1}} = \dfrac{n_{1}}{n_{1} + n_{2} + n_{3} + n_{n}}$ … $(2)$
where, $\chi_{n_{1}} =$ Mole fraction of gas $1$
$n_{1} , n_{2} , n_{3}$ and $n_{n}=$ Moles of all gases in mixture
Given, $P_{total} = 1$ atm.
And, $\dfrac{n_{H_2}}{n_{O_2}} = \dfrac{1}{4}$

Step 2: Solving using equations

$n_{H_2} = x$ and $n_{O_2} = 4x$
$Moles_{(total)} = 5x$
$\chi_{O_2} = \dfrac{4}{5}$
Putting above values in equation $(1)$ , we get:
$P_{partial} ({O_2}) = \dfrac{4}{5} \times 1$ atm
Now, $1$ $atm \approx 10^5$ $N/m^2$
Therefore, $P_{partial} ({O_2}) = \dfrac{4}{5} \times 1 \times 10^5$ $N/m^2$

Hence, $P_{O_{2}} = 8 \times 10^4$ is the partial pressure of dioxygen.

Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess 'partial charges'. The partial charge is:

0%
more than unit electronic charge
0%
equal to unit electronic charge
0%
less than unit electronic charge
0%
double the unit electronic charge

Explanation

Partial charges are always less than the unit electronic charge

$(1.6\times 10^{-19}C)$ .

According to kinetic theory of gases, the collisions between molecules of a gas:

0%
occur in a zig-zag path
0%
occur in a straight line
0%
change velocity and energy
0%
result in settling down of molecules

Liquids are similar to gases because:

0%
both possess the property of flowing and take the volume of the containers
0%
both diffuse and take the shape of the containers
0%
both are readily compressible and diffuse
0%
both are capable of infinite expansion

Explanation

Liquids and gases are similar in both shape and volume because they both have its' shape determined by its' surroundings.

$\Longrightarrow$ Both diffuses and take the shape of containers.

Which of the following assumptions is incorrect according to kinetic theory of gases?

0%
Particles of a gas move in all possible directions in straight lines
0%
All the particles, at any particular time, have same speed and same kinetic energy
0%
There is no force of attraction between the particles of a gas at ordinary temperature and pressure
0%
The actual volume of the gas is negligible in comparison to the empty space between them

Kinetic energy of a gas depends upon its:

0%
Molecular mass
0%
Atomic mass
0%
Equivalent mass
0%
None of these

Explanation

Kinetic energy

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

Hence, option D is correct.

As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?

0%
Increases
0%
Decreases
0%
Remains same
0%
Becomes half

Explanation

Gay Lussac's law states that at constant volume, pressure of a fixed amount of a gas varies directly with the temperature.

$\therefore P\propto T$ (at constant volume)

Thus, if temperature is increased at constant volume pressure increases.

'Equal volumes of all gases at the same temperature and pressure contain equal number of molecules'. This law is called as

0%
Boyle's law
0%
Charle's law
0%
Avogadro's law
0%
Gay Lussac's law

The kinetic theory of matter helped a great deal in understanding the behaviour of gases. Which of the following statements did not belong to it?

0%
When the gas is heated by raising its temperature the molecules move faster
0%
The molecules in a gas are always moving
0%
Gases are made up of small particles called molecules
0%
When molecules collide, they lose energy

Which of the following is correct?

0%
Van der waals radius of chlorine is bigger than nitrogen.
0%
Covalent radius of nitrogen is bigger than chlorine.
0%
Van der waals radius of chlorine is smaller than nitrogen.
0%
All are correct.

Which scientist discovered that the same amount of space was occupied by equal numbers of molecules of gases, irrespective of whether it was hydrogen or chlorine or fluorine?

0%
Gay-Lussac
0%
Avogadro
0%
Marconi
0%
Wohler

Explanation

Avogadro first discovered it.

It states that

$1$

$mole$ of any element or compound have

$6.022\times 10^{23}$ atoms or molecules present at any constant temperature and pressure.

Among the following substances, the lowest vapour pressure is exerted by:

0%
water
0%
alcohol
0%
ether
0%
mercury

Atmospheric pressures recorded in different cities are as follows:

Cities	Shimla	Bangalore	Delhi	Mumbai
$p$ in $N/m^{2}$	$1.01\times 10^{5}$	$1.2\times 10^{5}$	$1.02\times 10^{5}$	$1.21\times 10^{5}$

Consider the above data and mark the place at which liquid will boil first.

0%
Shimla
0%
Bangalore
0%
Delhi
0%
Mumbai

$3$ 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

$60$ torr respectively?

0%
$80$ torr
0%
$140$ torr
0%
$72$ torr
0%
$70$ torr

Explanation

We know that :

$P_{total}=p_P+p_Q=p^0_Px_P + p^0_Qx_Q$

where

$p^0_P$ and

$p^0_Q$ are partial pressures of pure substances.

$x_P$ and

$x_Q$ are mole fractions.

$x_P=\dfrac{3}{5}=0.6$

$x_Q = \dfrac{2}{5}=0.4$

hence

$P_{total}=p^0_Px_P + p^0_Qx_Q= 80\times 0.6 + 60\times 0.4 =72torr$

The vapour pressure of a solvent at

$293$ K is

$100$ mm Hg. Then the vapour pressure of a solution containing

$1$ mole of a strong electrolyte

$(AB_2)$ in

$99$ moles of the solvent at

$293$ K is:

(Assume complete dissociation of solute)

0%
$103$ mm Hg
0%
$99$ mm Hg
0%
$97$ mm Hg
0%
$101$ mm Hg

Explanation

Let us consider the problem:

$\frac{{({P_A}^0 - {\rm{ }}{P^0})}}{{{P_A}^0}} = i\frac{{({n_B})}}{{({n_A} + {n_B})}}$

Since,

$\frac{{\left( {100 - {P_A}} \right)}}{{100}} = 3{\rm{ }}\left[ {\frac{1}{{\left( {99 + 1} \right)}}} \right]$

Since,

$100 - {P_A} = 3$

Hence the solution is

${P_A} = 97 mm$ of

$Hg$ .

According to Avogadro's hypothesis, equal volumes of gases under the same conditions of temperature and pressure will contain:

0%
the same number of molecules
0%
different number of molecules
0%
the same number of molecules only if their molecular masses are equal
0%
the same number of molecules if their densities are equal.

Explanation

According to avogadro's hypothesis, equal volumes of gases under the same conditions of temperature and pressure will contain, the same no. of molecules.

For eg. One volume of hydrogen combines with one volume of chlorine to produce two volumes of HCl gas.

$H_2\\1vol$

$+$

$Cl_2=\\1 vol$

$2HCl\\2 vol$

What percent of a sample of nitrogen must be allowed to escape if its temperature, pressure and volume are to be changed from

$22{ 0 }^{ 0 }$ C,3atm and 1.65 liter to

$11{ 0 }^{ 0 }$ C,0.7atm and 1.0 liter respectively?

0%
81.8%
0%
71.8%
0%
76.8%
0%
86.8%

Explanation

${ P }_{ 1 }=3atm\quad { P }_{ 2 }=0.7atm\\ { T }_{ 1 }=273+220\quad { T }_{ 2 }=273+110\\ =493\quad \quad \quad \quad \quad \quad \quad \quad \quad =383K\\ { V }_{ 1 }=1.65\quad { V }_{ 2 }=?\\ \cfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\cfrac { { P }_{ 2 }{ V }_{ 1 } }{ { T }_{ 2 } } \\ \cfrac { 3\times 1.65 }{ 493 } =\cfrac { 0.7\times { V }_{ 2 } }{ 383 } \\ { V }_{ 2 }=\cfrac { 3\times 1.65\times 385 }{ 0.7\times 493 } \\ =5.5$

$\therefore$ %

$\quad to\quad escape=\cfrac { 5.5-1 }{ 5.5 } \times 100$

Find the lifting power of a

$100$ litre balloon filled with

$He$ at

$730\ mm$ and

$25^oC$ .

(Density of air

$=1.25\ g/L$ ).

0%
$125.23g$
0%
$95.28g$
0%
$109.28g$
0%
$104g$

Explanation

Since,

$PV = nRT$

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

$\therefore W=\dfrac{PVM}{RT} = \dfrac{730}{760} \times \dfrac{100\times 4}{0.082\times 298}g$

i.e., Wt. of

$He = 15.72\ g$

Wt. of air displaced

$=100\times 1.25\ g/L=125 \ g$

Lifting power of the balloon

$=125\ g - 15.72\ g = 109.28\ g$

If the absolute temperature of a gas is doubled and the pressure is reduced to one-half, the volume of the gas will

0%
remain unchanged
0%
be doubled
0%
increase four-fold
0%
be reduced to $1/4th$

Explanation

We know that

$\frac{P_1V_1}{T_1}$ =

$\frac{P_2V_2}{T_2}$

Volume of the gas = v

Pressure

$P_1$ = p

Pressure

$P_2$ =

$\frac{1p}{2}$

Temperature =

$T_1$ = t

Temperature =

$T_2$ = 2t

$V_2$ =

$P_1V_1T_2/T_1$

$( p\times v\times2t)$ /

$(2t\times p)$

= v

That means

$V_2 = V_1$

So it increases four-fold so C is the correct option.

An ideal solution has equal mol-fractions of two volatile components A and B in the vapour above the solution, the mol-fractions of A and B.

0%
are both $0.50$
0%
are equal but necessarily $0.50$
0%
are not very likely to be equal
0%
are $1.00$ and $0.00$ respectively

Which of the following has maximum vapour pressure at a given temperature?

Explanation

Vapour pressure increases with decreasing net molecular weight of a compound. Now among

$A,B,C$ and

$D$ in and

$D$ intermolecular

$H-$ bonding is present as a result their net molecular weight increases but in

$A$ and

$C$ due to presence of intermolecular

$H-$ bonding their net molecular weight is approximately equals to their absolute molecular weight. Now intermolecular

$H-$ bonding is more feasible in

$C$ due to presence of negative charge on

${ NO }_{ 3 }^{ \left( - \right) }$ ion.

But absolute molecular weight of

$A$ is less than

$C$ . So vapour pressure of

$A$ will be highest.

1101541_1028448_ans_471a893c9e2142179d5645ff29b506a0.png

Two liquids X and Y form an ideal solution. At

$300$ K, vapour pressure of the solution containing

$1$ mol of X and

$3$ mol of Y is

$550$ mm Hg. At the same temperature, if

$1$ mol of Y is further added to this solution, vapour pressure of the solution increases by

$10$ mmHg. Vapour pressure (in mm Hg) of X and Y in their pure states will be ________ respectively.

0%
$200$ and $300$
0%
$300$ and $400$
0%
$400$ and $600$
0%
$500$ and $600$

Explanation

Let

$P=x_xP^0_x+x_yP^0_y$

$(\dfrac{1}{1+3})P_x^0+(\dfrac{3}{1+3})P_y^0=550\;mm\;of\;Hg$

$(\dfrac{1}{1+4})P_x^0+(\dfrac{4}{1+4})P_y^0=560\;mm\;of\;Hg$

on solving for

$P_y^0$ and

$P_x^0$

We get,

$P_x^0=400\;mm\;of\;Hg$

$P_y^0=600\;mm\;of\;Hg$

If you are given Avogadro's number of atoms of a gas

$X$ . If half of the atoms are converted into

$X_{(g)}^+$ by energy

$\Delta H$ . The IE of

$X$ is :

0%
$\dfrac{2\Delta H}{N_A}$
0%
$\dfrac{2N_A}{\Delta H}$
0%
$\dfrac{\Delta H}{2N_A}$
0%
$\dfrac{N_A}{\Delta H}$

Explanation

Given no. of atoms = Avogadro's no. of atoms =

${N}_{A} = 6.023 \times {10}^{23}$

Given that

$\cfrac{{N}_{A}}{2}$ atoms are ionized, i.e.,

Ionization energy of

$\cfrac{{N}_{A}}{2}$ atoms of gas X =

$\Delta{H}$

$\therefore$ Ionization energy of 1 atom of gas X =

$\cfrac{\Delta{H}}{\left( \cfrac{{N}_{A}}{2} \right)} = \cfrac{2. \Delta{H}}{{N}_{A}}$

Hence, Ionisation energy of gas X is

$\cfrac{2. \Delta{H}}{{N}_{A}}$ .

A solution at

$20^o$ C is composed of

$1.5$ mol of benzene and

$3.5$ mol of toluene. If the vapour pressure of pure benzene of pure benzene and pure toluene at this temperature are

$74.7$ torr and

$22.3$ torr. respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively.

0%
$38.0$ torr and $0.589$
0%
$30.5$ torr and $0.389$
0%
$35.8$ torr and $0.280$
0%
$35.0$ torr and $0.480$

Explanation

Let us the problem.

According to the raoult's law is used which states that for a solution of volatile liquids, the partial vapour pressure of each liquid in the solution is directly proportional to its mole fraction.

${X_{benzene}} = \frac{{1.5}}{5} = 0.3$

${X_{toluence}} = \frac{{3.5}}{5} = 0.7$

${P_T} = 74.7 \times 0.3 + 0.7 \times 22.3$

$= 22.41 + 15.61$

$= 38.02\,torr$

${Y_{benzene}} = \frac{{22.41}}{{38.02}} = 0.589$

Under the same conditions how many

$ml$ of

$1\ M$

$KOH$ and

$0.5\ M\ H_2SO_4$ solutions, respectively, when mixed to form a total volume of

$100\ mL$ produce the highest rise in temperature?

0%
$67, 33$
0%
$33, 67$
0%
$40, 60$
0%
$50, 50$

What is the density of wet air with

$75\%$ relative humidity at

$1$ atm and

$300$ K?

[Given: vapour pressure of

$H_2O$ is

$30$ torr and average molar mass of air is

$29$ g

$mol^{-1}$ ]

0%
$1.614$ g/L
0%
$0.96$ g/L
0%
$1.06$ g/L
0%
$1.164$ g/L

Explanation

Let us consider the equation

${P_{wet\,air}} = \dfrac{{{P_d}}}{{{R_d}T}} + \dfrac{{{P_v}}}{{{R_v}T}} = \dfrac{{{P_d}{M_d} + {P_v}{M_v}}}{{RT}}$

Given that :

$T=$ temperature

$=300\;K$

$M_d=$ molar mass of dry air

$=29\;gmol^{-1}$

$M_v=$ molar mass of water vapourr

$=18\;gmol^{-1}$

$R=0.821\;L\;atm\;mol^{-1}K^{-1}$

$P_v=30\;torr=0.0394\;atm$

$P_d=P-P_v=1-0.0394=0.9606\;atm$

where

$P_d$ partial pressure of dry air

$R_d=$ specific gas constant for dry air

$R_v=$ Specific gas constant for water vapour

${P_{wet\,air}} = \dfrac{{0.9606 \times 29 + 0.0394 \times 18}}{{24.63}}$

$=1.164\;gL^{-1}$

Hence, the correct option is

$D$

Which of the following statements is/are wrong?

0%
At a given temperature the transitional K.E. of one mole of every gas is same which is equal to $3/2$ RT
0%
K.E. of a gas depends on its mass
0%
K.E. depends on the volume
0%
K.E. depends on the pressure

Under

$3$ atm,

$12.5$ litre of a gas weighs

$15$ gm. What will be the average speed of gaseous molecules?

0%
$8.028\times 10^4$ cm/sec
0%
$6.028\times 10^2$ cm/sec
0%
$8.028\times 10^5$ cm/sec
0%
$6.028\times 10^4$ cm/sec

Explanation

Given,

Pressure = 3atm

Volume = 12.5 litre

Weight = 15 g

For gas We have

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

$\implies 3×12.5=\dfrac{15}{m}×0.0821×T$

$T_m=30.45$

Now,

$U_{AV}=\sqrt{\dfrac{8RT}{πm}}$

$=\sqrt{\dfrac{8×8.314×107×30.4×7}{22}}$

$=8.028×10^4cm\ sec^{−1}$

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:

$PV=nRT$

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

$3.00\ L$ 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 glass container is sealed with a gas at

$0.800$ atm pressure and at

$25^o$ C. The glass container sustain pressure of

$2$ atm. Calculate the temperature to which gas can be heated before bursting the container.

0%
$432^o$ C.
0%
$472^o$ C.
0%
$372^o$ C.
0%
$572^o$ C.

Explanation

We have,

$p_1=0.8$ atm

$,p_2=2$ atm and

$T_1=298$ k

Thus,

According to the Gas equation,

$\dfrac{p_1}{T_1}=\dfrac{p_2}{T_2}$

$\dfrac{0.8}{298}=\dfrac{2}{T_2}$

$T_2=745K$

$T_2=472^0C$

Hence, the correct option is

$\text{B}$

A sample of water gas contains

$42\%$ by volume of carbon monoxide. If the total pressure is

$760$ mm of Hg, the partial pressure of carbon monoxide is :

0%
$380$ mm of Hg
0%
$319.2$ mm of Hg
0%
$38$ mm of Hg
0%
$360$ mm of Hg

Explanation

$1.$

$42\%$ of CO is released

$100\%$

$-$ 28g

$42\%$

$-$

$x$

$x = \dfrac{42\times28}{100}$

$x =11.76$

$2.$ for

$H_2O$

100%

$-$ 18g

58%

$-$

$x$

$x = \dfrac{58\times16}{100}$

$x = 10.44$

mole of CO

$= \dfrac{11.76}{28} = 0.42$

moles of

$H_2O$

$= \dfrac{10.44}{18} = 0.58$

Dalton's law (p)

$= {P_T}\times{X_Q}$

$= 760\times{\dfrac{0.42}{0.42+0.58}}$

$P = 319.2mm$

The correct option is B.

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

$(P_T)=72 torr$

At what relative humidity will

$Na_2SO_4$ be deliquescent (absorb moisture) when exposed to the air at

$0^o$ C?

Given:

$Na_2SO_4\cdot 10H_2O(s)\rightleftharpoons Na_2SO_4(s)+10H_2O(g); K_p=4.08\times 10^{-25}$ and vapour pressure of water

$0^o$ C

$=4.58$ Torr.

0%
Below $50.5\%$ but Above $30.5\%$
0%
Above $50.5\%$
0%
Above $30.5\%$
0%
Below $30.5\%$

Assume centre of sun to consist of gases whose average molecular weight isThe density and pressure of the gas are

$1.3g{ cc }^{ -1 }$ and

$1.12\times { 10 }^{ 9 }$ atm respectively. The temperature of sun is:

0%
$2\times { 10 }^{ 3 }K$
0%
$2\times { 10 }^{ 5 }K$
0%
$2\times { 10 }^{ 7 }K$
0%
$2\times { 10 }^{ 9 }K$

Explanation

We know tha,

$P=\dfrac{dRT}{m}$

Placing the values provided in the question, we get,

$T=\dfrac{Pm}{dR}$

$=\dfrac{1.12\times10^9×2}{1.3×1000×0.0821}$

$\implies T=2.076×10^7K$

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

$T=27^{0}C$

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.

To expel half the mass of air from a large flask at

${ 27 }^{ 0 }C$ , it must be heated to:

0%
$54^{ 0 }C$
0%
$177^{ 0 }C$
0%
$277^{ 0 }C$
0%
$327^{ 0 }C$

Explanation

To expel half the mass of air, we need to double the volume of air so that half of it remains inside the flask. The ideal gas equation contains variables of pressure, volume, no. of moles and temperature.
At the given conditions, we assume the pressure to be constant and the mass of air is not changing therefore no. of moles is also constant. Hence, we are left with volume and temperature variables.

According to Charles’ Law: V is directly proportional to T in Kelvin

Therefore, to double the Volume we need to double the Temperature in Kelvin.

We know that, $27^{0} C =300K$

$\implies T_2=600K =327^{0} C$

The elements which normally exist in the liquid state are :

0%
Bromine and iodine
0%
Mercury and chlorine
0%
iodine and mercury
0%
Bromine and mercury

Volume of an object is

$20$

$cm^3$ and the mass in

$50$ g. Density of water is

$1$ g

$cm^{-3}$ . Does the object float on water?

0%
True
0%
False

Explanation

If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.

Density =

$\frac{Mass}{Volume}$ =

$\frac{50}{20}$ =2.5 g/

$cm^3$

Hence, The object will sink.

If 2 moles of A and 3 moles of B are mixed to form an ideal solution vapour pressure of A and B are 120 and 180 mm of Hg respectively. then the composition of A and B in the Vapour phase when the first traces of vapour are formed in the above case is:

0%
$X^l A = 0.407$
0%
$X^l A = 0.8$
0%
$X^l A = 0.109$
0%
$X^l A = 0.307$

Explanation

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

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

$X^1_A=\dfrac{P^0_A\times A}{P_{total}}$

$=\dfrac{120\times \dfrac{2}{5}}{120\times \dfrac{2}{5}+180\times \dfrac{3}{5}}$

$=\dfrac{48}{48+108}$

$X^1_A=0.307$

The density of a gas is equal to:

(

$P =$ pressure,

$V=$ volume,

$T=$ temperature,

$R=$ gas constant,

$n=$ number of moles and

$M=$ molecular weight).

0%
$nP$
0%
$\dfrac{PM_w}{RT}$
0%
$\dfrac{P}{RT}$
0%
$\dfrac{M_w}{V}$

Explanation

According to ideal gas equation for monoatomic gas,

$PV=RT$

Volume

$=\dfrac { Mass/Molecular\quad weight }{ Density(D) }$

$\dfrac { P{ M }_{ w } }{ D } =RT$

$D=\dfrac { P{ M }_{ w } }{ RT }$

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:

$P_T=x_AP_A^o+x_BP_B^o$

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

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

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

$CH_4= 1.2\ atm$

Which system contains molecules with the same average kinetic energy as the molecules in

$10.0$ g of carbon dioxide at

$10^o$ C?

0%
$10$ g of $CO_2$ at $40^o$ C
0%
$20$ g of $CO_2$ at $20^o$ C
0%
$20$ g of $CO_2$ at $5^o$ C
0%
$40$ g of $CO_2$ at $10^o$ C

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.