MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 5

At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole percent which at

$80$ is in equilibrium with vapour containing 30 mole percent of benzene.

0%
24, 56
0%
26 ,54
0%
56 24
0%
34 ,76

Which of the following relationships for various gas laws is not correct?

0%
$V_{t} = V_{0} + \dfrac {V_{0}}{273}\times t$
0%
$\dfrac {V_{1}}{T_{1}} = \dfrac {V_{2}}{T_{1}} (constant\ P)$
0%
$\dfrac {P_{1}}{T_{1}} = \dfrac {P_{2}}{T_{1}} (constant\ V)$
0%
$\dfrac {P_{1}T_{1}}{V_{1}} = \dfrac {P_{2}T_{2}}{V_{2}}$

Explanation

According to ideal gas equation

$PV=nRT$

$\therefore { P }_{ 1 }{ V }_{ 1 }=nR{ T }_{ 1 }$ or

${ P }_{ 1 }{ V }_{ 1 }/{ T }_{ 1 }=nR$

${ P }_{ 2 }{ V }_{ 2 }=nR{ T }_{ 2 }$ or

${ P }_{ 2 }{ V }_{ 2 }/{ T }_{ 2 }=nR$

$\Rightarrow \cfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\cfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } }$

Hence

$\cfrac { { P }_{ 1 }{ T }_{ 1 } }{ { V }_{ 1 } } =\cfrac { { P }_{ 2 }{ T }_{ 2 } }{ { V }_{ 2 } }$ is incorrect

$4\ moles$ of an ideal gas at

$300\ K$ occupy volume of

$89.6\ L$ , then pressure of the gas will be :

0%
$2\ atm$
0%
$1\ atm$
0%
$1.099\ atm$
0%
$2.910\ atm$

Explanation

By Ideal gas equation,

$PV = nRT$

$P = \dfrac {4\times 0.0821\times 300}{89.6} = 1.099\ atm$ .

Which of the following values does not represent the correct value of

$R$ ?

0%
$8.314\ Pa\ m^{2} K^{-1} mol^{-1}$
0%
$8.314\times 10^{-2} bar\ L\ K^{-1} mol^{-1}$
0%
$0.0821\ J\ K^{-1} mol^{-1}$
0%
$8.314\ J\ K^{-1} mol^{-1}$

Explanation

The value of ideal gas constant in vaious units can be given as follow:

$R = 8.314\ Pa\ m^{3}\ K^{-1} mol^{-1}$

$R = 8.314\times 10^{-2} bar\ L\ K^{-1} mol^{-1}$

$R = 8.314\ J\ K^{-1} mol^{-1}$

$R = 0.0821\ L\ atm\ K^{-1} mol^{-1}$ .

Thus value given in option C is incorrect.

How many number of moles of nitrogen will be present in

$2.24\ L$ of nitrogen gas at

$STP?$

0%
$9.9$
0%
$0.099$
0%
$0.001$
0%
$1.00$

Explanation

$PV = nRT$

$P = 1\ atm, T = 273\ K, V = 2.24\ L$

$R = 0.0821\ L\ atm\ K^{-1} mol^{-1}$

$n = \dfrac {1\times 2.24}{0.0821\times 273} = 0.099$ moles.

The statements for laws of chemical combinations are given below. Mark the statement which is not correct.

0%
Matter can neither be created nor destroyed: Law of conservation of mass
0%
A compound always contains exactly the same proportion of elements by weight: Law of definite proportions
0%
When gases combine they do so in a simple ratio by weight: Gay Lussac's Law
0%
Equal volumes of gases at same temperature and pressure contain same number of molecules: Avogadro's Law

What volume in litres will be occupied by

$4.4$ g of

$CO_2$ at STP?

0%
$22.4\ L$
0%
$44.8\ L$
0%
$12.2\ L$
0%
$2.24\ L$

Explanation

$44\ g$ of

$CO_{2}$ occupies

$22.4\ L$

$4.4\ g$ of

$CO_{2}$ occupies

$\dfrac {22.4}{44}\times 4.4 = 2.24\ L$ .

The density of

$O_2$ is 16 kg/m

$^3$ at NTP. At what temperature its density will be 19? consider that the pressure remains constant.

0%
$39^oC$
0%
$43^oC$
0%
$57^oC$
0%
None of these

Explanation

As we know

$PM=dRT$

For

$O_2$ M is constant

Given P is constant

Thus

$d\alpha\cfrac{1}{T}\quad dT=constant$

at NTP,

$T=20°C=293K,d_1=16$

Given

$d_2=19,T_2?$

$d_1T_1=d_2T_2\\16\times293=19\times T_2\\ T_2=246.73K\Rightarrow -26.26°C$

$6 g$ . of the area of dissolved in

$90 g$ . of boiling water. The vapour pressure of the solution is:

0%
$744.8$ mm
0%
$758$ mm
0%
$761$ mm
0%
$760$ mm

Explanation

Vapour pressure

$P=P°x,\quad P°=$ Boiling point of pure solvent

$P=P°\cfrac { { n }_{ 1 } }{ { n }_{ 1 }+{ n }_{ 2 } }$

$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right) \quad \quad 90g{ H }_{ 2 }O=5mol={ n }_{ 2 };\quad 6g$ urea

$=0.1mol={ n }_{ 1 }$

$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right)$

$=760\times 0.0196=14.89$ Torr=lowered pressure

Now, 1 torr = 1 mm of Hg

Vapour pressure of solution

$=760-14.89\approx 744.8 mm\ of\ Hg$

Option A is correct.

Which of the following statement about Avogadro's hypothesis is correct?

0%
Under similar conditions of temperature and pressure, gases react with each other in simple ratio.
0%
Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules
0%
At NTP all gases contain same number of molecules
0%
Gases always react with gases only at the given temperature and pressure

The volume occupied by $88\ g$ of $CO_2$ at $30^{\circ}C$ and $1\ bar$ pressure will be:

0%
$5.05\ L$
0%
$49.8\ L$
0%
$2\ L$
0%
$55\ L$

Explanation

$w = 88\ g, M = 44\ g\ mol^{-1}, T = 273 + 30 = 303\ K$

$P = 1\ bar, V = ?$

$PV = nRT$ or

$V = \dfrac {nRT}{P} = \dfrac {w}{m}\dfrac {RT}{P}$

$V = \dfrac {88\times 0.0821\times 303}{44\times 1} = 49.8\ L$ .

To which of the following the Dalton's law of partial pressures is not applicable?

0%
$H_{2}$ and $He$
0%
$NH_{3}$ and $HCl$
0%
$N_{2}$ and $H_{2}$
0%
$N_{2}$ and $O_{2}$

Explanation

Daltons' law of partial pressure is applicable to non-reacting gases,

$NH_{3}$ and

$HCl$ are reacting gases, so Dalton's law will not be applicable for them.

$10\ L$ flask contains a gaseous mixture of

$CO$ and

$CO_{2}$ at a total pressure of

$2\ atm$ and

$298\ K$ . If

$0.20\ mole$ of

$CO$ is present, find its partial pressure.

0%
$0.49\ atm$
0%
$1.51\ atm$
0%
$1\ atm$
0%
$2\ atm$

Explanation

Here,

$V=10L$ ,

$n=0.20mole$ , and

$T=298K$

By Ideal gas equation

$PV=nRT$

$\therefore P = \dfrac {nRT}{V} = \dfrac {0.20\times 0.0821\times 298}{10} = 0.49\ atm$ .

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

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.

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

Equal masses of helium and oxygen are mixed in a container at

$25^{\circ}C$ . The fraction of the total pressure exerted by oxygen in the mixture of gases is :

0%
$1/3$
0%
$2/3$
0%
$1/9$
0%
$4/9$

Explanation

Mole fraction of

$O_{2} = \dfrac {\dfrac {w}{32}}{\dfrac {w}{32} + \dfrac {w}{4}} = \dfrac {w}{32}\times \dfrac {32}{9w} = \dfrac {1}{9}$
Partial pressure of oxygen

$= P \times x_{O_{2}}$

$= P_{total}\times \dfrac {1}{9}$ or

$\dfrac {1}{9}P_{total}$ .

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.

The kinetic energy of 4 moles of nitrogen gas at 127 is: (R=2 cal

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

0%
4400 cal
0%
3200 cal
0%
4800 cal
0%
1524 cal

Explanation

Answer:-

Kinetic theory of gases is defined as -

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

where,

n = no. of moles

$=4\ moles$ (given)

T = temperature

$= 127℃$ (given)

$= (273 + 127)K = 400K$

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

$\therefore \; K.E. = \cfrac{3}{2} \times 4 \times 2 \times 400 = 4800 \; cal$

$3$ 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 temperature of

$5$ mL of a strong acid increases by

$5^o$ C when

$5$ mL of a strong base is added to it. If

$10$ mL of each is mixed, temperature should increase by:

0%
$5^o$ C
0%
$10^o$ C
0%
$15^o$ C
0%
cannot be known

Explanation

Let the concentration of strong acid and a strong base is c M.

For the reaction

$H^++OH^- \rightarrow H_2O$

1 mol of acid reacts with 1 mol of base to give 1 mol of water and generated heat H J

Thus, on mixing 5 mL i.e

$5c \times 10^{-3}$ mol of each, amount of heat generated =

$H \times 5c \times 10^{-3}\ J$

Also, heat generated

$q=m.C_p \Delta T$

where m=mass of solution, C=specific heat of water,

$\Delta T$ =temperature change. let the density of solution =

$\rho\ g/mL$

For 5 mL of each acid and base:

$H \times 5c \times 10^{-3}\ J=\rho \times(5+5) \times C \times \Delta T_1$

$\Delta T_1=\frac{H5c \times 10^{-3}}{10C\rho}$

$\Delta T_1=\frac{Hc}{2C\rho} \times 10^{-3}$

Similarly for 10 mL of each acid and base solution :

$H \times 10c\ mJ=\rho \times(10+10) \times C \times \Delta T_2$

$\Delta T_2=\frac{10cH}{20C\rho} \times 10^{-3}$

$\Delta T_2=\frac{Hc}{2C\rho} \times 10^{-3}$

Thus

$\Delta T_1=\Delta T_2$

In general,

Reaction of strong acid with strong base is a neutralization reaction. The enthalpy change for neutralization of strong acid and strong base is constant an has a value of -57.1 kJ/mol.

$H^++OH^- \rightarrow H_2O; \Delta_{neut}H^0=-56.1 kJ/mol$

The temperature of 5 ml of a strong acid increases by

$5^{\circ}C$ when 5 ml of a strong base is added to it. If 10 ml of each is mixed, temperature should increase by

$5^{\circ}C$ .

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.

$H_2\\1vol$

$+$

$Cl_2=\\1 vol$

$2HCl\\2 vol$

The

$U-$ tube shown in the figure has a uniform cross section. A liquid is filled up to height

${ h }_{ 1 }$ and

${ h }_{ 2 }$ in the two arms and then they are allowed to move, Neglecting viscosity and surface tension, find the state of the liquid when the levels equalize in the two arms. Find the correct statement.

0%
The liquid will be at rest
0%
The liquid will be moving with acceleration
0%
The liquid will be moving with velocity

$\sqrt { \dfrac { g }{ a({ h }_{ 1 }+{ h }_{ 2 }+h) } }$
0%
The liquid will exert as force to the right on the tube

A balloon filled with

$N_2O$ is pricked with a sharper point and plunged into a tank of

$SO_2$ under the same pressure and temperature. The balloon will:

0%
enlarged
0%
shrink
0%
collapse
0%
remain unchanged

Explanation

According to Graham's law:

$\frac{{{\gamma _{{H_2}}}}}{{{\gamma _{C{H_4}}}}} = \sqrt {\frac{{{M_{C{H_4}}}}}{{{M_{{H_2}}}}}}$

Implies that

$= \sqrt {\frac{{16}}{2}}$

Implies that

$= \frac{4}{{\sqrt 2 }}$

$\frac{{{\gamma _{{H_2}}}}}{{{\gamma _{C{H_4}}}}} = 2\sqrt 2 {\gamma _{C{H_4}}}$

Hence the enlarged.

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.

For an ideal gas, a number of moles per liter in terms of its pressure P, gas constant R and temperature T is:

0%
PT/R
0%
PRT
0%
P/RT
0%
RT/P

Explanation

According to Ideal Gas Equation,

$PV =nRT$

P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature.

Therefore number of moles per litre,

$\frac{n}{v}$ =

$\frac{P}{RT}$

Therefore, C is the correct option.

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

$Hg$ .

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

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

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

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

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

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

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

$X^1_A=0.307$

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%
$50, 50$

$2\ L$ of

$SO_2$ gas at

$760$

$mm\ Hg$ is transferred to

$10$ L flask containing oxygen at a participate temperature, the particular pressure of

$SO_2$ in the flask is:

0%
$63.3 mm\ Hg$
0%
$152\ mm\ Hg$
0%
$760\ mm\ Hg$
0%
$1330\ mm\ Hg$

Explanation

Let us consider the problem :-

Putting

$2$ litre of any gas into a

$10$ litre flask. and temprature is constant.

Since pressure drop by a factor of

$\frac{10}{2}=5$

Hence the partial pressure of

$SO_2$

$\frac{{760\,\,mm\,\,Hg}}{5}$

$=152$ mm Hg.

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

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,

$U_{AV}=\sqrt{\dfrac{8RT}{πm}}$

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

$=8.028×10^4cm\ sec^{−1}$

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

$\text{B}$

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

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

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

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

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

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

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.

A gaseous mixture containing

$0.35 g$ of

$N_2$ and 5600 ml of

$O_2$ at STP is kept in a 5 litres flask at 300K. The total pressure of the gaseous mixture is :

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

Explanation

As we know that,

$\text{No. of moles} = \cfrac{\text{mass}}{\text{molar mass}}$

Mass of

${N}_{2} = 0.35 \; g$

Molar mass of

${N}_{2} = 28 \; g$

No. of moles of

${N}_{2} = \cfrac{0.35}{28} = 0.0125 \text{ mol}$

Volume of

${O}_{2} = 5600 \; mL$

Volume at NTP

$= 22400 \; mL$

No. of moles of

${O}_{2} = \cfrac{5600}{22400} = 0.25 \text{ mol}$

Given that :

Volume of flask

$\left( V \right) = 5 \; L$

Temperature of flask

$\left( T \right) = 300 \; K$

Now from ideal gas law :

$PV = nRT$

$\Rightarrow P = \cfrac{nRT}{V}$

Therefore,

Partial pressure of

${O}_{2}, \left( {P}_{{O}_{2}} \right) = \cfrac{0.25 \times 0.0821 \times 300}{5} = 1.2315 \; atm$

Partial pressure of

${N}{2}, \left( {P}_{{N}_{2}} \right) = \cfrac{0.0125 \times 0.0821 \times 300}{5} = 0.0615 \; atm$

$\therefore$ Total pressure,

$\left( P \right) = {P}_{{O}_{2}} + {P}_{{N}_{2}} = 1.2315 + 0.0615 = 1.293 \; atm$

The correct option is A.

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 $30.5\%$

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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