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CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 9 - MCQExams.com
CBSE
Class 11 Medical Chemistry
States Of Matter Gases And Liquids
Quiz 9
The number of $$H_{3}O^{+}$$ ions present in 10ml of water at $$25^{\circ}C$$ is:
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0%
$$6.023 \times 10^{-14}$$
0%
$$6.023 \times 10^{14}$$
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$$6.023 \times 10^{-19}$$
0%
$$6.023 \times 10^{19}$$
Explanation
No. of hydronium ions present in $$10$$ml water at $${ 25 }^{ 0 }C$$.
$$pH=7$$ at $${ 25 }^{ 0 }C$$
$$\left[ { H }^{ + } \right] ={ 10 }^{ -7 }M$$
No. of moles $$=$$ Concentration $$\times$$ Volume
$$=\left( { 10 }^{ -7 } \right) \left( 10 \right) \times { 10 }^{ -3 }$$
$$={ 10 }^{ -9 }$$moles
1 mole has $$6.023\times { 10 }^{ 23 }$$ ions of hydronium.
$$\Rightarrow \quad 6.023\times { 10 }^{ 23 }\times { 10 }^{ -9 }$$ ions in $$10$$ml water
$$\Rightarrow 6.023\times { 10 }^{ 14 }$$ ions are there in $$10$$ml water at $${ 25 }^{ 0 }C$$.
Hence, option B is correct.
The vapour pressure of a solution containing 5 g of a non electrolyte in 100 g of water at a particular temperature is $$2985 N{m}^{-2}$$. If the vapour pressure of pure water is $$3000 N{m}^{-2}$$, the molecular weight of solute is:
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$$60$$
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$$120$$
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$$180$$
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$$380$$
Explanation
we have formula of $$\text{Relative lowering of vapour pressure}$$ which is
$$\dfrac{P_o-P}{P_o}=X_{solute}=\dfrac{n_{solute}}{n_{solute}+n_{solvent}}\approx\dfrac{n_{solute}}{n_{solvent}}$$
Now, pute the values
$$\dfrac{3000-2985}{3000}=\dfrac{\dfrac{5}{M}}{\dfrac{100}{18}}$$
$$\dfrac{15}{3000}=\dfrac{5\times 18}{100\times M}$$
$$M=180$$
correct answer is C.
The temperature and pressure at which ice, liquid. water and water vapour can exist together are ?
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$${ 0 }^{ o }C,\quad 1 atm $$
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$${ 2 }^{ o }C,\quad 4.7atm$$
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$${ 0 }^{ o }C,\quad 4.7mm$$
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$${ -2 }^{ o }C,\quad 4.7mm$$
Explanation
The temperature and pressure at which ice, liquid, water and water vapour exist together at triple point of water.
Triple point of water is $$0^\circ C \ and \ 4.7mm$$
Hence, Option "C" is the correct answer.
Which of the following represents the Avogadro number?
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Number of molecule present in 1 L of gas at N.T.P
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Number of molecule present in 22.4 L of gas at N.T.P
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Number of molecule present in 22.4 of gas at 298K and 1 atm pressure
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Number of molecule present in one mole of gas at any temp and pressure
Explanation
Avogadro number is defined as the number of gas molecules present in one mole of gas at any pressure and temperature. It's value is $$6.023 \times 10^{23}$$. So the correct option is D
According to the equipartition principle, each degree of freedom contribute following amount to the translational kinetic energy of molecule: (hence k = Boltzmann constant)
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$$\dfrac{1}{2} kT$$
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$$kT$$
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$$\dfrac{3}{2} kT$$
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$$\dfrac{1}{3} kT$$
Explanation
According to the equipartition principle, each degree of freedom contributes an amount of $$\dfrac{1}{2}kT$$ to the kinetic energy of the molecule.
The solubility of a specific non-volatile salt is $$4\mathrm { g }$$ in $$100\mathrm { g }$$ of water at $$25 ^ { \circ } \mathrm { C } .$$ If $$2.0 g,$$ $$4.0\mathrm g$$ and $$6.0 g$$ of the salt added of $$100 g$$ of water at $$25 ^ { \circ } \mathrm { C } ,$$ in system $$X , Y$$ and $$Z .$$ The vapour pressure would be in the order:
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$$X < Y < Z$$
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$$X > Y > Z$$
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$$Z > X = Y$$
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$$x > Y = z$$
A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius $$R$$ and volume $$V$$. If $$T$$ is the surface tension of the liquid, then.
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Energy = $$4VT(\frac{1}{r} + \frac{1}{R})$$ is released
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Energy = $$3VT(\frac{1}{r} - \frac{1}{R})$$ is absorbed
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Energy = $$3VT(\frac{1}{r} - \frac{1}{R})$$ is released
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Energy is neither released nor absorbed
Explanation
Since intial volume=final volume
Let's say number of drops $$=N$$
Then energy released $$=T,\Delta A$$
$${A_i} = \left( {4\pi {r^2}} \right).n$$
$$n=$$number of drops
Total volume $$ = n.\left( {\frac{{4\pi }}{3}{r^3}} \right) = \left( {n.4\pi {r^2}} \right).\frac{r}{3}$$
$$\begin{array}{l} or\, \, ,{ A_{ i } }=\left( { n.4\pi { r^{ 2 } } } \right) =\frac { { 3V } }{ r } \\ { A_{ f } }=n.4\pi { R^{ 2 } }=\frac { { 3V } }{ R } \\ \therefore Energy\, \, released\, \, =T\left( { \frac { { 3V } }{ R } -\frac { { 3V } }{ r } } \right) \\ =3VT\left( { \frac { 1 }{ r } -\frac { 1 }{ R } } \right) \end{array}$$
If volume of $$2$$ moles of a gas at $$27^\circ C$$ is $$8.21 L$$ then pressure of this gas will be.
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$$0.6$$ atm
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$$6$$ atm
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$$60$$ atm
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$$600$$ atm
If $$500ml$$ of a gas $$A$$ at $$1000$$ torr, and $$1000ml$$ of gas $$B$$ at $$800$$ torr are placed in $$2L$$ container the final pressure will be -
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$$100$$ torr
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$$650$$ torr
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$$1800$$ torr
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$$2400$$ torr
Explanation
According to Boyle's Law (Modified)
$$P_1V_1+P_2V_2+........=P_fV_f$$
Given,
$$P_1=1000;\quad V_1=500$$ $$mL$$
$$P_2=800;\quad V_2=1000$$ $$mL$$
$$P_f=?;\quad V_f=2000$$ $$mL$$
$$\implies P_1V_1+P_2V_2=P_fV_f$$
$$\implies (1000)(500)+(800)(1000)=(P_f)(2000)$$
$$\implies \cfrac {1300}{2}=P_f$$
$$\implies P_f=650$$ $$torr$$
At $$250^{o}C$$ and $$1$$ atmosphere pressure, the vapour density of $$PCI_{5}$$ is $$57.9$$ . What will be the dissociation of $$PCI_{5}-$$
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$$1.00$$
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$$0.90$$
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$$0.80$$
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$$0.65$$
Explanation
When $$20\ KJ$$ of heat is removed from $$1.2\ kg$$ of ice originally at $$-15C^{o}$$, its new temperature is
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$$-18C^{o}$$
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$$-23C^{o}$$
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$$-26C^{o}$$
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$$-35C^{o}$$
Which of the following sets of variables give a staright line with negative slope when plotted ?
( P = Vapour pressure ; T = Temperature in K)
y- axis x -axis y -axis x-axis
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$$ P\quad \quad \quad T $$
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$$ log_{10}P\quad \quad \quad T $$
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$$ log_{10}P\quad \quad \quad \frac{1}{T} $$
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$$ log_{10}P \quad \quad \quad log_{10} \frac{1}{T} $$
When the temperature of 23ml of dry $$CO_{2}$$ gas is changed from $$10^\circ$$ to $$30^\circ$$ at constant pressure of 760 mm, the volume of gas becomes closest to which one of the following?
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7.7 ml
0%
21.5 ml
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24.6 ml
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69 ml
Equation of state of an ideal gas is/are
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$$pV=(1/3)mN$$
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$$pV=nRT$$
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$$p=\rho RT/M$$
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$$p=3Mk/2V$$
A liquid will not wet the surface of a solid if the angle of contact is
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$$0 ^ { \circ }$$
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$$45 ^ { \circ }$$
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$$60 ^ { \circ }$$
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$$> 90 ^ { \circ }$$
Explanation
Hint :- Contact angle is the measure of ability of a liquid to wet the surface
Explanation:-
A liquid will not wet the surface of a solid or poor wetting if its angle of contact is more than
$$90^0$$.It means meniscus must be convex.
For example mercury.
$${\textbf{Correct option: D}}$$
Which of the following is an anti-knocking compound, that has been phased out in many countries so far?
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Lead tetrachloride
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Tetra Ethyl Lead
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Ethyl
tetrachloride
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None of these
Explanation
Given :
Which of the following is an anti-knocking compound, that has been phased out in many countries so far?
Solution :
An anti-knocking agent is a chemical compound that, when added to gasoline, raises the octane value of the gasoline.
Tetra Ethyl Lead
is an anti-knocking compound, that has been phased out in many countries so far.
The Correct Opt = B
At $$300\ K$$, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen$$(N_{2})$$ at 4 bar.The molar mass of gaseous molecule is:
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$$224 g mol^{-1}$$
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$$112 g mol^{-1}$$
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$$56 g mol^{-1}$$
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$$28 g mol^{-1}$$
A container has $$SO_{2}$$ gas at $$2\ atm$$ pressure in a vessel of $$V\ L$$ capacity, if no. of moles of $$SO_{2}$$ are doubled in the same container at the same temperature and volume. Calculate new pressure in the container.
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$$1\ atm$$
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$$4\ atm$$
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$$2\ atm$$
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$$8\ atm$$
A solution $$X$$ of $$A$$ and $$B$$ contains $$30$$ mole $$\%$$ of $$A$$ & is in equilibrium with its vapour that contains $$40$$ mole $$\%$$ of $$B$$. The ratio of $$V.P$$ of pure, $$A$$ and $$B$$ will be?
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$$2:7$$
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$$7:2$$
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$$3:4$$
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$$4:3$$
The vapour pressure of water at $$20^0C$$ is 17.54 mm. When 20 g of non-ionic, substance is dissolved in 100 g of water, the vapour pressure is lowered by 0.30 mm.What is the molecular weight of the substances.
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210.2
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208.16
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215.2
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200.8
Consider the following statements
a. Kinetic energy of a molecule is zero at $$0^0C$$.
b. A gas in a closed container will exert much higher pressure due to gravity at the bottom than at the top
c. Between collisions, the molecules move in straight lines with constant velocities.
Choose the incorrect statement(s)
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a, b & c
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a & b
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b & c
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a & c
One gram molecule of any gas at $$NTP$$ occupies $$22.4\ L$$. This fact was derived from:
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Dalton's theory
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Avogardro's hypothesis
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Berzelius hypothesis
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Law of gaseous volume
Explanation
According to Avogadro's law, 1 mole of every gas occupies 22.4L at $$NTP$$.
From ideal gas equation,
$$PV=nRT$$. . . . . . .(1)
At $$NTP$$,
$$T=273K$$
$$P=1atm$$
$$n=1$$
$$R=0.0821atm L/K mol$$
from equation (1),
$$V=\dfrac{nRT}{P}$$
$$V=\dfrac{1\times 0.0821\times 273}{1}=22.4L$$
The correct option is B.
The sample of neon gas heated from 300 K to 390 K, percentage increases in K.E is:
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$$10\%$$
0%
$$20\%$$
0%
$$30\%$$
0%
$$40\%$$
Explanation
Kinetic energy of gas is given by $$K.E = \dfrac{3}{2}nRT$$
therefore, kinetic energy is directly proportional to temperature.
so, if we find out percentage change in temperature , we can get percentage change in kinetic energy .
percentage change in temperature $$=\dfrac{\Delta T } {T}$$
$$= \dfrac{390 - 300}{90} \times 100$$
$$= \dfrac{90}{300}\times 100$$
$$= 30$$ %
hence, percentage change in kinetic energy of neon gas is 30%
$$273 ml$$ of a gas at STP was taken to $$27^oC$$ and $$600 mm$$ pressure. The final volume of the gas would be:
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273 mL
0%
300 mL
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380 mL
0%
586 mL
Explanation
We know that, for constant no. of moles of gas,
$$\dfrac{P_1 V_1}{T_1}$$ = $$\dfrac{P_2 V_2}{T_2}$$
Given,
initial conditions,
$$P_1 = 760mm$$ (STP)
$$V_1 = 273mL$$
$$T_1 = 273K$$ (STP)
Final conditions,
$$P$$
$$= 600mm$$
$$T$$
$$= 300K$$
Thus,
$$V_2 = \dfrac{P_1 V_1}{T_1} \times \dfrac{T_2}{P_2}$$ = $$\dfrac{760 \times 273 \times 300}{273 \times 600} = 380mL$$
Hence, option C is correct.
Equal amounts of two gases of molecular mass 4 and 40 are mixed. the pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is :
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0.55 atm
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0.11 atm,
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1 atm
0%
o.12 atm
Explanation
Assume 40 g each of two gases is present.
The number of moles of lighter gas are $$\displaystyle \frac {40}{4} = 10 $$.
The number of moles of heavier gas are $$\displaystyle \frac {40}{40} = 1 $$.
The mole fraction of the lighter gas is $$\displaystyle \frac {10}{10+1} = 0.9090 $$.
The partial pressure of the lighter gas is $$\displaystyle 0.9090 \times 1.1 = 1 \: atm $$.
Mole fraction of A vapours above the solution in mixture of A and B ($$X_A=0.4$$)will be
[Given : $${ P }_{ A }^{ 0 }=100mmHg\quad and\quad { P }_{ B }^{ 0 }=200mmHg$$]
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0.4
0%
0.8
0%
0.25
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none of these
If there unreactive gases having partial pressure $${P_A},{P_B}$$ and $${P_C}$$ and there moles are $$1,\,2$$ and $$3$$ respectively then their total pressure will be :
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$$P = {P_A} + {P_B} + {P_C}$$
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$$P = \cfrac{{{P_A} + {P_B} + {P_C}}}{6}$$
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$$P = \cfrac{{\sqrt {{P_A} + {P_B} + {P_C}} }}{3}$$
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None of these
Explanation
According to Dalton's law of partial pressure,
Total pressure = sum of partial pressures of individual gases
$$P_{total} = P_a + P_b + P_c$$
Therefore, option A is correct.
6 g of urea is dissolved in 90 g of boiling water.The vapour preasure of the solution is:
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$$744.8 mm$$
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$$758 mm$$
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$$761 mm$$
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$$760 mm$$
The conversion of oxygen to ozone occurs to the extent of 15% the mass of ozone that can be prepared from 67.2 L of oxygen at 1 atm and 273 K will be:
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14.4 gm
0%
96 gm
0%
640 gm
0%
64 gm
Explanation
$$3 O_2 \rightarrow 2 O_3$$
As per stoichiometric equation, $$3 × 22.4 L$$ of oxygen is converted to $$2 × 22.4 L$$ of ozone. But given 15 % of oxygen is converted to ozone.
That means 67.2 L of oxygen used $$\dfrac{15}{100}\times 67.2 \times \dfrac{2}{3} = 6.72\ L$$ ozone is produced.
The mass of ozone present in $$22.4 L = 48\ g$$
$$10.08\ L$$ of ozone means $$= \dfrac{48}{22.4} \times 6.72 =14.4\ g $$
An $$\alpha$$ - particle is projected towards the following nucleus with same kinetic energy in different experiment the distance of closet approach is maximum for
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Na(Z=11)
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Ca(Z=20)
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Ag (Z=47)
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Au (Z=79)
Explanation
K.E $$ = \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q_{1} q_{2}}{r} $$
$$q_{1} $$ = change on $$1st$$ particle
$$q_{2} $$ = charge on second particle
$$r$$ = distance between them or distance of closet approach
Assume $$q_{1} $$ charge on alpha particle
$$ = \alpha_{q} $$
$$q_{2} $$ = charge on nuclei given in question
$$ r = \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q_{1} q_{2}}{K.E} $$
K.E is same for all nuclei. But $$q_{2} $$ is maximum for gold $$(+79) $$. Hence $$r$$ will be maximum for gold.
The vapour pressure of two liquids are $$15000$$ and $$30000$$ in a unit. When equimolar solution of liquids is.. The the mole fraction of A and B in vapour phase will be:
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$$\dfrac{2}{3}, \dfrac{1}{3}$$
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$$\dfrac{1}{3}, \dfrac{2}{3}$$
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$$\dfrac{1}{2}, \dfrac{1}{2}$$
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$$\dfrac{1}{4}, \dfrac{3}{4}$$
Rain drops acquire spherical shape due to
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viscosity
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surface tension
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friction
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elasticity
Explanation
The Surface tension pulls the surface of the drop equally at all points thus produces the spherical shape having the minimum surface area
The reflective lowering of sapour pressure of a solution containing $$6\ g$$ of urea dissolved in $$90\ g$$ of water is ?
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$$0.0196$$
0%
$$0.05$$
0%
$$1.50$$
0%
$$0.01$$
Which is not correct in terms of kinetic theory of gases-
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Gases are made up of small particles called molecules
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The molecules are in random motion
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When molecules collide, they lose energy
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When the gas is heated, the molecules moves faster
Explanation
Gases are made up of small particles called molecules is a postulate of kinetic theory of gases, rest all are assumptions of kinetic theory of gases.
Hence, Option "A" is the correct answer.
A box of $$1L$$ capacity is divided into two equal compartments by a thin partition which are filled with $$2 g\ \mathrm { H } _ { 2 } \text { and } 16 \mathrm { g } \ \mathrm { CH } _ { 4 }$$ respectively. The pressure in each compartment is recorded as $$P$$ atm. The total pressure when partition is removed will be:
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$$P$$
0%
$$2P$$
0%
$$P/2$$
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$$P/4$$
The vapour pressure of a solution at $$100^{o}C$$ containing $$3\ g$$ of cane sugar in $$70\ g$$ of water? (vapour pressure of water at $$100^{o}C=760\ mm$$ of $$Hg$$) is)
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$$758\ mm$$ of $$Hg$$
0%
$$730.36\ mm$$ of $$Hg$$
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$$400\ mm$$ of $$Hg$$
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$$700\ mm$$ of $$Hg$$
The volume of a gas is held constant while its temperature is raised. The pressure of the gas exerts on the walls of the container increases because:
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The masses of the molecules increases
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Each molecule loses more kinetic energy when it strikes the wall
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The molecules collide with the wall with relatively greater momentum
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The molecules strikes the walls more often
Explanation
Volume is kept constant and temperature increased.
We have relation, $$PV=nRT$$
$$P\propto T$$
Volume constant
So, $$P\propto T$$ pressure increases as temperature increases.
As we know the speed or average molecular speed depend on temperature.
As temperature increases molecules collide with wall with greater force or momentum.
Also temperature increases, molecular vibrations increases. So molecule will strike with wall more often.
So option C and D.
A is incorrect because there is no change in masses of molecule when we increase temperature.
The vapour pressure of pure $$CHCl_{3}$$ and $$CH_{2}Cl_{2}$$ are $$200$$ and $$41.5\ atm$$ respectively. The weight of $$CHCl_{3}$$ and $$CH_{2}Cl_{2}$$ are respectively $$11.9\ g$$ and $$17\ gm$$. The vapour pressure of solution will be
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$$80.5$$
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$$79.5$$
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$$94.3$$
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$$105.5$$
Explanation
$$P_{CHCl_{3}}{0} = 200\ atm\ P_{CH_{3}Cl_{2}}{0} = 41.5\ atm$$
$$n = \dfrac {11.9}{119}\ n = \dfrac {17}{85}$$
$$= .1\ = .2$$
$$P_{T} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B}$$
$$= 200x \dfrac {0.1}{.3}\ + \dfrac {41.5\times 0.2}{.3}$$
$$= 94.33$$.
A pre-weighed vessel was filled with oxygen at $$NTP$$ and weighed. It was then evacuated, filled with $$SO_{2}$$ at the same temperature and pressure and again weighed. The weight of oxygen is ______________.
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the same as that of $$SO_{2}$$
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$$\dfrac{1}{2}$$ that of $$SO_{2}$$
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twice that of $$SO_{2}$$
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$$\dfrac{1}{4}$$ that of $$SO_{2}$$
Explanation
At constant temperature and pressure, an equal volume of gas contains an equal number of moles.
$$\therefore$$ Moles of $$O_{2}=$$ Moles of $$SO_{2}$$
$$1\ mole\ O_{2}=32\ g$$
$$1\ mole\ SO_{2}=64\ g$$
$$\therefore wt$$ of $$O_2=\dfrac{1}{2}$$ that of $$SO_{2}$$
Option $$B$$ is correct.
If water vapor comprises 3.5 percent of an air parcel whose total pressure is 1,000 mb, the water vapor pressure would be:
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1,035 mb
0%
35 mb
0%
350 mb
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965 mb
The vapour pressure of pure $$M$$ and $$N$$ are $$700$$mm of $$Hg$$ and $$450 mm$$ of $$Hg$$ respectively.which of the following option is correct?
Given :$$X_N,X_M$$- mole fraction of $$N$$ and $$M$$ in liquid phase
$$Y_N,Y_M$$= mole fraction of $$N$$ and $$M$$ in vapour phase.
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$$X_M-X_N>Y_M-Y_N$$
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$$\dfrac{X_M}{X_N}>\dfrac{Y_M}{Y_N}$$
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$$\dfrac{X_M}{X_N}<\dfrac{Y_M}{Y_N}$$
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$$\dfrac{X_M}{X_N}=\dfrac{Y_M}{Y_N}$$
Explanation
Solution:- (C) $$\dfrac{X_M}{X_N}<\dfrac{Y_M}{Y_N}$$
Mole fraction of more voltatile component increase in vapour phase, i.e.,
$$P_N^o<P_M^o$$
On increasing the altitude at constant temperature, vapour pressure of liquid
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increases
0%
decreases
0%
remains the same
0%
depends upon climate
Explanation
Vapour pressure
is a
temperature
dependent phenomena and it does not change unless
temperature
is changed.
On increasing altitude
, the outside
pressure
decreases and to maintain that the rate of evaporation and condensation also changes which keeps
vapour pressure constant
. So answer is option C.
A closed vessel contains equal number of nitrogen and oxygen molecules at a pressure of $$p\ mm$$. If nitrogen is removed from the system then the pressure will be:
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$$p$$
0%
$$2p$$
0%
$$p/2$$
0%
$$p^2$$
Explanation
At constant $$V, P\propto n$$
$$\therefore$$ if no. of moles is halved then pressure will also reduce to $$p/2$$.
Option C is correct.
For a solution formed by mixing liquids $$L$$ and $$M$$, the vapour pressure of $$L$$ plotted against the mole fraction of $$M$$ in solution is shown in the following figure. Here $$x_L$$ and $$x_M$$ represent mole fractions of $$L$$ and $$M$$ respectively in the solution. The correct statement $$\left(s\right)$$ applicable to this system is $$\left(are\right)$$
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The point $$Z$$ represents vapour pressure of pure liquid $$M$$ and Raoults law is obeyed from $$X_{L}=0$$ to $$X_{L}=1$$.
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Attractive intermolecular interactions between $$L-L$$ in pure liquid $$L$$ and $$M - M$$ in pure liquid $$M$$ are stronger than those between $$L - M$$ when mixed in solution.
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The point $$Z$$ represents vapour pressure of pure liquid $$M$$ and Raoults law is obeyed when $$X_L \to 0$$.
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The point $$Z$$ represents vapour pressure of pure liquid L and Raoults law is obeyed when $$X_L\to 1$$
Explanation
We know from Raoult's law,
$$P=P_L^ox_L+{P_M}^ox_M$$
Vapour Pressure of liquid,
$$P_L={P_L}^ox_L=P_L^o(1-x_M)$$ at $$x_M\rightarrow 0$$ or $$x_L \rightarrow 1$$
$$\therefore P_L={P_L}^o$$ This follows the graph at point Z
Option D is correct.
The statement B is also correct because the attraction in like molecules are more than that in unlike molecules.
Option B is also correct.
Ans- B, D.
Select the incorrect order of boiling point between the following compounds :
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$$N_3H < CH_3N_3$$
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$$Me_2SO_4 < H_2SO_4$$
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$$Me_3BO_3 < B(OH)_3$$
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$$BF_3 < BI_3$$
Explanation
$$\rightarrow$$ Boiling point of $$HN_3 > CH_3N_3$$ as in hydrazoic acid ($$HN_3$$) intermolecular H-bonding occurs and its bond energy is higher than weak dipole-dipole interactions present in $$CH_3N_3$$.
$$\rightarrow$$ Boiling point of $$BI_3$$ is greater than that of $$BF_3$$ as molecular weight of $$BI_3$$ is higher than that of $$BF_3$$.
$$\rightarrow$$ Due to intermolecular H-bonding in $$H_2SO_4$$ its boiling point is higher than $$Me_2SO_4$$.
$$\rightarrow$$ Similarly due to intermolecular H-bonding in $$B(OH)_3$$ its boiling point is higher. than that of $$Me_3BO_3$$.
Which of the given laws of chemical combination is satisfied by the figure?
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Law of multiple proportion
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Gay Lussac's law
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Avogadro law
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Law of definite proportion
What is air?
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Stone
0%
Element
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Compound
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Mixture
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.
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If both assertion and reason are true and reason is the correct explanation of assertion.
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If both assertion and reason are true but reason is not the correct explanation of assertion.
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If both assertion and reason are false.
1 mole of chlorine and 1 mole of hydrogen were taken in a 10 L evacuated flask along with a little charcoal. The flask was then irradiated with light until the reaction was complete. Subsequently, 5L of water was introduced into the flask and the flask was cooled to $$27^{o}C$$. The pressure exerted by the system is approximately equal to:
[Aqueous tension at $$27^{o}C = 26 \ mm$$]
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98.5 mm
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26 mm
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52 mm
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76 mm
Explanation
Since, $$HCl$$ is highly soluble in water, no gas is present in the flask. The pressure exerted is by water vapour which is equal to aqueous tension i.e., 26 mm at $$27^oC$$.
According to the kinetic theory of gases, in an ideal gas, between two successive collisions, a gas molecule travels:
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in a wavy path
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in a straight line path
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with an accelerated velocity
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in a circular path
Explanation
According to the postulates of kinetic theory of gases, the molecules of a gas move in a straight line between two successive collisions. They change their path only when they suffer collisions with the other molecules or with the walls of the container.
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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