{"id":17583,"date":"2021-02-02T09:50:27","date_gmt":"2021-02-02T04:20:27","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=17583"},"modified":"2021-02-03T12:58:49","modified_gmt":"2021-02-03T07:28:49","slug":"mcq-questions-for-class-11-chemistry-chapter-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/mcq-questions-for-class-11-chemistry-chapter-5\/","title":{"rendered":"MCQ Questions for Class 11 Chemistry Chapter 5 States of Matter with Answers"},"content":{"rendered":"

Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 5 States of Matter with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers<\/a> are prepared as per the Latest Exam Pattern. Students can solve these States of Matter Class 11 MCQs Questions with Answers and assess their preparation level.<\/p>\n

States of Matter Class 11 MCQs Questions with Answers<\/h2>\n

Solving the States of Matter Multiple Choice Questions of Class 11 Chemistry Chapter 5 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on States of Matter Class 11 with answers pave for a quick revision of the Chapter thereby helping you to enhance subject knowledge. Have a glance at the MCQ of Chapter 5 Chemistry Class 11 and cross-check your answers during preparation.<\/p>\n

Question 1.
\nThree containers A, B, C of equal volume contain oxygen, neon and methane respectively at same temperature and pressure. The increasing order of their masses is
\n(a) A < B < C
\n(b) B < C < A
\n(c) C < A < B
\n(d) C < B < A<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) C < B < A
\nExplanation:
\nUnder similar conditions of temperature and pressure, equal volumes of different gases contain equal number of moles.
\nTherefore, Masses of O2<\/sub>, Ne and CH4<\/sub> will be in the ratio 32 : 20 : 16
\nSo, increasing order of their mass is oxygen > neon > methane<\/p>\n<\/details>\n

\n

Question 2.
\nA gas will approach ideal behaviour at
\n(a) Low temperature, low pressure
\n(b) Low temperature, high pressure
\n(c) High temperature, low pressure
\n(d) High temperature, high pressure<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) High temperature, low pressure
\nExplanation:
\nAt high temperature and low pressure, the gas volume becomes large and both intermolecular force as well as the molecular volume can be neglected. Under this condition postulate of kinetic theory applies appropriately and gas approaches ideal behavior.<\/p>\n<\/details>\n

\n

Question 3.
\nContainers A and B have same gas. Pressure, volume and temperature of A are all twice those of B. The ratio of number of molecules of A and B is
\n(a) 1 : 2
\n(b) 2 : 1
\n(c) 1 : 4
\n(d) 4 : 1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2 : 1
\nExplanation:
\nFor B, P1<\/sub> = P, V1<\/sub> = V, T1<\/sub> = T.
\nFor A, P1<\/sub> = 2P, = 2P, V2<\/sub> = 2P,T1<\/sub> = 2T
\nApplying ideal gas equation, PV = nRT.
\n(P1<\/sub>V1<\/sub>)\/ (n1<\/sub>RT1<\/sub>) = (P2<\/sub>V2<\/sub>)\/ (n2<\/sub>RT2<\/sub>)
\nor (PV)\/(n1<\/sub>RT) = (2p \u00d7 2V)\/(n2<\/sub>R(2T))
\nor (n2<\/sub>\/n1<\/sub>) = (2\/1)<\/p>\n<\/details>\n

\n

Question 4.
\nAccording to kinetic theory of gases,in an ideal gas,between two successive collisions a gas molecule travels
\n(a) In a circular path
\n(b) In a wavy path
\n(c) In a straight line path
\n(d) With an accelerated velocity<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) In a straight line path
\nExplanation:
\nln between two successive collisions, no force is acting on the gas molecules. Resultantly they travel with uniform velocity during this interval and hence, moves along a straight line.<\/p>\n<\/details>\n

\n

Question 5.
\nWhen did substances exist in different crystalline forms the phenomenon is called :
\n(a) Allotropy
\n(b) Polymorphism
\n(c) Polymerization
\n(d) Isomorphism<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Polymorphism
\nExplanation:
\nPolymorphism, in crystallography, the condition in which a solid chemical compound exists in more than one crystalline form; the forms differ somewhat in physical and, sometimes, chemical properties, although their solutions and vapours are identical.<\/p>\n<\/details>\n

\n

Question 6.
\nSl unit of pressure is :
\n(a) Pascal
\n(b) torr
\n(c) mm of Hg
\n(d) none of the above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Pascal
\nExplanation:
\nPressure is the effect of a force applied to an area of any surface. The basic unit of pressure is obtained from combining base units which are force over area.
\nPressure = (Force\/Area)
\nUnit of force is: Newton(N)
\nUnit of Area is: m\u00b2(metre)
\nP = N\/m\u00b2
\nAnd the unit of pressure in SI system is Pascal which is denoted by Pa.
\n1 Pa = 1 (N\/m\u00b2)<\/p>\n<\/details>\n

\n

Question 7.
\nIf the pressure of a gas is increased then its mean free path becomes:
\n(a) 0
\n(b) Less
\n(c) More
\n(d) Infinity<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Less
\nExplanation:
\nAs gas pressure increases mean free path of the gas decreases.
\nMean free path is the distance travelled by a gas molecule between two successive collisions.
\nSo, as pressure increases number of collisions also increases. Hence, mean free path decreases.\u200b<\/p>\n<\/details>\n

\n

Question 8.
\n1 atmosphere is equal to:
\n(a) 1 torr
\n(b) 760 cm
\n(c) 760 mm
\n(d) 76 torr<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 760 mm
\nExplanation:
\nStandard atmosphere, unit of pressure, equal to the mean atmospheric pressure at sea level. It corresponds to the pressure exerted by a vertical column of mercury (as in a barometer) 760 mm (29.9213 inches) high. One standard atmosphere, which is also referred to as one atmosphere, is equivalent to 101,325 pascals, or newtons of force per square metre (approximately 14.7 pounds per square inch).<\/p>\n<\/details>\n

\n

Question 9.
\nGrahams law refers to :
\n(a) Boiling point of water
\n(b) Gaseous Diffusion
\n(c) Gas Compression
\n(d) Volume changes of gases<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Gaseous Diffusion
\nExplanation:
\nGraham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. This formula can be written as:
\n(Rate1<\/sub>\/ Rate2<\/sub>) = (M2<\/sub>\/ M1<\/sub>)(1\/2)<\/sup>
\nwhere:
\nRate1<\/sub> is the rate of effusion for the first gas. (volume or number of moles per unit time).
\nRate2<\/sub> is the rate of effusion for the second gas.
\nM1<\/sub> is the molar mass of gas 1
\nM2<\/sub> is the molar mass of gas 2.<\/p>\n

Grahams law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Thus, if the molecular weight of one gas is four times that of another, it would diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate of the other (heavier gases diffuse more slowly). A complete theoretical explanation of Grahams law was provided years later by the kinetic theory of gases. Grahams law provides a basis for separating isotopes by diffusion a method that came to play a crucial role in the development of the atomic bomb<\/p>\n<\/details>\n

\n

Question 10.
\nThe rise or fall of a liquid within a tube of small bore is called :
\n(a) Surface Tension
\n(b) Capillary Action
\n(c) Viscosity
\n(d) Formation of Curvature<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Capillary Action
\nExplanation:
\nCapillarity, rise or depression of a liquid in a small passage such as a tube of small cross-sectional area, like the spaces between the fibres of a towel or the openings in a porous material. Capillarity is not limited to the vertical direction. Water is drawn into the fibres of a towel, no matter how the towel is oriented.<\/p>\n

Liquids that rise in small-bore tubes inserted into the liquid are said to wet the tube, whereas liquids that are depressed within thin tubes below the surface of the surrounding liquid do not wet the tube. Water is a liquid that wets glass capillary tubes; mercury is one that does not. When wetting does not occur, capillarity does not occur.<\/p>\n

Capillarity is the result of surface, or interfacial, forces. The rise of water in a thin tube inserted in water is caused by forces of attraction between the molecules of water and the glass walls and among the molecules of water themselves. These attractive forces just balance the force of gravity of the column of water that has risen to a characteristic height. The narrower the bore of the capillary tube, the higher the water rises. Mercury, conversely, is depressed to a greater degree, the narrower the bore.<\/p>\n<\/details>\n

\n

Question 11.
\nThe rates of diffusion of gases are inversely proportional to square root of their densities . This statement refers to :
\n(a) Daltons Law
\n(b) Grahams Law
\n(c) Avogadros Law
\n(d) None of the Above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Grahams Law
\nExplanation:
\nGrahams law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight. In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore the rate of diffusion of different gases is inversely proportional to the square root of their mass densities.
\nr \u03b1 \u221a(1\/d)
\nand r \u03b1 \u221a (1\/M)<\/p>\n<\/details>\n

\n

Question 12.
\nCooling is caused by :
\n(a) Evaporation
\n(b) Convection
\n(c) Conduction
\n(d) none of the above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Conduction
\nExplanation:
\nEvaporation is a cooling process because when liquid turns to gas. When a liquid evaporates, its molecules convert from the liquid phase to the vapor phase and escape from the surface. Heat drives this process. In order for the molecule to leave the liquid surface and escape as a vapor, it must take heat energy with it. The heat that it takes with it comes from the surface from which it evaporated. Since the molecule is taking heat with it as its leaving, this has a cooling effect on the surface left behind.<\/p>\n<\/details>\n

\n

Question 13.
\nIf helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is:
\n(a) 2 : 1
\n(b) 1 : 2
\n(c) 3 : 5
\n(d) 4 : 1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 2 : 1
\nExplanation:
\nAccording to Grahams law
\n(r1<\/sub>\/ r2<\/sub>) = \u221a (M2<\/sub>\/M1<\/sub>)
\n(rHe<\/sub>\/rCH4<\/sub>) = \u221a (16\/4)
\n= (2\/1)<\/p>\n<\/details>\n

\n

Question 14.
\nEqual masses of ethane and hydrogen are mixed in an empty container at 25\u00b0C . The fraction of total pressure exerted by hydrogen is
\n(a) 1 : 2
\n(b) 1 : 1
\n(c) 01 : 16
\n(d) 15 : 16<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 15 : 16
\nExplanation:
\nLet 30g of both are mixed.
\nMoles of H2<\/sub> = (30\/2) =15
\nMoles of C2<\/sub>H6<\/sub> = (30\/30) = 1
\nMole fraction of H2<\/sub> = (15)\/(1 + 15) = (15\/16)
\nWhich is also the fraction of total pressure executed by H2<\/sub><\/p>\n<\/details>\n

\n

Question 15.
\nThe volume of 2.8 g of carbon monoxide at 27\u00b0C and 0.0821 atm is
\n(a) 30 L
\n(b) 3 L
\n(c) 0.3 L
\n(d) 1.5 L<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 30 L
\nExplanation:
\nAccording to the ideal gas equation, we have
\nPV = nRT
\n\u200bPV = (w\/M) RT
\n\u200bV = \u200b (w\/M) (RT\/P)
\nGiven values are:
\nw = 2.8 g
\nM = Molar mass of CO = 28 g mol-1<\/sup>
\nT = 27\u00b0C = (273 + 27) = 300 K
\nP = 0.821 atm
\nR = 0.0821 L atm mol-1<\/sup> K-1<\/sup>
\nPutting the values in the formula we get :
\nV = (2.8 g \/28 g mol-1<\/sup>) \u00d7 (0.0821 L atm mol-1<\/sup> K-1<\/sup>) \u00d7 (300 K)\/(0.821 atm)
\n= 3 L<\/p>\n<\/details>\n

\n

Question 16.
\nAccording to kinetic theory of gases,in an ideal gas,between two successive collisions a gas molecule travels
\n(a) In a circular path
\n(b) In a wavy path
\n(c) In a straight line path
\n(d) With an accelerated velocity<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) In a straight line path
\nExplanation:
\nln between two successive collisions, no force is acting on the gas molecules. Resultantly they travel with uniform velocity during this interval and hence, moves along a straight line.<\/p>\n<\/details>\n

\n

Question 17.
\nStandard conditions are :
\n(a) 0\u00b0C and 14. 7mm
\n(b) 32\u00b0F and 76 cm
\n(c) 273\u00b0C and 760 mm
\n(d) 4\u00b0C and 76 m<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 32\u00b0F and 76 cm
\nExplanation:
\nStandard conditions for temperature and pressure are standard sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data.
\nUntil 1982, STP was defined as a temperature of 273.15 K (0 \u00b0C, 32 \u00b0F) and an absolute pressure of exactly 1 atm (101.325 kPa).
\nSince 1982, STP is defined as a temperature of 273.15 K (0 \u00b0C, 32 \u00b0F) and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).<\/p>\n<\/details>\n

\n

Question 18.
\nThe internal resistance to the flow of a liquid is called :
\n(a) Surface Tension
\n(b) Diffusion
\n(c) Viscosity
\n(d) Osmosis<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Viscosity
\nExplanation:
\nViscosity is a measure of a fluids resistance to flow. It describes the internal friction of a moving fluid. A fluid with large viscosity resists motion because its molecular makeup gives it a lot of internal friction. A fluid with low viscosity flows easily because its molecular makeup results in very little friction when it is in motion.
\nGases also have viscosity, although it is a little harder to notice it in ordinary circumstances.<\/p>\n<\/details>\n

\n

Question 19.
\nWhen did substances exist in different crystalline forms the phenomenon is called :
\n(a) Allotropy
\n(b) Polymorphism
\n(c) Polymerization
\n(d) Isomorphism<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Polymorphism
\nExplanation:
\nPolymorphism, in crystallography, the condition in which a solid chemical compound exists in more than one crystalline form; the forms differ somewhat in physical and, sometimes, chemical properties, although their solutions and vapours are identical.<\/p>\n<\/details>\n

\n

Question 20.
\nThe temperature above which the gas cannot be liquified by pressure alone is called :
\n(a) Melting Point
\n(b) Critical Temperature
\n(c) Transition Temperature
\n(d) Absolute Zero<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Critical Temperature
\nExplanation:
\nThe critical temperature of a gas is the temperature below which it can be liquefied by application of pressure.<\/p>\n<\/details>\n

\n

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Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 5 States of Matter with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers are prepared as per the Latest Exam Pattern. Students can solve these States of Matter Class 11 MCQs Questions with Answers and …<\/p>\n

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