{"id":17566,"date":"2023-02-15T12:00:44","date_gmt":"2023-02-15T06:30:44","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=17566"},"modified":"2023-02-15T16:32:13","modified_gmt":"2023-02-15T11:02:13","slug":"mcq-questions-for-class-11-chemistry-chapter-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/mcq-questions-for-class-11-chemistry-chapter-1\/","title":{"rendered":"MCQ Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry with Answers"},"content":{"rendered":"

Students are advised to practice the NCERT MCQ Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 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 Some Basic Concepts of Chemistry Class 11 MCQs Questions with Answers and assess their preparation level.<\/p>\n

Some Basic Concepts of Chemistry Class 11 MCQs Questions with Answers<\/h2>\n

Solving the Some Basic Concepts of Chemistry Multiple Choice Questions of Class 11 Chemistry Chapter 1 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on Some Basic Concepts of Chemistry 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 1\u00a0Chemistry Class 11 and cross-check your answers during preparation.<\/p>\n

Formula of normality<\/a> in chemistry is one of the expressions to measure the concentration of a chemical solution<\/p>\n

Question 1.
\nFormula of Ferric Sulphate is:
\n(a) FeSo4<\/sub>
\n(b) Fe (So4<\/sub>)3<\/sub>
\n(c) Fe2<\/sub> (So4<\/sub>)3<\/sub>
\n(d) Fe2<\/sub>So4<\/sub><\/p>\n

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

Answer: (c) Fe2<\/sub> (So4<\/sub>)3<\/sub>
\nExplanation:
\nIron (III) sulfate (or ferric sulfate), is the chemical compound with the formula Fe2<\/sub>(SO4<\/sub>)3<\/sub>. Usually yellow, it is a salt and soluble in water.<\/p>\n<\/details>\n

\n

Question 2.
\nApproximate atomic weight of an element is 26.89. If its equivalent weight is 8.9, the exact atomic weight of element would be
\n(a) 26.89
\n(b) 8.9
\n(c) 17.8
\n(d) 26.7<\/p>\n

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

Answer: (a) 26.89
\nExplanation:
\nAtomic weight = (Equivalent weight \u00d7 Valency)
\n= (8.9 \u00d7 3) = 26.7
\n(Valency = (26.89)\/(8.9) \u2248 3).<\/p>\n<\/details>\n

\n

Question 3.
\nThe total number of atoms represented by the compound CuSO4<\/sub>. 5H2<\/sub>O is
\n(a) 27
\n(b) 21
\n(c) 5
\n(d) 8<\/p>\n

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

Answer: (b) 21
\nExplanation:
\n21 atoms are present in the compound CuSO4<\/sub>. 5H2<\/sub>O.<\/p>\n<\/details>\n

\n

Question 4.
\nAn atom is 10 times heavier than 1\/12th of mass of a carbon atom (C – 12). The mass of the atom in a.m.u. is
\n(a) 10
\n(b) 120
\n(c) 1.2
\n(d) 12<\/p>\n

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

Answer: (a) 10
\nExplanation:
\n(1\/2)th if mass of carbon atom weighs exact as 1 gm as from (1 \u00d7 12)\/ (12) = 1 amu.
\nTherefore 10 times of this would be = (10 \u00d7 1) = 10 gms.
\nHence 10 g would be the molar mass of that atom.<\/p>\n<\/details>\n

\n

Question 5.
\n81.4 g sample of ethyl alcohol contains 0.002 g of water. The amount of pure ethyl alcohol to the proper number of significant figures is
\n(a) 81.398 g
\n(b) 71.40 g
\n(c) 91.4 g
\n(d) 81 g<\/p>\n

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

Answer: (a) 81.398 g
\nExplanation:
\nPure ethyl alcohol
\n= (81.4 – 0.002)
\n=81.398.<\/p>\n<\/details>\n

\n

Question 6.
\nWhich of the following halogen can be purified by sublimation
\n(a) F2<\/sub>
\n(b) Cl2<\/sub>
\n(c) Br2<\/sub>
\n(d) I2<\/sub><\/p>\n

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

Answer: (d) I2<\/sub>
\nExplanation:
\nSublimation is going directly from the solid to vapor state without passing through the liquid state. The classic demonstration of sublimation is iodine crystals. Heat them at one end of a sealed tube with the other end cooled. We get a beautiful violet vapor and can watch the iodine crystals form from the vapor in the cool end. Let us Wait for some time and all the solid iodine will disappear in the hot end and reappear as beautiful black crystals in the cold end.<\/p>\n<\/details>\n

\n

Question 7.
\n1 mol of CH4<\/sub> contains
\n(a) 6.02 \u00d7 1023<\/sup> atoms of H
\n(b) 4 g atom of Hydrogen
\n(c) 1.81 \u00d7 1023<\/sup> molecules of CH4<\/sub>
\n(d) 3.0 g of carbon<\/p>\n

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

Answer: (b) 4 g atom of Hydrogen
\nExplanation:
\n1 mole of CH4<\/sub> contains 4 mole of hydrogen atom i.e. 4g atom of hydrogen.<\/p>\n<\/details>\n

\n

Question 8.
\nThe prefix zepto stands for
\n(a) 109<\/sup>
\n(b) 10-12<\/sup>
\n(c) 10-15<\/sup>
\n(d) 10-21<\/sup><\/p>\n

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

Answer: (d) 10-21<\/sup>
\nExplanation:
\n1 zepto =10-21<\/sup><\/p>\n<\/details>\n

\n

Question 9.
\nWhich has maximum number of atoms?
\n(a) 24 g of C (12)
\n(b) 56 g of Fe (56)
\n(c) 27 gof Al (27)
\n(d) 108 g of Ag (108)<\/p>\n

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

Answer: (a) 24 g of C (12)
\nExplanation:
\nNumber of atoms = (number of moles \u00d7 Avogadros number (N A) )
\n\u21d2 Number of atoms in 24 g C
\n= (24\/12) \u00d7 N A= 2N A
\nNumber of atoms in 56 g of Fe
\n= (56\/56) N A = N A Number of atoms in 27 g of A1
\n= (27\/27) N A = N A Number of atoms in 108 g of Ag
\n= (108\/108)N A = N A
\nHence, 24 g of carbon has the maximum number of atoms.<\/p>\n<\/details>\n

\n

Question 10:
\nIrrespective of the source, pure sample, of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of
\n(a) Conservation of Mass
\n(b) Multiple Proportions
\n(c) Constant Composition
\n(d) Constant Volume<\/p>\n

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

Answer: (c) Constant Composition
\nExplanation:
\nThe H : O ratios in water is fixed, irrespective of its source. Hence it is law of constant composition<\/p>\n<\/details>\n

\n

Question 11.
\nHaemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (At. wt. of Fe = 56) present in one molecule of haemoglobin is
\n(a) 6
\n(b) 1
\n(c) 4
\n(d) 2<\/p>\n

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

Answer: (c) 4
\nExplanation:
\nBecause 100 gm Hb contains = 0.33 gm Fe
\nTherefore, 67200 gm Hb = (67200 \u00d7 0.33)\/ (100 gm)
\nFe gm atom of Fe = (672 \u00d7 0.33)\/(56)
\n= 4.<\/p>\n<\/details>\n

\n

Question 12.
\nThe -ve charged particles is called :
\n(a) Anion
\n(b) Cation
\n(c) Radical
\n(d) Atom<\/p>\n

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

Answer: (a) Anion
\nExplanation:
\nA charged particle, also called an ion, is an atom with a positive or negative charge.
\nThis happens whenever something called an ionic bond forms.
\nTwo particles that have different numbers of electrons (the smallest particle in an atom which is negative) start reacting to each other.
\nThe particle that has the greater amount of electrons takes the other particles electrons.
\nOne becomes positive because it lost an electron, and the other negative because it got another electron.
\nThe two particles become attracted to each other and mix together, making a new kind of particle.<\/p>\n<\/details>\n

\n

Question 13.
\nWhich of the following contains same number of carbon atoms as are in 6.0 g of carbon (C – 12) ?
\n(a) 6.0 g Ethane
\n(b) 8.0 g Methane
\n(c) 21.0 g Propane
\n(d) 28.0 g CO<\/p>\n

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

Answer: (b) 8.0 g Methane
\nExplanation:
\n6 g carbon
\nMoles of carbon = (6\/12) = 0.5 mol
\nNumber of carbon atoms
\n= 0.5 \u00d7 NA<\/sub> =0.5NA<\/sub> (NA<\/sub> is Avogadro number)
\n6 g ethane (C2<\/sub>H6<\/sub> two atoms of C per mole)
\nMoles = (6\/30) = 0.2 mol
\nNumber of carbon atoms = 0.2 \u00d7 2 \u00d7 NA<\/sub> = 0.4 NA<\/sub>
\n(Number of carbon atoms = moles of compound X number of C atoms per mol \u00d7 Avogadro number)
\n8 g methane (CH4<\/sub>)
\nMoles = (8\/16) = 0.5 mol
\nNumber of carbon atoms = 0.5 \u00d7 1 \u00d7 NA<\/sub> = 0.5 NA<\/sub>
\n21 g propane (C3<\/sub>H8<\/sub>)
\nMoles = (21\/44) =0.48 mol
\nNumber of carbon atoms = 0.48 \u00d7 3 \u00d7 NA<\/sub> = 1.44 NA<\/sub>
\n28 g CO
\nMoles = (28\/28) =1 mol
\nNumber of carbon atoms = 1 \u00d7 1 \u00d7 NA<\/sub> = NA<\/sub><\/p>\n<\/details>\n

\n

Question 14.
\nThe density of a gas is 1.78 gL-1<\/sup> at STP. The weight of one mole of gas is
\n(a) 39.9 g
\n(b) 22.4 g
\n(c) 3.56 g
\n(d) 29 g<\/p>\n

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

Answer: (a) 39.9 g
\nExplanation:
\nMolar gas volume at STP is:
\n1 mole = 22400 cm\u00b3 = 22.4 litres
\nDensity = (mass \/ volume)
\nDensity = 1.78 g\/litre
\nVolume = 22.4 litres
\nMass = (volume \u00d7 density)
\n(1.78 \u00d7 22.4) = 39.872 g<\/p>\n<\/details>\n

\n

Question 15.
\nMolarity of 0.2 N H2<\/sub>SO4<\/sub> is
\n(a) 0.2
\n(b) 0.4
\n(c) 0.6
\n(d) 0.1<\/p>\n

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

Answer: (d) 0.1
\nExplanation:
\nMolarity = (number of moles of solute \/ volume of solution in litres)
\nHere number of moles = (given mass of solute \/ molar mass)
\nwhereas Normality = ( Number of gram equivalent \/ volume of solution in liter )
\nwhere gram equivalent = ( mass of solute \/ equivalent mass )
\nConsider an example of H2<\/sub>SO4<\/sub> whose molar mass = 98 g per mole
\nConsider a Solution containing 0.98 g of sulphuric acid in 100 ml.
\nVolume = 100 ml = 0.1 l
\nThen,
\nNumber of moles = (0.98 \/ 98)
\nNumber of moles = 0.01
\nHence molarity = (0.01 \/ 0.1) = 0.1 M
\nHence Molarity = 0.1 M
\nNow sulphuric acid is dibasic therefore
\nIts equivalent weight = (98 \/ 2)
\nHence equivalent weight = 49
\nSo the gram equivalent = (0.98 \/ 49) = 0.02
\nNow Normality = (0.02 \/ 0.1)
\nHence the Normality is equal to 0.2 N.
\nThus for H2<\/sub>SO4<\/sub> (i.e. dibasic) Normality is 0.2 N and molarity is 0.1 M.<\/p>\n<\/details>\n

\n

Question 16.
\nAny charged particle is called:
\n(a) Atom
\n(b) Molecule
\n(c) Ion
\n(d) Mixture<\/p>\n

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

Answer: (c) Ion
\nExplanation:
\nA charged particle, also called an ion, is an atom with a positive or negative charge.This happens whenever something called an ionic bond forms. Two particles that have different numbers of electrons (the smallest particle in an atom which is negative) start reacting to each other. The particle that has the greater amount of electrons steals the other particles electrons. One becomes positive because it lost an electron, and the other negative because it got another electron. The two particles become attracted to each other and mix together, making a new kind of particle.<\/p>\n<\/details>\n

\n

Question 17.
\nThe balancing of equations is based upon which of the following law?
\n(a) Law of Multiple Proportions
\n(b) Law of Conservation of Mass
\n(c) Boyles Law
\n(d) Law of Reciprocal Proportions<\/p>\n

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

Answer: (b) Law of Conservation of Mass
\nExplanation:
\nBalanced chemical equation: A chemical equation in which the number of atoms of reactants and the number of atoms of products is equal is called a balanced equation. Every chemical equation should be balanced because:
\ni) According to the law of conservation of mass, atoms are neither created not destroyed in chemical reactions.
\nii) It means the total mass of the products formed in a chemical reaction must be equal to the mass of reactants consumed.<\/p>\n<\/details>\n

\n

Question 18.
\nIrrespective of the source, pure sample, of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of
\n(a) Conservation of Mass
\n(b) Multiple Proportions
\n(c) Constant Composition
\n(d) Constant Volume<\/p>\n

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

Answer: (c) Constant Composition
\nExplanation:
\nThe H : O ratios in water is fixed, irrespective of its source. Hence it is law of constant composition<\/p>\n<\/details>\n

\n

Question 19.
\nA chemical formula based on actual number of molecule is called ____ formula:
\n(a) Structural
\n(b) Molecular
\n(c) Empirical
\n(d) None<\/p>\n

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

Answer: (b) Molecular
\nExplanation:
\nMolecular formulas indicate the simple numbers of each type of atom in a molecule, with no information on structure. For example, the empirical formula for glucose is CH2<\/sub>O (twice as many hydrogen atoms as carbon and oxygen), while its molecular formula is C6<\/sub>H12<\/sub>O6<\/sub> (12 hydrogen atoms, six carbon and oxygen atoms).<\/p>\n<\/details>\n

\n

Question 20.
\nApproximate atomic weight of an element is 26.89. If its equivalent weight is 8.9, the exact atomic weight of element would be
\n(a) 26.89
\n(b) 8.9
\n(c) 17.8
\n(d) 26.7<\/p>\n

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

Answer: (d) 26.7
\nExplanation:
\nAtomic weight = (Equivalent weight \u00d7 Valency)
\n=(8.9 \u00d7 3) = 26.7
\n(Valency = (26.89)\/(8.9) \u2248 3).<\/p>\n<\/details>\n

\n

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Students are advised to practice the NCERT MCQ Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 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 Some Basic Concepts of Chemistry Class 11 …<\/p>\n

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