CBSE Solutions for Class 11 English

જવાબ : Four

જવાબ : 1 a.m.u. or 1 u = 1/12 th mass of an atom of carbon 12.

જવાબ : 0.5 ml of NaOH means 0.5 mole (20.0 g) of NaOH, 0.5M of NaOH means that 0.5 mole (20.0g) of NaOH are dissolved in 1L of its solution.

જવાબ : Equal volumes of all gases under the conditions of same temperature and pressure contain the same number of molecules.

જવાબ : (i) Reliability of measuring instrument.
(ii) Skill of the person making the measurement.

જવાબ : Molar mass of water is 18 g/mol.
Number of oxygen atoms is 18 g of water = 6.02 x 10²³

જવાબ : Molecular formula = (Empirical formula)n where n is positive integer.

જવાબ : It states that matter can neither be created nor destroyed.

જવાબ : The compound is glucose. Its molecular formula is C₆H₁₂O₆, while empirical formula is CH₂O.

જવાબ : According to Dalton’s atomic theory, an atom is the ultimate particle of matter which cannot be further divided.

જવાબ : This is due to the presence of dust particles.

જવાબ : It is defined as the standard of reference chosen to measure a physical quantity.

જવાબ : SI unit of molarity = mol dm^-3

જવાબ : The coefficients of reactant and product involved in a chemical equation represented by the balanced form, are known as stoichiometric coefficients.
For example, N₂(g) + 3H₂(g) ———–> 2 NH₃(g)
The stoichiometric coefficients are 1, 3 and 2 respectively.

જવાબ : (a) Yes, the statement is true. (b) According to the law of multiple proportions (c), Hydrogen and oxygen react to form water and hydrogen peroxide H₂ + 1/2O₂ → H₂O H₂ + O₂ → H₂O₂ Masses of oxygen which combine the fixed mass of hydrogen are in the ratio 16:32 or 1

જવાબ : The above experiment proves Gay-Lussac’s law which states that gases combine or produced in a chemical reaction in a simple whole-number ratio by volume provided that all gases are the same temperature and pressure.

જવાબ : Molecular mass of Ca₃(PO₄) = 3*40+2*31+8*16 =310 Mass per cent of Ca = 3*40/310*100 = 38.71% Mass per cent of P = 2*31/310*100 = 20% Mass per cent of O = 8*16/310 = 41.29%

જવાબ : Molarity is the number of moles of solute dissolved in 1 litre of the solution. Molality is the number of moles of solute present in 1kg of the solvent.

જવાબ : The symbol for the SI unit of the mole is mol. A mole is defined as the amount of substance that contains as many entities as there are atoms in 12g carbon.

જવાબ : Two significant figures should be present in this. Since the least number of significant figures from the given figure is 2 (in 2.5 and 3.5).

જવાબ : 1 mole of carbon atom = 12g= 6.022 × 10²³ atoms.

જવાબ : Average atomic mass = 99.985*1+0.015*2/100 =099.985*1+0.015*2/100 =1.00015u

જવાબ : (i) B will be the limiting reagent as it gives a lesser amount of product. (ii) Let B is completely consumed 4 mol B gives 3 mol C 6 mol B will give 3/4 *6 mol C =4.5 mol C

જવાબ : Mass does not change as the temperature changes. Therefore, the molality of a solution does not change. Molality = moles of solute/ weight of solvent (in g) *1000

જવાબ : 3 molal solution of NaOH = 3 moles of NaOH dissolved in 1000g water 3 mole of NaOH = 3*40g = 120g Density of solution = 1.110gmL-1 Volume = mass/density = 1120g/1.110gmL-1 =1.009L Molarity of the solution = 3/1.009 = 2.97M

જવાબ : 1 mol of gas occupies = 22.7L Volume at STP atomic mass of Zn = 65.3u From the above equation, 65.3g of Zn when reacts with HCl produces = 22.7L H2 at STP Therefore, 32.65g of Zn when reacts with HCl will produce = 22.7 * 32.65/65.3 =11.35L of H2 at STP

જવાબ : (i) Molecular mass of H₂O = 2(1.008 amu) + 16.00 amu=18.016 amu
(ii) Molecular mass of CO₂= 12.01 amu + 2 x 16.00 amu = 44.01 amu
(iii) Molecular mass of CH₄= 12.01 amu + 4 (1.008 amu) = 16.042 amu

જવાબ : 1 mole of Mn0₂, i.e., 55 + 32 = 87 g Mn0₂ react with 4 moles of HCl, i.e., 4 x 36.5 g = 146 g of HCl

જવાબ :

જવાબ : Molar mass of Na₂C0₃= 2 x 23 + 12 + 3 x 16 = 106g mol^-1 0.50 mol Na₂C0₃ means 0.50 x 105 g = 53 g 0. 50 M Na2C03 means 0.50 mol, i.e., 53 g Na₂C0₃ are present in 1 litre of the solution

જવાબ : H₂ and 0₂ react according to the equation 2H₂(g) + 0₂ (g) ——>2H₂O (g)  Thus, 2 volumes of H₂ react with 1 volume of 0₂ to produce 2 volumes of water vapour. Hence, 10 volumes of H₂ will react completely with 5 volumes of 0₂ to produce 10 volumes of water vapour.

જવાબ :

જવાબ : (i) 34.2 (ii) 10.4 (iii) 0.0460 (iv) 2810

જવાબ : (i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5.

જવાબ :

જવાબ : The digits in a properly recorded measurement are known as significant figures. It is also defined as follows. The total numbers of figures in a number including the last digit whose value is uncertain is called number of significant figures.

જવાબ :  S.I. unit of mass is kilogram (kg).

જવાબ : Pressure is the force (i.e., weight) acting per unit area But weight = mg

જવાબ :

જવાબ :

જવાબ :  (i) 1 mole of C₂H₆ contains 2 moles of carbon atoms
.•. 3 moles of C₂H₆ will C-atoms = 6 moles
(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms
.•. 3 moles of C₂H₆ will contain H-atoms = 18 moles

જવાબ :

જવાબ : Empirical formula mass of Fe₂0₃ = 2 x 55.85 + 3 x 16.00 = 159.7 g mol^–1
Hence, molecular formula is same as empirical formula, viz.,Fe₂0₃.

જવાબ : 1 mole of CuS0₄ contains 1 mole (1 g atom) of Cu Molar mass of CuS0₄= 63.5 + 32 + 4 x 16 = 159.5 g mol^-1 Thus, Cu that can be obtained from 159.5 g of CuS0₄ = 63.5 g

જવાબ : When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other bear a simple ratio to one another is the law of multiple proportions.

જવાબ : No: of moles of HCl taken = MV/1000 = 0.76*250/1000 = 0.19

જવાબ : (i) Moles of oxygen = 1.6/32 = 0.05mol

જવાબ :

જવાબ : Mole fraction of H2O = No; of moles of H2O/ Total no: of moles (H2O+NaOH)

જવાબ :

જવાબ : The balanced equation for the combustion of carbon in dioxygen/air is

(i) In air, combustion is complete. Therefore,C0₂ produced from the combustion of 1 mole of carbon = 44 g.(ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting reactant. Hence,C0₂ produced = 22 g.(iii) Here again, dioxygen is the limiting reactant. 16 g of dioxygen can combine only with 0.5 mole of carbon.C0₂ produced again is equal to 22 g.

જવાબ : 0.375 M aqueous solution means that 1000 mL of the solution contain sodium acetate = 0.375 mole

જવાબ : Mass percent of 69% means that 100 g of nitric acid solution contain 69 g of nitric acid by mass.
Molar mass of nitric acid HNO₃= 1 + 14 + 48 = 63 gmol^-1

જવાબ :

જવાબ :

જવાબ : (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
.•. 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be
left unreacted. Hence, B is the limiting reagent while A is the excess reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
.•. 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant.
(iii) No limiting reagent.
(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.
(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent.

જવાબ :

જવાબ : (a) 1 gram atom of Mg = 24g
2.5 gram atoms of Mg = 24 x 2.5 = 60g
(b) 1 gram atom of N = 14g;
14g of N = 1 gram atom 1
1.4g of N = 1/14 x 1.4 = 0.1 gram atom.

જવાબ :

જવાબ : 1 molecule of oxygen (O₂) = 2 atoms of oxygen
1 molecule of ozone (O₃) = 3 atoms of oxygen

જવાબ : Molecular mass of oxygen = 32 u

જવાબ : 1 mole of CuSO_{4}CuSO4​ contains 1 mole of Cu.

જવાબ : The law of multiple proportions rays that if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of other element are in a ratio of small whole numbers. e.g. hydrogen and oxygen can combine to form water as well as hydrogen peroxide.

જવાબ : Postulates of Dalton’s atomic theory –

જવાબ :

જવાબ : Let the required volume of 10M HCl be V liters.

જવાબ : Initial volume, V₁ = 2L

જવાબ :

જવાબ : 2KClO₃ à 2KCl + 3O₂

જવાબ :

જવાબ :

(i → e)

(ii → d)

(iii → b)

(iv → g)

(v → c)

(vi → f)

(vii → a)

(viii → i)

જવાબ :

A → b

B → c

C → a

D → e

E → d