<> (c) V1/V2 = T1/T2 22.4/V2 =273/546 V2 = 44.8 litres (d) Mass of 1 mole Cl2 gas =35.5 x 2 =71 g, (a) The molecular mass of (Mg(NO3)2.6H2O = 256.4 g % of Oxygen = 12 x 16/256 = 75%, (b) The molecular mass of boron in Na2B4O7.10H2O = 382 g % of B = 4 x 11/382 = 11.5%, (a) 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be = 100 x 63/252 = 25 g, (b) If 252 g of ammonium dichromate produces Cr2O3 = 152 g So, 63 g ammonium dichromate will produce = 63 x 152/252 = 38 g, Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled So,Volume of steam produced at 760mm Hg and 2730C = 4.48 à 2 = 8.96litre, V1/V2 = n1/n2 So, no. (a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules. It is defined as exactly 6.022 140 76 × 10 23 particles, which may be atoms, molecules, ions, or electrons.. Mass = 2 V.D Empirical formula weight = V.D/3 So, n = molecular mass/ Empirical formula weight = 6 Hence, the molecular formula is A6B6, Atomic ratio of N = 87.5/14 = 6.25 Atomic ratio of H= 12.5/1 = 12.5 This gives us the simplest ratio as 1:2 So, the molecular formula is NH2, Element % at. Topics of interest include the biodiversity, distribution, biomass, and populations of organisms, as well as cooperation and competition within and between species. of moles in 256 g S8 = 1 mole So, no. of atoms in 1 molecule of S = 8 So, no. of moles = 0.8/16 = 0.05 moles, (a) 6.023 x 10 23 atoms of oxygen has mass = 16 g So, 1 atom has mass = 16/6.023 x 1023 = 2.656 x 10-23 g (b) 1 atom of Hydrogen has mass = 1/6.023 x 1023 = 1.666 x 10-24 (c) 1 molecule of NH3 has mass = 17/6.023 x1023 = 2.82 x 10-23 g (d) 1 atom of silver has mass = 108/6.023 x 1023 =1.701 x 10-22 (e) 1 molecule of O2 has mass = 32/6.023 x 1023 = 5.314 x 10-23 g (f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g, (a) 0.1 mole of CaCO3 has mass =100(molar mass) x 0.1=10 g (b) 0.1 mole of Na2SO4.10H2O has mass = 322 x 0.1 = 32.2 g (c) 0.1 mole of CaCl2 has mass = 111 x 0.1 = 11.1g (d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g, 1molecule of Na2CO3.10H2O contains oxygen atoms = 13 So, 6.023 x1023 molecules (1mole) has atoms=13 x 6.023 x 1023 So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 1023 =7.8 x 1023, 3.2 g of S has number of atoms = 6.023 x1023 x 3.2 /32 = 0.6023 x 1023 So, 0.6023 x 1023 atoms of Ca has mass=40 x0.6023×1023/6.023 x 1023 = 4g, (a) No. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8. b) N2means 1 molecule of nitrogen and 2N means two atoms of nitrogen. volume of SO2 = 11.2 litre, More Resources for Selina Concise Class 10 ICSE Solutions, Filed Under: ICSE Tagged With: Mole Concept, Mole Concept and Stoichiometry, Selina Class 10 Chemistry Solutions, Selina ICSE Solutions, Selina ICSE Solutions for Class 10 Chemistry, Selina ICSE Solutions for Class 10 Chemistry - Sulphuric Acid, Selina ICSE Solutions for Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry, stoichiometry, ICSE Previous Year Question Papers Class 10, Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry, Selina ICSE Solutions for Class 10 Chemistry, Selina ICSE Solutions for Class 10 Chemistry - Sulphuric Acid, Concise Mathematics Class 10 ICSE Solutions, Concise Chemistry Class 10 ICSE Solutions, Concise Mathematics Class 9 ICSE Solutions, Essay on Child Trafficking | Child Trafficking Essay for Students and Children in English, Conversation Between Vegetable Seller and Customer | General, Funny and Dialogue Conversation Between Vegetable Seller and Customer, Global Warming Solutions Essay | Essay on Global Warming Solutions for Students and Children in English, Essay on TV Addiction | TV Addiction Essay for Students and Children in English, Essay on Dowry System | Dowry System Essay for Students and Children in English, Essay on Organ Trafficking | Organ Trafficking Essay for Students and Children in English, Ownership Certificate | Format and Application Process of Ownership Certificate, Essay on Mobile Addiction | Mobile Addiction Essay for Students and Children in English. of molecules under similar conditions of temperature and pressure. Stoichiometry. It is equal to 22.4 dm3. Unit II: Structure of Atom 14 Periods Bohr's model and its limitations, concept of … Using the concept of stoichiometry, the amount of product that results from a chemical of carbon atoms in 10-12 g = 10-12 x 6.023 x1023/12 = 5.019 x 1010 atoms. (d) The relative molecular mass of an compound is the number that represents how many times one moleculae of the substance is heavier than 1/12 of the mass of an atom of carbon-12. of moles of D = 1 because volume is 22.4 litre so, mass of N2O = 1 44 = 44 g, (a) Element % Atomic mass Atomic ratio Simple ratio K 47.9 39 1.22 2 Be 5.5 9 0.6 1 F 46.6 19 2.45 4 so, empirical formula is K2BeF4, (b) 3CuO + 2NH3 â 3Cu + 3H2O + N2 3 V 2 V 3 V 1V 3 x 80 g of CuO reacts with = 2 x 22.4 litre of NH3 so, 120 g of CuO will react with = 2x 22.4 x 120/80 x 3 = 22.4 litres, (a) The molecular mass of ethylene(C2H4) is 28 g No. It also shows Made by expert teachers. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles. Molecular mass of KNO3 = 101 g 63 g of HNO3 is formed by = 101 g of KNO3 So, 126000 g of HNO3 is formed by = 126000 x 101/63 = 202 kg Similarly,126 g of HNO3 is formed by 170 kg of NaNO3 So, smaller mass of NaNO3 is required. The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12; its symbol is mol. Instructors are permitted to make and distribute copies for their classes. 4. 13.68, 40 g of NaOH contains 6.023 x 1023 molecules So, 4 g of NaOH contains = 6.02 x1023 x 4/40 = 6.02 x1022 molecules, The number of molecules in 18 g of ammonia= 6.02 x1023 So, no. (a) 444 g is the molecular formula of (NH4)2 PtCl6 % of Pt = (195/444) x 100 = 43.91% or 44%, (b) simple ratio of Na = 42.1/23 = 1.83 = 3 simple ratio of P = 18.9/31 = 0.609 = 1 simple ratio of O = 39/16 = 2.43 = 4 So, the empirical formula is Na3PO4. (b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume. b) The value of avogadro’s number is 6.023 à 1023 c) The molar volume of a gas at STP is 22.4 dm3 at STP. 3. (c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. (b) Since molecular mass is 2 times the vapour density, so Mol. of moles of Cl = x/2 (since V is directly proportional to n) No. 22400 cm3 of CO has mass = 28 g So, 56 cm3 will have mass = 56 x 28/22400 = 0.07 g, 18 g of water has number of molecules = 6.023 x 1023 So, 0.09 g of water will have no. (Northern Arizona University) and Raymond Chang, this success guide is written for use with General Chemistry. of particles in s1 mole = 6.023 x 1023 So, particles in 0.1 mole = 6.023 x 10 23 x 0.1 = 6.023 x 1022, b) 1 mole of H2SO4 contains =2 x 6.023 x 1023 So, 0.1 mole of H2SO4 contains =2 x 6.023 x 1023 x0.1 = 1.2×1023 atoms of hydrogen, c) 111g CaCl2 contains = 6.023 x 1023 molecules So, 1000 g contains = 5.42 x 1024 molecules, (a) 1 mole of aluminium has mass = 27 g So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g (b) 0.1 mole of HCl has mass = 0.1 x 36.5 (mass of 1 mole) = 3.65 g (c) 0.2 mole of H2O has mass = 0.2 x 18 = 3.6 g (d) 0.1 mole of CO2 has mass = 0.1 x 44 = 4.4 g, (a) 5.6 litres of gas at STP has mass = 12 g So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6 = 48g(molar mass) (b) 1 mole of SO2 has volume = 22.4 litres So, 2 moles will have = 22.4 x 2 = 44.8 litre, (a) 1 mole of CO2 contains O2 = 32g So, CO2 having 8 gm of O2 has no. 40 So, 342g(molar mass) of sugar will cost=342×40/1000=Rs. According to Avogadros law: Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules. Mass of O2) x 0.5=16 g, (c) No. That ratio by weight of hydrogen and oxygen is 1:8. APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry. (iv) The volume of D = 5.6 4 = 22.4 dm3, so the number of molecules = 6 x 1023 because according to mole concept 22.4 litre volume at STP has = 6 x 10 23 molecules, (v) No. So, 20 litre nitrogen contains x molecules So, 10 litre of chlorine will contain = x à 10/20=x/2 mols. - Atomic and molecular masses, mole concept, molar mass, percentage composition, empirical and molecular formulae - Chemical equations and stoichiometry Unit 2: States of Matter Now we have to find out the volume at standard pressure. of moles in 10g of CaCO3 = 10/100(mol. Chlorination of ww is used to improve effluent quality 1. (b) Avogadro’s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules. %PDF-1.4 (c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12. (ii) The empirical formula is CH3 (iii) The empirical formula mass for CH2O = 30 V.D = 30 Molecular formula mass = V.D 2 = 60 Hence, n =mol. '�"��� ��*g���JmL�v�3�d���` �Ms0��5�MV����]3'|�M����K�!Nz&�%"�@���L�%��IS��)� %�쏢 ��g� F��&��h��_m�u0�:��* a'�㊇��#4���Nѯ���(Qώ�+Π��[��g�γV���[n����:��no3"c,�)�V7S2��K���\׀�M��G������_�;�:��ǝ��۸�E. Formula mass/empirical formula mass= 2 So, molecular formula = (CH2O)2 = C2H4O2, The relative atomic mass of Cl = (35 x 3 + 1 x 37)/4=35.5 amu, So, mass of CO2 = 22 kg (b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. We have to find out the volume of one liter of unknown gas at standard temperature 273 K. V1= 1 L T1 = 546 K V2=?      T2 = 273 K V1/T1 = V2/ T2 V2 = (V1 x T2)/T1      = (1 L x 273 K)/546 K = 0.5 L. We have found out the volume at standard temperature.
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