{"id":29078,"date":"2023-03-13T01:26:33","date_gmt":"2023-03-12T19:56:33","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=29078"},"modified":"2023-03-14T09:52:55","modified_gmt":"2023-03-14T04:22:55","slug":"new-simplified-chemistry-class-10-icse-solutions-mole-concept","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/new-simplified-chemistry-class-10-icse-solutions-mole-concept\/","title":{"rendered":"New Simplified Chemistry Class 10 ICSE Solutions – Mole Concept"},"content":{"rendered":"
ICSE Solutions<\/a>Selina ICSE Solutions<\/a>ML Aggarwal Solutions<\/a><\/p>\n Viraf J Dalal Chemistry Class 10 Solutions and Answers<\/strong><\/p>\n Simplified Chemistry<\/a>English<\/a>Maths<\/a>Physics<\/a>Chemistry<\/a>Biology<\/a><\/p>\n Practice Questions<\/strong><\/span><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> <\/p>\n Relative Molecular Mass :<\/strong> \u201cIt is the number that represents how many times one molecule of a substance is heavier than one atom of hydrogen whose weight has been taken unity or 1\/12 of 6<\/sub>C12<\/sup>. Applications Of Avogadro\u2019S Law :<\/strong><\/p>\n 1. Determines the atomicity of a gas :<\/strong> 2. Determines the molecular formula of a gas.<\/strong> 3. Determines the relation between molecular weight and vapour density.<\/strong> 4. Explains Gay-Lussac\u2019s Law :<\/strong> 5. Determines relationship between gram molecular mass and gram molecular volume :<\/strong> Additional Problems<\/strong><\/span><\/p>\n Q.1. Lussac\u2019S Law<\/strong><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Q.2. Mole Concept – Avogadro\u2019S Law – Avogadro\u2019S Number<\/strong> Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Answer: Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span>
\nWhat volume of oxygen would he required to burn completely 400 ml of acetylene (C2<\/sub>H2<\/sub>) ? Calculate the volume of CO2<\/sub> formed.
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\n2500 cc of oxygen was burnt with 600 cc of ethane (C2<\/sub>H6<\/sub>). Calculate the volume of unused oxygen and the volume of carbon dioxide formed, after writing the balanced equation :
\nEthane + Oxygen \u2192 Carbon dioxide + Water vapour
\nAnswer:
\n
\n
\n<\/strong><\/span><\/p>\n
\n80 cm3<\/sup> of methane are mixed with 200 cm3<\/sup> of pure oxygen at similar temperature and pressure. The mixture is then ignited. Calculate the composition of resulting mixture if it is cooled to initial room temperature and pressure.
\nCH4<\/sub> + 2O2<\/sub> \u2192 CO2<\/sub> + 2H2<\/sub>O
\nAnswer:
\n
\n<\/strong><\/span><\/p>\n
\nCalculate the volume of HCl gas formed and chlorine gas required when 40 mL of methane reacts completely with chlorine at S.T.P.
\nCH4<\/sub> + 2Cl2<\/sub> \u2192 CH2<\/sub>Cl2<\/sub> + 2HCl
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nWhat volume of propane is burnt for every 500 cm3<\/sup> of air used in the reaction under the same conditions ?
\nC3<\/sub>H8<\/sub>\u00a0+ 5O2<\/sub> \u2192 3CO2<\/sub> + 4H2<\/sub>O
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\n450 cm3<\/sup> of nitrogen monoxide and 200 cm3<\/sup> of oxygen are mixed together and ignited. Calculate the composition of resulting mixture.
\n2NO + O2<\/sub> \u2192 2NO2<\/sub>
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\n24 cc marsh gas (CH4<\/sub>) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc of which 58 cc were unchanged oxygen. Which law does their experiment supports ? Explain with calculations.
\nAnswer:
\n
\n<\/strong><\/span>
\nAvogadro\u2019s Law :<\/strong> \u201cUnder the same conditions of temperature and pressure equal volumes of all gases contain die same number of molecules.\u201d
\nIf we assume that 7 litre of oxygen gas contains \u2018n\u2019 molecules of the gas then by Avogadro\u2019s Law :<\/p>\n\n
\nRelative Molecular Mass :<\/strong> \u201cIt is the number that represents how many times one molecule of a substance is heavier than one atom of hydrogen whose weight has been taken unity or 1\/12 of 6<\/sub>C12<\/sup>.
\nAvogadro\u2019s Number :<\/strong> \u201cThe number of atoms present in 12 g (gm atomic mass) of 6<\/sub>C12<\/sup> is called Avogadro\u2019s number.
\nNa or L = 6.023 \u00d7 1023<\/sup><\/p>\n
\nMole :<\/strong> Mole is the mass of substance containing particles equal to Avogadro\u2019s number i.e. 6.023 \u00d7 1023<\/sup>.
\nGram Atom :<\/strong> \u201cThe relative atomic mass of an element expressed in grams is called gram atom.
\nGram Mole :<\/strong> \u201cThe relative atomic mass of a substance expressed in grams is called gram mole.
\nMolar Volume :<\/strong> Volume occupied by one mole of any gas at STP is called molar volume.<\/p>\n\n
\nAtomicity :<\/strong> The number of atoms present in one molecule of that element is called atomicity.
\nMonoatomic :<\/strong> Elements which have one atom in their molecules e.g. Helium, Neon.
\nDiatomic :<\/strong> Elements which have two atoms in their molecule e.g. Hydrogen, nitrogen, oxygen.
\ne.g. of Determination of atomicity of a gas :<\/strong>
\n
\nA molecules of nitrogen contains two atoms : atomicity – Diatomic<\/p>\n
\nMolecular formula :<\/strong> A chemical formula which gives the actual or exact number of atoms of the elements present in one molecule of a compound.
\n<\/p>\n
\nMolecular weight:<\/strong> It is the ratio of the weight of 1 molecule of a substance to the weight of one atom of hydrogen.
\n
\nVapour Density :<\/strong> It is the ratio of the mass of a certain volume of gas or vapour of the mass of the same volume of hydrogen.
\nMol. wt. = 2\u00d7 vapour density<\/p>\n
\nGay-Lussac\u2019s Law<\/strong> is explained by Avogadro\u2019s Law which states \u201cUnder similar conditions of temperature and pressure, equal volumes of different gases have same number of molecules.\u201d
\nSince substances react in simple ratio of number of molecules, volumes of gaseous reactants and products will also bear a simple ratio to one another. This is what Gay-Lussac\u2019s Law says.
\n<\/p>\n
\nGram molecular mass is the relative molecular mass of a substance expressed in grams. It is also called gram molecule of that element.
\nGram molecular volume : The volume occupied by e.g. molecular wt. of a gas at s.t.p.
\n<\/p>\n
\nNitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formed when nitrogen reacts with 6 litres of hydrogen. All volumes measured at s.t.p.
\nAnswer:
\n
\n<\/strong><\/span><\/p>\n
\n2500 cc of oxygen was burnt with 600 cc of ethane [C2<\/sub>H6<\/sub>]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed.
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\n20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure. What will be the volume and composition of the mixture of the gases when they are cooled to room temperature.
\nAnswer:
\n
\n<\/strong><\/span><\/p>\n
\n224 cm3<\/sup> of ammonia undergoes catalytic oxidation in presence of Pt to given nitric oxide and water vapour. Calculate the volume of oxygen required for the reaction. All volumes measured at room temperature and pressure.
\nAnswer:
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\n
\n<\/strong><\/span><\/p>\n
\nAcetylene [C2<\/sub>H2<\/sub>] burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burn 50 cm3<\/sup> of acetylene. [Assume air contains 20% oxygen].
\nAnswer:
\n
\n<\/strong><\/span><\/p>\n
\nOn igniting a mixture of acetylene [C2<\/sub>H2<\/sub>] and oxygen, 200 cm3<\/sup> of CO2<\/sub> is collected at s.t.p. Calculate the volume of acetylene & O2<\/sub> at s.t.p. in the original mixture.
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nAmmonia is formed from the reactants nitrogen and hydrogen in presence of a catalyst under suitable conditions. Assuming all volumes are measured in litres at s.t.p. Calculate the volume of ammonia formed if only 10% conversion has taken place.
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\n100 cc. each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temp. Calculate the composition of the resultant mixture. [Water gas contains CO and H2<\/sub> in equal ratio]
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nCalculate the following : [all measurement at s.t.p. or as stated in the problem]<\/strong><\/p>\n
\nThe mass of 2.8 litres of C02<\/sub>. [C = 12, O = 16]
\nAnswer:
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\n
\n<\/strong><\/span><\/p>\n
\nThe volume occupied by 53.5g of Cl2<\/sub>. [Cl = 35.5]
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nThe number of molecules in 109.5 g of HCl. [H = 1, Cl = 35.5]
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nThe number of<\/p>\n\n
\n
\n<\/strong><\/span><\/p>\n
\nThe mass of (Na) sodium which will contain 6.023 \u00d7 1023<\/sup> atoms. [Na = 23]
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nThe no. of atoms of potassium present in 117 g. of K. [K = 39]
\nAnswer:
\n
\n<\/strong><\/span><\/p>\n
\nThe number of moles and molecules in 19.S6 g. of Pb (NO3<\/sub>)2<\/sub>. [Pb = 207, N = 14, O = 16]
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nThe mass of an atom of lead [Pb = 202].
\nAnswer:
\n
\n<\/strong><\/span><\/p>\n
\nThe number of molecules in 1 1\/2 litres of water. [density of water 1.0 g.\/cm3<\/sup>. \u2014 \u2234 mass of water = volume \u00d7 density]
\nAnswer:
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\n
\n<\/strong><\/span><\/p>\n
\nThe gram-atoms in 88.75 g of chlorine [Cl = 35.5]
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nThe number of hydrogen atoms in 0.25 mole of H2<\/sub>SO4<\/sub>.
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nThe gram molecules in 21 g of nitrogen [N = 14]
\nAnswer:
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\n<\/strong><\/span><\/p>\n
\nThe number of atoms in 10 litres of ammonia [N = 14, H = 1]
\nAnswer:
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\n
\n<\/strong><\/span><\/p>\n
\nThe number of atoms in 60 g of neon [Ne = 20]
\n