{"id":2308,"date":"2023-10-02T12:06:28","date_gmt":"2023-10-02T12:06:28","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2308"},"modified":"2023-10-02T12:15:13","modified_gmt":"2023-10-02T12:15:13","slug":"week-9-ss1-second-term-chemistry-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-9-ss1-second-term-chemistry-notes\/","title":{"rendered":"Week 9 – SS1 Second Term Chemistry Notes"},"content":{"rendered":"

\u00a0WEEK 9
\n<\/strong>GAY LUSSAC’S LAW: It<\/strong> states that when gasses react, they do so in volumes and these volumes are in simple ratio to one another and to volume of the product if gaseous provided the temperature and pressure remain constant.
\n\t\t\t<\/strong>Gay Lussac observed as follows:
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0H2<\/sub> + O2<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02H2<\/sub>O
\n<\/strong>Volume\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2
\n<\/strong>Ratio\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02 : 1 : 2
\n<\/strong>He noticed that the combining volumes as well as the volumes of the products were related by simple ratios of whole number, provided they are gases.
\nEXERCISE
\n<\/strong><\/p>\n

    \n
  1. \n
    2O cm3<\/sup> of CO are sparked with 20 cm3<\/sup> of Oxygen. If all the volumes of gases are measured at s.t.p, calculate the volume of the residual gases after sparking.\n<\/div>\n

    \"\"\/2CO (g)<\/sub> + O2(g)<\/sub> 2CO2(g)<\/sub>
    \n\t\t\t\t\t<\/strong>SOLUTION
    \n<\/strong><\/li>\n<\/ol>\n

    \"\"\/2CO2 (g)<\/sub> + O2 (g)<\/sub><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02CO2 (g)
    \n<\/sub><\/strong>Combining volume: \u00b7 2\u00a0\u00a0\u00a0\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0:\u00a0\u00a0\u00a0\u00a02
    \nVol. before sparking: 20cm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a020cm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0–
    \nVol. during sparking: 20cm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010cm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a020cm3<\/sup>
    \n\t\tVol. after sparking:\u00a0\u00a0\u00a0\u00a0 –\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010cm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a020cm3<\/sup>\u00a0\u00a0\u00a0\u00a0
    \nResidual gases = Unreacted gas + Product formed\u00a0\u00a0\u00a0\u00a0
    \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010\u00a0\u00a0\u00a0\u00a0+\u00a0\u00a0\u00a0\u00a020 = 30cm3<\/sup>
    \n\t\t2. 100cm3<\/sup> of Nitrogen gases mixed together with 150 cm3 of Hydrogen. The mixtures were
    \nmade to react at low temperature at which the product cannot dissociate.
    \n I. which of the two gases was in excess and by what volume?
    \n ii. What was the volume of NH3<\/sub> produced?
    \n\"\"\/Equation for the reaction; N2<\/sub> + 3H2<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02NH3<\/sub>
    \n\t\t3. What is the volume of Oxygen required to burn completely 45cm3<\/sup> of Methane gas (CH4<\/sub>)?
    \n\"\"\/Equation for the reaction: CH4<\/sub> (g) + 2O2<\/sub> (g) \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0CO2<\/sub> (g) + 2H2<\/sub>O (g)
    \nSOLUTION BY GAY-LUSSAC’S LAW<\/strong>
    \n\t\t1 Volume of CH4<\/sub> requires 2 volumes of oxygen
    \ni.e. 1cm3<\/sup> of CH4<\/sub> requires 2cm3<\/sup> of oxygen
    \n\"\"\/ 45cm3<\/sup> of CH4<\/sub> requires\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02 x 45 = 90 cm3<\/sup>\u00a0\u00a0\u00a0\u00a0
    \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1
    \nAlternatively:
    \n<\/strong>By mole concept, from the equation, 1 mole of CH4<\/sub> requires 2 moles of O2,
    \n<\/sub>(1 X 22.4) dm3<\/sup> of CH4<\/sub> requires (2x 22.4) dm3<\/sup> of O2<\/sub>.
    \n22.4 dm3<\/sup> of CH4<\/sub><\/strong> require 44.8 dm3<\/sup> of O2<\/sup>.
    \n\"\"\/Hence, 45cm3<\/sup> require 44.8 x 45 = 90 cm3<\/sup>
    \n\t\t 22. 4
    \nAVOGADRO’S LAW:<\/strong> By an Italian Professor Avogadro (1776 \u2013 1856). It states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules<\/p>\n

    \u00a0AVOGARO’S LAW AND GAY LUSSAC’S LAW.
    \n<\/strong>Avogadro’s law is used to convert the volume of gases into the number of molecules contained by those gases i.e.
    \n\"\"\/Equation\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02H2 (g)<\/sub> + O2(g)<\/sub> 2H2<\/sub>O(g)<\/sub>
    \n\t\tVolume\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02 1 2
    \nGay Lussac 2 : 1 2
    \nAvogadro\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02mols 1molecules\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2molecules.
    \nTherefore, Gay Lussac’s law can be re-stated that when gasses react, they do so in small whole numbers of molecules of reactant to produce small whole numbers of products.
    \nRELATIVE VAPOUR DENSITY OF GASSES.
    \n<\/strong>It is the number of times a given volume of a gas is as heavy as the same volume of Hydrogen gas at a particular temperature and pressure.
    \n\"\"\/V.D = Mass of a given volume of a gas
    \n Mass of an equal volume of Hydrogen gas
    \nRelationship <\/strong>between V.D and R.M.M.
    \n<\/strong><\/em>Applying Avogadro’s law into the formula of VD,
    \n\"\"\/V.D = Mass of a given molecule of gas
    \n Mass of an equal molecule of H2<\/sub>
    \n\t\tWhere the molecule is 1,
    \n<\/strong>\"\"\/V.D = Mass of 1 molecule of a gas
    \n Mass of 1 molecule of H2<\/sub>
    \n\t\tHowever, H2<\/sub> is diatomic,
    \n<\/strong>\"\"\/V.D = Mass of 1 molecule of a gas
    \n Mass of 2 atoms of H2<\/sub>
    \n\t\t\"\"\/ Mass of 1 molecule of a gas
    \n 2(mass of 1atom of H2<\/sub>)
    \n\"\"\/ 2 x V.D = mass of 1 molecule of a gas
    \n Mass of 1 atom of H2<\/sub>
    \n\t\tNOTE: R. M. M= mass of 1 molecule of a gas
    \n Mass of 1 atom of H2<\/sub>
    \n\t\tSubstitute this equation
    \nHence 2 x V.D = R.M.M
    \nV.D = R.M.M\/2
    \n\t\t\t<\/strong>GRAHAM’S LAW OF DIFFUSION OF GASSES
    \n<\/strong>Graham in 1833 discovered that a less dense gas can diffuse faster than a denser gas, so the density of the gas determines the rate of diffusion of a gas.
    \n<\/strong>\"\"\/\"\"\/The law states that at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density.
    \n<\/strong>\"\"\/i.e. R \u03b1
    \n<\/strong>\"\"\/ Where R = rate of diffusion
    \n= density (Greek letter rho)
    \nFor two gases (say 1 and 2)
    \n\"\"\/\"\"\/\"\"\/\"\"\/
    \n\t\t R1<\/sub>\u03b1 and R2<\/sub>\u03b1 \u2082\"\"\/\"\"\/
    \n\t\t\t\t<\/strong>On multiplication<\/p>\n

    \u00a0R1 <\/sub><\/strong> =
    \nR2 <\/sub><\/strong>=
    \n\t\t\t<\/strong>Note: The rate of a gas is directly proportional to the square root of its molecular mass.
    \nR1 <\/sub><\/strong>= M1<\/sub><\/strong>
    \n\t\tR2 <\/sub><\/strong>= M2 <\/sub><\/strong>where M = Molecular mass<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0Note: In calculation, M where R is not given, it can be calculated as follows:
    \n\"\"\/R = Volume
    \n Time
    \nCALCULATIONS
    \n<\/strong>1.400 cm3<\/sup> of a gas A diffuses through a porous partition in 5 seconds and 200cm3<\/sup> of a gas B diffuses in 20 seconds under the same condition of temperature and pressure:
    \n<\/strong><\/p>\n

      \n
    1. Calculate the rates of diffusion of gases A and B
      \n<\/strong><\/li>\n
    2. \n
      Which gas is denser?
      \n<\/strong><\/div>\n

      SOLUTION\n<\/li>\n<\/ol>\n

        \n
      1. \n
        RA<\/sub> = Vol = 400 = 80 cm3<\/sup><\/strong>\n\t\t\t\t<\/div>\n

        \"\"\/ Time 5
        \nRB<\/sub> =200 = 10 cm3<\/sup><\/strong>
        \n\t\t\t\t\"\"\/ 20.\n<\/li>\n

      2. Gas B is Denser\n<\/li>\n<\/ol>\n

        2.A gas X diffused through a porous partition at the rate of 2.5 cm3<\/sup>\/s . Under the same conditions, hydrogen diffused at the rate of 10cm3 <\/sup>\/s . Calculate the RMM of the gas. (H = 1) Ans = 32
        \n<\/strong>
        \n\u00a0APPLICATION OF MOLE CONCEPT IN CHEMICAL REACTION<\/strong>
        \n\t\tThe mole concept can be applied to the following types of calculates which are based upon a
        \nbalanced chemical equation.<\/p>\n

        \u00a0Calculation involving mass-mass relationship
        \nExercise 1: Calculate the mass of calcium oxide formed when 1.5g of calcium is completely burnt in oxygen
        \n(Ca =40, 0 = 16)
        \n2Ca(s)<\/sub> + 02<\/sub>g 2Ca 0(s)<\/sub> answer = 2.1g of Ca0(s) <\/p>\n

        \u00a0EXERCISE 2
        \n<\/strong>Calculate the mass of oxygen gas formed when 10g of potassium trioxonitrate (v) (potassium nitrate) is heated strongly.
        \nEquation for the reaction
        \n2KN03(s)<\/sub> 2KN02(s)<\/sub> + 02<\/sub>(g)
        \nAnswer = 1.58g of 02<\/sub>(g)<\/p>\n

        \u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0EXERCISE 3
        \n<\/strong>When 1.4g of impure calcium trioxocarbonate (calcium carbonate) reacts with hydrochloric acid, 0.01mole of carbon (iv) oxide (carbon dioxide) gas was evolved calculate
        \ni.\u00a0\u00a0\u00a0\u00a0Percentage purity
        \nii.\u00a0\u00a0\u00a0\u00a0Percentage impurity of calcium carbonate (Ca =40, C=12, 0=16, H=1)<\/p>\n

        \u00a0SOLUTION:
        \n<\/strong>Ca<\/sub>C03<\/sub>(s) + 2HCl (aq)<\/sub>CaCl2 (aq)<\/sub> + H2 (aq)<\/sub> + C02<\/sub> (g)
        \n1 mole C02<\/sub> = 1 mole Ca<\/sub>C03<\/sub>
        \n\t\t1 mole = (40 + 12 + 16 x 3) Ca<\/sub>C03<\/sub>
        \n\t\t1 mole C02 = 100 Ca<\/sub>C03<\/sub>
        \n\t\t:.0.01 mole C02 = 100 x 0.01 Ca<\/sub>C03<\/sub>
        \n\t\tMas of pure = 1g Ca<\/sub>C03<\/sub>
        \n\t\tTotal mass of Ca<\/sub>C03<\/sub> = 1.4g
        \nMas of pure Ca<\/sub>C03<\/sub> = 1.04g
        \nMass of impure CaC03
        \n= 1.4 \u2013 1.0 = 0.4g <\/p>\n

        \u00a0i.\u00a0\u00a0\u00a0\u00a0Percentage purity
        \n\u00a0\u00a0\u00a0\u00a0= mass of prime x 100
        \n\u00a0\u00a0\u00a0\u00a0 Total mass of Ca<\/sub>C03 1<\/sub>
        \n\t\t\u00a0\u00a0\u00a0\u00a0= 1.0 x 100
        \n\u00a0\u00a0\u00a0\u00a0 1.4
        \n\u00a0\u00a0\u00a0\u00a0= 71.4%<\/p>\n

        \u00a0EXERCISE 2
        \n<\/strong>Calculate the volume of ammonia gas formed at s.t.p and r.t.p when 0.01g of hydrogen reacts
        \nwith nitrogen gas
        \n(N14, H=1 M.V = 22.4drm3<\/sup> at s.t.p and 24dm3<\/sup> at r.t.p )<\/p>\n

        \u00a0SOLUTION:
        \n<\/strong>Equation for the reaction
        \nN2 (g)<\/sub> + 3H2 (g)<\/sub> ______2NH3 (g<\/sub>)<\/p>\n

        \u00a0Answer:
        \n0.075dm3<\/sup> NH3<\/sub> at s.t.p
        \n8dm3<\/sup> NH3<\/sup> at r.t.p<\/p>\n

        \u00a0EXERCISE 3
        \n<\/strong>Calculate the volume of oxygen gas at s.t.p and r.t.p needed to burn 1:20g of magnesium,
        \naccording to the equation below
        \n2mg(s<\/sub>) + 02(g) _________2mg0(s)<\/sub>
        \n\t\t(Mg = 24, 0 = 16, Mr. = 22.4dm3<\/sup> at s.t.p, 24dm3<\/sup> at r.t.p)<\/p>\n

        \u00a0Answer:
        \n= 0.56dm3<\/sup> 02<\/sub> at s.t.p
        \n= 0.56dm3<\/sup> 02<\/sub> at r.t.p<\/p>\n

        \u00a0EXERCISE 4:
        \n<\/strong>The complete combustion of methane in oxygen is represented by the equation
        \nCH4(g)<\/sub> + 02(g)<\/sub>) _____ C02<\/sub> (g) + 2H2<\/sub>0(g)<\/sub>
        \n\t\tIf 1000cm3<\/sup> of methane was completely combusted in oxygen at s.t.p
        \nCalculate
        \ni.\u00a0\u00a0\u00a0\u00a0The mole of methane combusted
        \nii.\u00a0\u00a0\u00a0\u00a0The volume of oxygen used for the combustion
        \niii.\u00a0\u00a0\u00a0\u00a0The mass of carbon (iv) C02<\/sub> produced<\/p>\n

          \n
        • \n
          The number of water molecule produced.\n<\/div>\n<\/li>\n<\/ul>\n

          \u00a0SOLUTION:<\/strong>i.\u00a0\u00a0\u00a0\u00a0Answer = 0.04 mole of CH4<\/sub><\/p>\n

          \u00a0ii.\u00a0\u00a0\u00a0\u00a0Answer = 2000cm3<\/sup> of 02<\/sub><\/p>\n

          \u00a0iii.\u00a0\u00a0\u00a0\u00a0Answer = 1.96g of C02<\/sub><\/p>\n

          \u00a0iv.\u00a0\u00a0\u00a0\u00a0Answer = 5.38 x 1022<\/sup> molecule of water <\/p>\n

          \u00a0GAS VOLUME \u2013 GAS VOLUME RELATIONSHIP
          \n<\/strong>
          \n\u00a0EXERCISE 1:
          \n<\/strong>40cm3<\/sup> of nitrogen gas (N2<\/sub>) reacts with 60cm3<\/sup> of hydrogen gas to form ammonia gas. Calculate
          \nthe volume of unused and volume of ammonia gas formed at the same temperature and pressure
          \nequation for the reaction<\/p>\n

          \u00a0N2 (g)<\/sub> + 3H2 (g)<\/sub> ________2NH3 (g)<\/sub>
          \n\t\tIV\u00a0\u00a0\u00a0\u00a0 3vols\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2vols
          \n40cm3<\/sup>\u00a0\u00a0\u00a0\u00a0 60cm3<\/sup>\u00a0\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0\u00a0<\/p>\n

          \u00a03vols 3H2 (g) <\/sub>= 60cm3<\/sup>
          \n\t\t1 vol = 60\/3<\/sub> = 20cm3<\/sup>
          \n\t\t:. 40cm3<\/sup> 60
          \n 1 x 20cm3<\/sup> 3 x 20 = 0cm3<\/sup><\/p>\n

          \u00a0
          \n\u00a020cm3<\/sup> 60cm3<\/sup> ___________2 x 2 = 40cm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
          \n:.Volume of unused gas = nitrogen
          \n40cm3n 20cm3 = 20cm3
          \n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
          \nEXERCISE 2
          \n<\/strong>100cm3 <\/sup>of sulphur (iv) oxide (sulphur dioxide-so2<\/sub>) gas reacts with 80cm3<\/sup> of oxygen gas to produce sulphur (vi) oxide (sulphur trioxide). Calculate the volume of the resulting gas and the volume of unused gas measured at the same temperature and pressure.
          \nEquation for the reaction 2S02 (g)<\/sub> + 02(g)<\/sub> __________2S03 (g)<\/sub><\/p>\n

          \u00a0Answer:
          \nVolume of unused 02<\/sub> = 30cm3<\/sup>
          \n\t\tVolume of resulting gas 2S03<\/sub> = 100cm3<\/sup>
          \n\t\tMASS \u2013 LIQUID VOLUME CALCULATIONS
          \n<\/strong>1.\u00a0\u00a0\u00a0\u00a0What volume of 2M HCl will be needed to react complete by with 4.0g of calcium.<\/p>\n

          \u00a0SOLUTION:
          \n<\/strong>Equation for the reaction Ca (g)<\/sub> + 2HCl (aq)<\/sub> ______CaCl2 (aq)<\/sub> + H2(g<\/sub>)
          \n1 mole Ca(s)<\/sub> __________ 2 moles HCl(aq)<\/sub>
          \n\t\t40g Ca(s)<\/sub>______________ 2 moles HCl(aq)<\/sub>
          \n\t\t1g Ca(s<\/sub> _____________ 2 mole HCl
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 40
          \n4.0g of Ca(s)<\/sub> = 2 x 4 mole HCl
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a040
          \n\u00a0\u00a0\u00a0\u00a0= 0.2 mole HCl
          \nConcentration of HCl = 2M
          \nMole = cone in mol\/dm3<\/sup> x Vol in cm3<\/sup>
          \n\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1000<\/p>\n

          \u00a00.2 mole = 2 x V
          \n\t\t\u00a0\u00a0\u00a0\u00a0 1000\u00a0\u00a0\u00a0\u00a0
          \n0.2 x 1000 = 2 x V
          \n\t\t\u00a0\u00a0\u00a0\u00a0 1000
          \n:. V = 0.2 x 1000
          \n\t\t2
          \nV = 1000Cm3<\/sup><\/p>\n

          \u00a0
          \n\u00a02.\u00a0\u00a0\u00a0\u00a0Calculate the mass of calcium which will complete by react with 500cm3<\/sup> of 0.1 MHCl<\/p>\n

          \u00a0SOLUTION:
          \n<\/strong>Equation for the reaction
          \nCa(s)<\/sub> + 2 HCl (aq) <\/sub>_________CaCl2<\/sub> (aq) + H2 (q)<\/sub>
          \n\t\tMole = cone mol\/dm3<\/sup> x Vol in cm3<\/sup>
          \n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01000
          \n= 0.1 x 500
          \n\t\t 1000
          \n= 0.05 mole<\/p>\n

          \u00a0From the equation of reaction Ca(s)<\/sub> +2HCl (aq) <\/sub>_______ CaCl2 (aq) <\/sub>+H2(q)<\/sub>
          \n\t\t2 moles of HCl (aq)<\/sub> 1 mole of Ca(s)<\/sub>
          \n\t\t2 moles HCl (aq)<\/sub> = 40g of Ca(s)<\/sub>
          \n\t\t1 mole HCl (aq)<\/sub> = 40 moles HCl
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20
          \n:. 0.05 molHCl = 40 x 0.05 mole C
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0
          \n\u00a0\u00a0\u00a0\u00a0= 1g of Ca<\/p>\n

          \u00a0THE MOLE FRACTION AND MOLE PERCENT
          \n<\/strong>
          \n\u00a0DEFINITION
          \n<\/strong>The mole fraction can be defined as the number of moles of a particular substance in a mixture divided by the total number of moles of all the substances present in the mixture.<\/p>\n

          \u00a0Example
          \nA mixture of 1 moles of chloroform and 3 moles of ethanol were kept in a measuring cylinder calculate
          \ni.\u00a0\u00a0\u00a0\u00a0The mole fraction of chloroform and ethanol<\/p>\n

          \u00a0ii.\u00a0\u00a0\u00a0\u00a0The mole percent of chloroform and ethanol.<\/p>\n

          \u00a0SOLUTION
          \n<\/strong>Total number of moles of mixture = 1 + 3 = 4 moles <\/p>\n

          \u00a0ia.\u00a0\u00a0\u00a0\u00a0The mole fraction of chloroform = 1
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0
          \nb.\u00a0\u00a0\u00a0\u00a0The mole fraction of ethanol = 3\/4<\/p>\n

          \u00a0iia.\u00a0\u00a0\u00a0\u00a0The mole percent of chloroform = \u00bc x 100
          \n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 25%
          \nb.\u00a0\u00a0\u00a0\u00a0The mole percent of chloroform = 3 x 100 \u00a0\u00a0\u00a0\u00a0= 75%
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4 1\u00a0\u00a0\u00a0\u00a0<\/p>\n

          \u00a0
          \n\u00a0
          \n\u00a0
          \n\u00a0NOTE:
          \n<\/strong>i.\u00a0\u00a0\u00a0\u00a0When the masses of the substances in the mixture are given in grammes they are converted to moles by dividing with their relative molecular masses before the mole fractions are calculated.<\/p>\n

          \u00a0ii.\u00a0\u00a0\u00a0\u00a0When the volumes of gasses at stated temperature and pressure are given, they are converted to moles before the mole fractions are calculated. <\/p>\n

          \u00a0EXAMPLE 2
          \n<\/strong>46g of ethanol was mixed with 36g of water in a reaction vessel, calculate:
          \ni.\u00a0\u00a0\u00a0\u00a0The mole fraction of water and ethanol<\/p>\n

          \u00a0ii.\u00a0\u00a0\u00a0\u00a0The mole percent of water and ethanol.<\/p>\n

          \u00a0SOLUTION:
          \n<\/strong>Rmm of water (H20<\/sup>)
          \n= (1 x 2) + 16 = 18glmol
          \nRmm of ethanol (C2<\/sub>H50<\/sup>H)
          \n= (12 x 2 + 1 x 5 + 16 + 1) = 46g\/mol
          \nNo of mole of H2<\/sub>0 molecules
          \n\u00a0\u00a0\u00a0\u00a0= 36`
          \n\t\t\u00a0\u00a0\u00a0\u00a0 18 = 2 moles
          \nNo of molecules of C2<\/sub> H5<\/sub>OH
          \n = 46
          \n\t\t 46\u00a0\u00a0\u00a0\u00a0 = 1 moles
          \nTotal number of moles present in the mixture = 2 + 1 = 3 moles
          \nia.\u00a0\u00a0\u00a0\u00a0Mole fraction of water 2
          \n\t\t 3
          \nb.\u00a0\u00a0\u00a0\u00a0Mole fraction of ethanol = 1
          \n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3
          \niia.\u00a0\u00a0\u00a0\u00a0The mole percent of H2<\/sub>0
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 2 x 100\u00a0\u00a0\u00a0\u00a0
          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3 1\u00a0\u00a0\u00a0\u00a0= 66.7%
          \nb.\u00a0\u00a0\u00a0\u00a0The mole percentage of C2<\/sub>H5<\/sub>0H
          \n\u00a0\u00a0\u00a0\u00a0= 1 x100
          \n\u00a0\u00a0\u00a0\u00a0 3 1\u00a0\u00a0\u00a0\u00a0= 33.3%<\/p>\n

          \u00a0NOTE:
          \n<\/strong>That one mole of a gas (the relative formula mass) will always tale up a volume of 24dm3<\/sup> and 24000cm3<\/sup> at room temperature and pressure (r.t.p)<\/p>\n

          \u00a0This means that 28g of N2<\/sub> will take up a volume of 24dm3<\/sup> as will 71g of Cl2 <\/sub>also
          \nR.f.m of N2<\/sub> = 14 x 2 = 28
          \nR.fm of Cl2<\/sub> = 35.5 x 2 = 71
          \nFormula
          \nVol of gas (dm3<\/sup>) = mass of gas (g)
          \n\t\t 24dm3 Rf.m of gas <\/p>\n

          \u00a0= 8
          \n\t\t 2\u00a0\u00a0\u00a0\u00a0= 4 moles<\/p>\n

          \u00a0
          \n\u00a0
          \n\u00a0Vol of gas = moles x 24dm3<\/sup>
          \n\t\t= 4 x 24 = 96dm3<\/sup><\/p>\n

          \u00a0This calculation shows that 8g of hydrogen will take up a volume of 96dm3<\/sup><\/p>\n

          \u00a0
          \n\u00a0QUESTION
          \n<\/strong>1.What volume is taken up by 10g of Ne?<\/sub><\/p>\n

          \u00a02.\u00a0\u00a0\u00a0\u00a0What volume is taken up by 56g of N2?<\/sub>\u00a0\u00a0\u00a0\u00a0
          \n3.\u00a0\u00a0\u00a0\u00a0What volume is taken up by 14.2g of Cl2?<\/sub><\/p>\n

          \u00a0
          \n\u00a0\"\"\/4.\u00a0\u00a0\u00a0\u00a0If 0.1 mole of AgN03<\/sub> reacts with HCl acid, what mass of AgCl could be produced according to the equation AgN03 (aq) <\/sub>+ HCl (aq) _____<\/sub>AgCl(s)<\/sub> + HN03 (aq)<\/sub>
          \n\t\t\u00a0\u00a0\u00a0\u00a0(Ag = 108, Cl = 35.5, H = 1, N = 14, 0 = 16)<\/p>\n

          \u00a0Answer = 14.35g of AgCl<\/p>\n

          \u00a0
          \n\u00a05.\u00a0\u00a0\u00a0\u00a0What mass of zinc metal would be required to react with dilute HCl to produce 0.5dm3<\/sup> of H2<\/sub> gas at s.t.p according to the equation below?
          \n\u00a0\u00a0\u00a0\u00a0Zn(s)<\/sub> + 2HCl (aq) <\/sub>________ZnCl2 (aq)<\/sub> + H2 (g)<\/sub>
          \n\t\t(Zn) = 65, H = 1, Cl = 35.5, G.m.v = 22.4dm3<\/sup> at s.t.p)<\/p>\n

          \u00a0Answer = 1.45g of Zn
          \n<\/strong>
          \n\u00a06.\u00a0\u00a0\u00a0\u00a0Find the number of molecules of 02<\/sub> needed to convert 5.60dm3<\/sup> of S02<\/sub> gas measured at s.t.p to form S03
          \n\u00a0\u00a0\u00a0\u00a0(G.m.v = 22.4dm3<\/sup>, NA<\/sup>= 6.02 x 1023<\/sup>molecules)
          \n\u00a0\u00a0\u00a0\u00a0
          \nAnswer =1.505 x 1023<\/sup> molecules<\/strong><\/p>\n

          \u00a07.\u00a0\u00a0\u00a0\u00a0When 10g of Na<\/sub>0H is dissolved in 1000cm3 of water, what will e the molar, concentration of the solution formed?
          \n\u00a0\u00a0\u00a0\u00a0Answer = 0.25 moles
          \n<\/strong>
          \n\u00a05.\u00a0\u00a0\u00a0\u00a0A gaseous mixture consist of 500cm3<\/sup> of hydrogen 250cm3<\/sup> of nitrogen and 1000cm3 of oxygen at s.t.p<\/p>\n

          \u00a0i.\u00a0\u00a0\u00a0\u00a0The mole fraction of each component<\/p>\n

          \u00a0<\/p>\n

            \n
          1. \n
            The mole percent of the gaseous mixture\n<\/div>\n<\/li>\n<\/ol>\n

            \u00a0
            \n\u00a0
            \n\u00a0
            \n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

            \u00a0WEEK 9 GAY LUSSAC’S LAW: It states that when gasses react, they do so in…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,196],"tags":[],"class_list":["post-2308","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2308","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2308"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2308\/revisions"}],"predecessor-version":[{"id":2309,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2308\/revisions\/2309"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2308"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2308"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}