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Chemistry Test - 3
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Chemistry Test - 3
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  • Question 1/10
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    A sample of calamine ore (ZnCO3) contains clay as an impurity which is 37.5%. Clay impurity loses some amount of its weight as moisture on prolonged heating. If the total loss in weight of ore on heating (due to CO2 and H2O) is 28%, then find the percentage loss in the weight of the clay. (Given atomic masses: Zn=65,C=12,O=16).

    Solutions

    A sample of calamine ore (ZnCO3) contains clay as impurity which is 37.5%.
    Let the total mass of the sample be 100g
    mZnCO3=62.5g and
    mclay=37.5g
    ZnCO3ZnO+CO2.
    125g of ZnCO3 corresponds to 44g of CO2.
    62.5g corresponds to 22g of CO2.
    mH2O+mCO2=28g
    mH2O=2822=6g
    % loss in the weight of clay =637.5×100=16%

    Hence option A is correct.
  • Question 2/10
    1 / -0

    2.76 g of silver carbonate on being strongly heated yields a residue weighing:

    Solutions

    When silver carbonate is strongly heated, a residue of silver is obtained.
    The reaction is as follows:
    Ag2CO3(s)Heat2Ag(s)+CO2(g)+12O2(g)
    The molecular weight of silver carbonate and silver is 276g/mol and 108g/mol respectively. Thus, when 276g (or 1mole ) of silver carbonate is heated, 2×108=216g(2moles) of silver is obtained. When 2.76g of silver carbonate is heated, the amount of residue (silver) obtained is 216276×2.76=2.16g.

    Hence option A is correct.
  • Question 3/10
    1 / -0

    We have two solutions, one is 1 L of 0.1 M NaCl and the other is 2 L of 0.2 M CaCl2. Using only these two solutions, what maximum volume of a solution can be prepared, which has [Cl−]=0.34M? 

    Given: Both electrolytes are strong.

    Solutions

    o.i MNaCl
    NaClNa++Cl
    I mol of NaCl=1 mol of Cl
    In 1 ltr solution, o.t mol of NaCl
    =0.1mol of Cl
    0.2MCaCl22ltr
    CaCl2Ca++2Cl
    I mol of CaCl2=2 mol of Cl
    In 1 ler solution, 0.2 mol of CaCl2=0.2×2mol of Cl
    In 2 ltr solution =0.8 mol of Cl
    Total moles of Cl
    =00.8+0.1=0.9mol
    Desired Molarity of Cl=0.34M
    So, volume of solution = 0.9mol0.34M=2.65 litr

    Hence option D is correct.
  • Question 4/10
    1 / -0

    There are two oxides of sulphur. They contain 50 % and 60% of oxygen, respectively, by weights. The ratio of weights of sulphur, which combines with 1 g of oxygen, is:

    Solutions

    In the first oxide, the percentages of O and S are 50% and 50% respectively.
    Thus, 1g of oxygen will combine with 1g of sulphur.
    In the second oxide, the percentages of O and S are 60% and 40%, respectively.
    Thus, 1g of oxygen will combine with 23g of sulphur.
    The ratio of weights of sulphur which combine with 1 g of oxygen =1:23=3:2

    Hence option D is correct.
  • Question 5/10
    1 / -0

    In the reaction:
    Na2S2O3+4Cl2+5H2ONa2SO4+8HCl, the equivalent weight of Na2S2O3 will be (M= molecular weight of Na2S2O3) :
    Solutions

    Na2S2O3Na2SO4
    The total change in oxidation number =4×2=8 ENa2S2O3=mol.wtV.f=M8

    Hence option B is correct.
  • Question 6/10
    1 / -0

    Calcium carbonate (10g) reacts with aqueous HCl(0.3mol) according to the reaction,
    CaCO3(s)+2HCl(aq)CaCl2(aq)+CO2(g)+H2O(l)The number of moles of CaCl2 formed in the reaction is:
    Solutions

    The balanced reaction is
    CaCO3(s)+2HCl(aq)CaCl2(aq)+CO2(g)+H2O(l)10g calcium carbonate corresponds to 10100=0.1mole (Molecular weight of CaCO3=100g/mol).
    It will react with 0.2molHCl.
    However, 0.3molHCl is present.
    Hence, calcium carbonate is the limiting reagent.
    0.1mol of calcium carbonate will form 0.1mol of calcium chloride.

    Hence option A is correct.
  • Question 7/10
    1 / -0

    Find the mass of pure hydrogen chloride gas dissolved per litre of its solution if the normality of the solution is 0.15N. [At. Wt. : H=1,Cl=35.5]

    Solutions

    The solution is 0.15N. Hydrogen chloride (HCl) contains one replaceable H atom.
    Molar mass of HCl=(1+35.5)=36.5u
    The expression for the equivalent weight of an acid is as given below.
    Equivalent weight=Molecular mass/Number of replaceable H
    HCl contains one replacable hydrogen. Substitute values in the above expression.
    =36.51=36.5u
    The expression for the normality of solution is as given below.
    =Solute in grams per litre of solution/Equivalent weight in grams
    Solute in grams per litre of solution = Normality Equivalent weight in grams.
    The mass of pure hydrogen chloride gas dissolved per litre of its solution
    =(0.15×36.5)g=5.475g/l

    Hence option A is correct.
  • Question 8/10
    1 / -0

    45.4 L of dinitrogen reacted with 22.7L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below:

    2N2(g)+O2(g)→2N2O(g)

    Which law is being obeyed in this experiment?

    Solutions

    The ratio of the volumes of dinitrogen, oxygen and nitrous oxide is 45.4:22.7:45.4.
    This is in the simple whole number ratio 2 : 1 : 2
    This satisfies the Gay Lussac's law of combining volume of gases.

    Hence option A is correct.
  • Question 9/10
    1 / -0

    Sulphuric acid is produced when sulphur dioxide reacts with oxygen and water in the presence of a catalyst according to the reaction:
    2SO2(g)+O2(g)+2H2O(l)2H2SO4
    If 5.6mol of SO2 reacts with 4.8 mol of O2 and an excess of water, what is the maximum number of moles of H2SO4 that can be obtained?
    Solutions

    The balanced reaction is given below:
    2SO2+O2+2H2O2H2SO4
     5.6 4.8 (No. of moles given) 
    5.624.81 (Ratio of moles given and stoichiometry coefficient) =2.8=4.8
    As 2.8 is lower than 4.8,SO2 is the limiting reagent and product will form according to number of moles of SO2 (given).
    So, moles produced of H2SO4=22×5.6=5.6mol

    Hence option D is correct.
  • Question 10/10
    1 / -0

    The mass percent of calcium, phosphorus and oxygen in calcium phosphate, Ca3(PO4)2 is :

    Solutions

    Molecular formula of calcium
    phosphate is Ca3(PO4)2
    Its molar mass is 3(40)+2(31)+8(16)=310g/mol
    Percent of calcium =120310×100=38.71%
    Percent of phosphorus =62310×100=20.0%
    Percent of oxygen =128310×100=41.29%

    Hence option A is correct.
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