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Some Basic Concepts of Chemistry Test - 45

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Some Basic Concepts of Chemistry Test - 45
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  • Question 1
    1 / -0
    A mixture of calcium and magnesium carbonates weighing 1.4 g was strongly heated until no further loss of weight was perceived. The residue weighed 0.76 g. What percentage of MgCO3MgCO_3 was present in the mixture?
    Solution
    CaCO3+MgCO3CaO+MgOCaCO_3 + MgCO_3 \rightarrow CaO + MgO
    let mass of CaCO3CaCO_3 = x g 
    and mass of MgCO3MgCO_3 = 1.4x1.4-x g

    moles of CaCO3CaCO_3 = x100\dfrac{x}{100} = moles CaOCaO produced


    moles of MgCO3MgCO_3 = x84\dfrac{x}{84} = moles MgOMgO produced

    mass of  CaOCaO produced = moles * no. of moles = (x)56100\dfrac{(x) * 56}{100}

    mass of  MgOMgO produced = moles * no. of moles = (1.4x)40100\dfrac{(1.4-x) * 40}{100}

    so, total weight of product = (1.4x)56100\dfrac{(1.4-x) * 56}{100} (1.4x)40100\dfrac{(1.4-x) * 40}{100} = 0.76
    so, value of x = 1.356 g
    percentage of CaCO3CaCO_3 = 1.3561.4100\dfrac{1.356}{1.4} * 100 = 96.8 %
    percentage of MgCO3MgCO_3 = 3.14 %
  • Question 2
    1 / -0
    The number of g-atom of oxygen in 6.02×10246.02\times { 10 }^{ 24 } COCO molecule is
    Solution
    One molecule of COCO contains one oxygen atom
    6.02×1024\therefore 6.02\times { 10 }^{ 24 }\quad molecules of COCO contain 6.02×10246.02\times { 10 }^{ 24 } oxygen atoms
    6.02×10236.02\times { 10 }^{ 23 } atoms of oxygen \equiv 1g1g atom of oxygen
    6.02×1024\therefore 6.02\times { 10 }^{ 24 } atoms of oxygen \equiv 10g10g atom of oxygen.
  • Question 3
    1 / -0
    Law of constant composition is same as the law of:
    Solution
    The law of constant composition is not the same as the law of conversation of mass or the law of conservation of energy, as these are related to the mass and energy of a system. It is also not related to the law of multiple proportions, which applies when two or more elements/ compounds have multiple ways of combining into different compounds.
    The Law of Constant Composition, discovered by Joseph Proust, is also known as the Law of Definite Proportions.
    According to the law of constant composition - "A chemical compound always contains its component element in the fixed ratio by mass." Hence option D is correct.
  • Question 4
    1 / -0
    Modern shoe soles are made of
  • Question 5
    1 / -0
    If 100 gm of salt solution contains 20 g of salt dissolved in it, the percentage of the mass of the solution is________.
    Solution
    Given, 100gm100gm solution contains 20g20g salt in it
    \therefore Percentage of mass in the solution= MassofsaltdissolvedMassofsolution×100\cfrac {Mass\quad of\quad salt\quad dissolved}{Mass\quad of\quad solution}\times 100
    =20gm100gm×100=\cfrac {20gm}{100gm}\times 100
    =20=20%
  • Question 6
    1 / -0
    How many grams of NaOHNaOH will be required to prepare 500 g500\ g solution containing 1010% wwNaOH\dfrac {w}{w}NaOH solution?
    Solution
    In 500 g500\ g of solution of 10% w/w NaOHNaOH present.

    Mass of the solute =10×500100=50 g= \dfrac {10\times 500}{100} = 50\ g

    So, 50 g50\ g of NaOHNaOH will be required to prepare 500 g500\ g of 1010% ww NaOH\dfrac {w}{w}\ NaOH solution.
  • Question 7
    1 / -0
    5.0 g5.0\ g sample of urea when heated with NaOBrNaOBr and NaOHNaOH gives 1120 ml1120\ ml of nitrogen at STP. The percentage purity of the sample is:
    Solution
    NH2CONH2+3NaOBr+2NaOHN2+3NaBr+Na2CO3+3H2O{ NH }_{ 2 }CO{ NH }_{ 2 }+3NaOBr+2NaOH\longrightarrow { N }_{ 2 }+3NaBr+{ Na }_{ 2 }{ CO }_{ 3 }+3{ H }_{ 2 }O

    \because 22400 ml22400\ ml of N2{N}_{2} is obtained from 60 g60\ g of urea at S.T.P.

    \therefore 1120 ml1120\ ml of N2{N}_{2} will be obtained from

    60×112022400g\Rightarrow \cfrac { 60\times 1120 }{ 22400 } g of urea =3 g=3\ g or urea

    Now,
    \because 5 g5\ g sample contains 3.0 g3.0\ g of urea

    \therefore 100g100g sample will contain 3.0×1005.0=60g\cfrac { 3.0\times 100 }{ 5.0 } =60g of urea

    Hence, the purity of urea sample is 6060%
  • Question 8
    1 / -0
    Sodium reacts with excess oxygen to form sodium oxide.A student wants to prepare 1.24 g of sodium oxide.While doing the calculations, he uses atomic number of sodium instead of obtained due to this mistake?
  • Question 9
    1 / -0
    The number of gram molecule of oxygen in 6.02×10246.02 \times 10^{24} CO molecules is :
    Solution
    Given :

    6.02×10246.02\times 10^{24} COCO molecules

    We know,

    11 mole COCO equivalent to 6.023×10236.023\times 10^{23} molecules of COCO

    6.023×10236.023\times 10^{23} molecules of COCO contain 6.023×10236.023\times 10^{23} atoms of OO

    Then, 6.02×10246.02\times 10^{24} COCO molecules contain 6.02×10246.02\times 10^{24} atoms of OO

    \Rightarrow no. of gram atoms of O=6.02×10246.02×1023O=\cfrac {6.02\times 10^{24}}{6.02\times 10^{23}}

    =10gm=10gm atoms  1\longrightarrow 1

    As oxygen is a diatomic molecule.

    No. of gm of molecules of oxygen=10gmatoms2atoms/molecule=5gm\cfrac {10gm\quad atoms}{2\quad atoms/molecule}=5gm molecules

    Therefore, the correct option is B.

  • Question 10
    1 / -0
    A solution has 0.05 MMg+20.05\ M - Mg^{+2} and 0.05MNH30.05M - NH_{3}. Calculate the minimum concentration of NH4ClNH_{4}Cl required to prevent only precipitation of MgMg as Mg(OH)2KspMg(OH)_{2}\cdot K_{sp} of Mg(OH)2=9×1012Mg(OH)_{2} = 9\times 10^{-12} and KbK_{b} of NH3=1.8×105NH_{3} = 1.8\times 10^{-5}.
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