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Some Basic Concepts of Chemistry Test - 42

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Some Basic Concepts of Chemistry Test - 42
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  • Question 1
    1 / -0
    What could be better than a dozen (12)(12) donuts? How about a baker's dozen (13)(13) of donuts? Another large unit of measurement is known as Avogadro's number (6.022×1023)(6.022\times 10^{23}).
    What is TRUE about Avogadro's number?
    Solution
    Avagadro constant i.e. 6.023×10236.023 \times 10^{23} particles is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by 1 mole1\space mole
  • Question 2
    1 / -0
    How many molecules of water are present in a 0.25 mole0.25\ mole of H2OH_2O?
    Solution
    1 mole1\space mole of a compound has 6.023×10236.023\times 10^{23} molecules in it.
    1 mole=\Rightarrow 1\space mole =6.023×10236.023\times 10^{23} molecules
    0.25 mole=x 0.25\space mole = x molecules
    x=(6.0234)×1023\Rightarrow x = (\dfrac{6.023}{4}) \times 10^{23}=1.5×1023= 1.5\times 10^{23} molecules of water.
  • Question 3
    1 / -0
    What is the mass of 6.022×10236.022\times { 10 }^{ 23 } molecules of NH3{NH}_{3}?
    Solution
    1 mole1\space mole of NH3NH_3 has 6.023×10236.023 \times 10^{23} molecules of NH3NH_3.
    So, the mass of 6.023×10236.023 \times 10^{23} molecules of NH3NH_3. is the mass of 1 mole1\space mole of NH3NH_3  i.e. molar mass of NH3.NH_3. 
    Molar mass =14.01+3(1.01)=17.04 g= 14.01 + 3(1.01) = 17.04 \space g
  • Question 4
    1 / -0
    What is the mass of 6.022×10236.022\times { 10 }^{ 23 } formula units of (NH4)2SO4{({NH}_{4})}_{2}{SO}_{4}?
    Solution

    1 mole1\space mole of (NH4)2SO4(NH_4)_2SO_4 has 6.023×10236.023 \times 10^{23} formula units.

    So, the mass of 6.023×10236.023 \times 10^{23} formula units of (NH4)2SO4(NH_4)_2SO_4. is molar mass of (NH4)2SO4(NH_4)_2SO_4

    Molar mass of (NH4)2SO4(NH_4)_2SO_4 =2(14+4)+32+4×16=132 g= 2(14 + 4) + 32 + 4\times 16 = 132 \space g

  • Question 5
    1 / -0
    Referring to the molecular sketch and table above, choose the option that best describes the percentage of carbon in the molecule, in terms of both atomic composition and mass-percent composition.

    Solution
    The molecular formula of compound =C4H9NO3= C_4H_9NO_3
    %\% of carbon atom =44+9+1+3=417×10025%= \dfrac{4}{4+9+1+3} = \dfrac{4}{17} \times 100 \approx 25\%

    %\% of carbon by mass =4×1248+9+14+48=48119×100=40.33%= \dfrac{4\times 12}{48+9+14+48} = \dfrac{48}{119} \times 100 = 40.33\%
  • Question 6
    1 / -0
    The atomic mass of copper is 64 amu64\ amu, while the atomic mass of oxygen is 16 amu16\ amu.
    Let =copper= copper and =oxygen= oxygen.
    Which of the following would represent a compound that was 5050% copper, by mass?

    Solution
    Given, compound has 50%50\% copper, by mass.
    Let Molar mass of compound =M= M 
    \Rightarrow Mass of copper 50100×M=0.5M\dfrac{50}{100} \times M = 0.5 M
    No. of moles of copper =0.5M64;= \dfrac{0.5M}{64} ;
    No. of moles of oxygen =0.5M16;= \dfrac{0.5M}{16} ;
    Ratio of copper to oxygen 1:41:4
    CuO4\Rightarrow CuO_4
  • Question 7
    1 / -0
    Consider the figure below, in which the blue circles represent atoms of copper metal, atomic mass 63.5 amu63.5\ amu, while the red circles represent atoms in zinc metal, atomic mass 65.4 amu65.4\ amu.
    Here are four statements concerning this system. Identify the correct statement or statements.
    I. This is a single-phase mixture, with a copper-to-zinc atom ratio of 3:13 : 1.
    II. This is two-phase mixture, with a copper-to-zinc atom ratio of 8:18 : 1.
    III. The mixture is 8080% copper, by mass.
    IV. This mixture could be an alloy of copper and zinc.

    Solution
    Here, no. of atoms of Cu=30 atomsCu = 30\space atoms
    No. of atoms of Zn=10 atomsZn = 10\space atoms
    Single phase mixture is a homogenous mixture with same proportions of atoms.
    Here, the ratio of copper to zinc is 3:13:1
    So, it is a two phase mixture.
    % Cu\Rightarrow \% \space Cu by mass =3×63.53×63.5+65.4=190.5255.9=0.744×100=74%80%= \dfrac{3\times 63.5}{3\times 63.5 + 65.4} = \dfrac{190.5}{255.9} = 0.744\times 100 = 74\% \approx 80\% by mass
    The mixture can be alloy of zinc.
  • Question 8
    1 / -0
    How many molecules of methane are present in a sample that contains 1.0×10101.0\times { 10 }^{ 10 } moles of methane?
    Solution

    1 mole1\space mole of metane has 6.023×10236.023 \times 10^{23} molecules of methane.

    1 mole=1\space mole = 6.023×10236.023 \times 10^{23} molecules

    1010 molesx10^{10}\space moles \rightarrow x


    x=x =  (6.023×1023)×1010=(6.023 \times 10^{23})\times 10^{10} =  6.023×10336.023 \times 10^{33} molecules 


    So, 1010 moles10^{10}\space moles has approximately 6.0×10336.0 \times 10^{33} molecules 

  • Question 9
    1 / -0
    Equal masses of oxygen, hydrogen, and methane are taken in a container in identical condition. Find the ratio of volumes of the gases.
    Solution
    The molecular weights of a given mixture of gases are
    1. Oxygen = 32
    2. Hydrogen = 2
    3. Methane = 16
     
    Since masses are equal, assume mass as x,
     
    So By using formula,  no of moles = massmolecularweight\dfrac{mass }{molecular weight}
    nO2=x32n_{O_2} = \dfrac{x}{32}   : nH2=x2n_{H_2} = \dfrac{x}{2}  : nCH4=x16n_{CH_4} = \dfrac{x}{16}  

    ratio is 1:16:21 : 16 : 2
  • Question 10
    1 / -0
    How many grams of CaWO4CaWO_4 would contain the same mass of tungsten that is present in 569g of $$FeWO_4$ $? (W=184)
    Solution
    Let the mass of CaWO4CaWO_4 be w g. As given,
    Mass of W in w g of CaWO4CaWO_4 = mass of W in 569g of FeWO4FeWO_4

    Moles of W in CaWO4×CaWO_4 \times at. mass of W = Moles of W in FeWO4×FeWO_4 \times at. mass of W

    As both CaWO4CaWO_4 and FeWO4FeWO_4 contains 1 atom of W each,
    ∴ moles of CaWO4× CaWO_4 \times  at. mass of W = Moles of FeWO4×FeWO_4 \times at. mass of W
     =  w288×184\dfrac{w}{288} \times 184 = (569/304)×184 (569/304) × 184
    = W=539.05g.W = 539.05 g.
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