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Some Basic Concepts of Chemistry Test - 59

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Some Basic Concepts of Chemistry Test - 59
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which  of the following weights the  most?
    Solution

  • Question 2
    1 / -0
    AB2AB_2 and A2B2A_2B_2 are two compounds of the elements A and B. 0.25 mole of each of these compounds weights 9 g and 16 g respectively. Find the atomic masses of A and B. 
    Solution

  • Question 3
    1 / -0
    Hemoglobin contains 0.334 % of iron by weight. The molecular weight of hemoglobin is approximately 67200.The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of hemoglobin is 
    Solution

  • Question 4
    1 / -0
    The critical density of the gas CO2CO_2 is 0.44 g cm3g\ cm^{–3} at a certain temperature. If rr is the radius of the molecule, r3r^3 in cm3cm^3 is approximate: 

    [N is Avogadro number]
    Solution
    Critical density of a gas =0.44gcm3= 0.44 gcm^{-3}
    Molar mass of CO2CO_2 = 44 g/mole

    Density=mass/volumeDensity = mass /volume
    Mass of N molecules = 44 g of CO2CO_2    N=6.022×1023N =6.022\times 10^{23}

    Mass of 1 molecule =44N= \dfrac{44}{N}
    The volume of 1 molecule =43πr3=\cfrac {4}{3}\pi r^3
    Putting the values,
    0.44=44N43πr3 0.44 = \dfrac{\dfrac{44}{N}}{\cfrac {4}{3}\pi r^3}

    r3=100πN r^3 = \dfrac{100}{\pi N}
  • Question 5
    1 / -0
    The solubility of potassium chloride in water at 20oC20^oC is 34.7g34.7g in 100g100g of water. The density of solution is 1.3g/ml1.3g/ml. Calculate the % of mass/volume concentration of potassium chloride.
    Solution

  • Question 6
    1 / -0
    In N7 14\overset { 14 }{ \underset { 7 }{ N }  } if mass attributed to electron were doubled & the mass attributed to protons were halved, the atomic mass would become approximately:-
    Solution

  • Question 7
    1 / -0
    The weight of 1 mole of a gas of density is 0.1784gL1atNTPis {\text{0}}{\text{.1784}}{\kern 1pt} {\kern 1pt} {\text{g}}{\kern 1pt} {{\text{L}}^{ - 1}}{\text{at}}{\kern 1pt} {\text{NTP}}{\kern 1pt} {\text{is}}{\kern 1pt}  -
    Solution

  • Question 8
    1 / -0
    Find the concentration of glucose in blood which have osmotic pressure π=7.7 atm\pi = 7.7 \ atm at T=25oCT = 25^oC
    Solution
    π=CRT\pi = CRT
    7.7=C×0.082×2987.7 = C \times 0.082 \times 298
    C=7.724.44×100=0.31C = \dfrac{7.7}{24.44} \times 100 = 0.31
  • Question 9
    1 / -0
    1212 gm of an alkaline earth metal gives 14.814.8g of its nitride. The atomic mass of metal is:
    Solution
    Let the alkaline earth metal be AA.

    3A+N2A3N2\therefore 3A+N_{2}\rightarrow A_{3}N_{2}

    Molecular mass of compoundAtomic mass of metal\Rightarrow \dfrac{Molecular\ mass\ of\ compound}{Atomic\ mass\ of\ metal}

    =Wt. of compoundWt. of metal=\dfrac{Wt.\ of\ compound}{Wt.\ of\ metal}

    Let the atomic mass of metal be xx.

    \rightarrow Molecular mass of compound=3x+28=3x+28

    3x+283x=14.812\Rightarrow \dfrac{3x+28}{3x}=\dfrac{14.8}{12}

    x=40\Rightarrow x=40
  • Question 10
    1 / -0
    A gas mixture contains 50%50\% helium and 50%50\% methane by volume. What is the percent by weight of methane in the mixture?
    Solution
    Solution contain He+CH4He+CH_4

    %\% by volume is same as %\% by moles.

    Molecular wt. of He=4 g/molHe=4\ g/mol

    Molecular wt. of CH4=16 g/molCH_4=16\ g/mol

    Assuming 1 mole of HeHe and 1 mole CH4CH_4

    % wt\%\ wt of CH4=wt of CH4Total wt×100=164+16×100=80.0%CH_4=\dfrac{wt\ of\ CH_4}{Total\ wt}\times 100=\dfrac{16}{4+16}\times 100=80.0\%
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