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Mole Concept Test - 1

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Mole Concept Test - 1
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  • Question 1
    1 / -0

    The number of hydrogen molecules in 4.032 g is

    Solution

    Atomic mass of Hydrogen = 1.0079 u
    This implies that 1 mole of Hydrogen atoms weighs 1.0079 g.
    Therefore, 4.032g of hydrogen atoms will have 4.032 / 1.0079 mole atoms = 4 mole atoms
    1 molecule of hydrogen has 2 atoms.4 moles of hydrogen atoms will be there in 2 mole molecules of hydrogen.
    Hence, the number of hydrogen molecules in 4.032 g is 2 × 6.023 × 023.

     

  • Question 2
    1 / -0

    A solution is prepared by dissolving 2.6 g of H2SO4 in 40 g of water. Which of the following is the correct value of mole fraction of H2SO4?

    Solution

    The molar mass of H2SO4 is 98 g and mass of water is 40.0 g as its assumed density is 1.00 g mL–1. Thus, the number of moles of H2SO4 in the solution is (2.6 g)/(98 g/mol) = .0265 mol. The number of moles of water is (40 g)/(18 g mol–1) = 2.222 mol. Then, the mole fraction of H2SO4 is (.0265 mol)/(0.0265 + 2.222) mol = .0265/2.2485 = 0.012.

     

  • Question 3
    1 / -0

    What is the mass in grams of 2.5 moles of slaked lime [Ca(OH)2]?
    (Atomic masses: Ca = 40 g, O = 16 g and H = 1 g)

    Solution

    Mass = Molar mass × Number of moles
    Molar mass of Ca(OH)2 = 40 + 2(16 + 1)
    = 40 + 34 = 74 g
    Therefore, mass of Ca(OH)2 = 74 × 2.5 = 185 g

     

  • Question 4
    1 / -0

    Calculate the volume occupied by 2.9 g of butane at STP.

    Solution

    PV = nRT to this:V = nRT / P
    Substitute:V = [ (2.9 g / 58 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm
    V = 1.12 L (to three significant figures)

     

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