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Some Basic Concepts Of Chemistry Test 5

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Some Basic Concepts Of Chemistry Test 5
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  • Question 1
    1 / -0

    Vapour density of a gas is 22. Its molecular mass will be:​

    Solution

    Molecular mass = 2 x vapour density
    = 2 x 22
    = 44
    Therefore, molecular mass of a gas will be 44.

  • Question 2
    1 / -0

    What is the mass of 0.20 mole of C2H5OH (ethanol)?

    Solution

    Molecular Mass of Ethanol
    = 2(12) + 5 (1) + 16 + 1 
    = 24 + 5 + 17 
    = 46
    Now,
    Moles (n) = Given Mass (m)/Molecular Mass (M)
    Thus, to = nM
    = (0.2) (46)
    = 9.2

  • Question 3
    1 / -0

    Consider the reaction between hydrogen and oxygen gases to form water. Which of the following is/are not conserved in the reaction?
    2H2(g) + O2(g) → 2H2O(l)​

    Solution

    In the equation both side of reaction moles of atoms is conserved ie, both reactant and product contain 4H and 2O but there is 3 (2+1) moles of reactant molecules turns into 2 moles of product molecules. 

  • Question 4
    1 / -0

    Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is C(s) + H2(g) + O2(g) → C12H22O11 (s)

    Solution

    C = 84/12 = 7 mole

    H2 = 12 g = 6 mole

    O2 = 56/22.4 = 5/2 mole

    12C + 11H2 + 11/2 O2 → C12H22O11

    L.R. = O2

    11/2 mole O2 produce 1 mole sucrose 5/2 mole O2 will for 5/11 mole sucrose mass of sucrose = 5/11 × (mol. mass)

    = 5/11 × 342

    = 155.45 g

  • Question 5
    1 / -0

    Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are

    Solution

    Let atomic weight of x = Mx

    atomic weight of y = My

    we know,

    mole = weight /atomic weight

    a/c to question,

    mole of xy2 = 0.1

    so,

    0.1 = 10g/( Mx +2My)

    Mx + 2My = 100g -------(1)

    for x3y2 ; mole of x3y2 = 0.05

    0.05 = 9/( 3Mx + 2My )

    3Mx + 2My = 9/0.05 = 9× 20 = 180 g ---(2)

    solve eqns (1) and (2)

    2Mx = 80

    Mx = 40g/mol

    and My = 30g/mole

  • Question 6
    1 / -0

    Number of atoms present in 52 g of helium is:
    (NA is Avogadro’s constant Gram atomic mass of He is 4 g)​

    Solution

    Gram atomic mass of helium is 4 g so number of atoms present in 52 g of helium
    52/4 × NA = 13 NA

  • Question 7
    1 / -0

    The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would be

    Solution

    As vapour density of mixture is 25,

    Molar mass of mixture is 2×25=50.

    Now, assume mole % of B in mixture is x.

  • Question 8
    1 / -0

    Atomic mass of Cl is 35.5 g. Calculate the mass of 4.50 moles of chlorine gas, Cl2.​

    Solution

    Mass of 1 mole of Clis 71 g. Hence mass of 4.50 moles is 71 X 4.50 = 319.5 g

  • Question 9
    1 / -0

    Atomic mass of bromine is 80 g. The mass of four moles of molecular bromine (Br2) is:​

    Solution

    Mass of one mole of Br2  is 160g. Hence mass of four moles of molecular bromine is 4 X 160 = 640 g.

  • Question 10
    1 / -0

    Number of nitrogen atoms present in 1.4 g of N2

    Solution

    Given weight of N₂ gas = 1.4 g

    Molar mass of N₂ gas = 28 g

    So, mole = given mass/ molar mass

    ⇒ mole = 1.4/28 = 1/20 mole

    Now, number of molecules = mole × avogadro number

    ⇒ number of molecules = 1/20 × 6.022 × 10²³

    ⇒ number of molecules = 3.011 × 10²²

    Now, we are asked for number of atoms. In N₂, there are 2 atoms, so to obtain number of atoms we will multiply with 2 in number of molecules.

    ⇒ number of atoms = 2 × 3.011 × 10²²

    ⇒ number of atoms = 6.022 × 10²²

  • Question 11
    1 / -0

    Volume of 17g of NH3 at N.T.P. will be​

    Solution

    Molar mass of NHis 17g. So the volume of 17g of NH3 i.e 1 mole NH3 at N.T.P. will be 22.4L.

  • Question 12
    1 / -0

    Which of the following has maximum number of moles?​

    Solution

    1g hydrogen has the maximum number of moles because atomic mass of hydrogen is smallest among all.

  • Question 13
    1 / -0

    Molar mass of F2 is 38 g. How many atoms are present in 0.147 mole of F2?​

    Solution

    One molecule of F2  contains 2 atoms of fluorine.

    1 mol of fluorine contains 2 X 6.023 X 1023 atoms

    So, 0.147 moles will contain 0.147 X 2 X 6.023 X 1023 = 1.76 X 1023 atoms of flourine.

  • Question 14
    1 / -0

    Mass of 6.022 x 1022 molecules of CO2 is about:​

    Solution

    1 mole of a substance is equal to its molecular mass expressed in terms of grams.
    Gram molecular mass of CO2 = 12 + (16 Χ 2) = 12 + 32 = 44 g
    Mass of 6.023 X1022 molecules of CO2 is 44g/10 = 4.4g.

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