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Some Basic Concepts Of Chemistry Test 15

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Some Basic Concepts Of Chemistry Test 15
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  • Question 1
    1 / -0

    0.48 g of a sample of a compound containing boron and oxygen contains 0.192 g of boron and 0.288 gof oxygen. What will be the percentage composition of the compound?

    Solution

    Mass of compound taken = 0.48 g
    Mass of boron in sample = 0.192 g
    Mass of oxygen in sample = 0.288 g
    Percentage of B = 0.192/0.48 x 100 = 40%
    Percentage of O = 0.288/0.48 x 100 = 60%

  • Question 2
    1 / -0

    A compound contains two elements 'x' and 'Y' in the ratio of 50% each. Atomic mass of 'x' is 20 and 'Y' is 40. What can be its simplest formula?

    Solution

    Percentage of element 'X' = 50;
    Mole ratio = 50/20 = 2.5
    Percentage of element 'Y' = 50, mole ratio = 50/40 = 1.25
    For 'X' simple ratio = 2.5/1.25 = 2
    For 'Y' simple ratio = 1.25/1.25 = 1
    Formula = X2Y

  • Question 3
    1 / -0

    The empirical formula of a compound is CH2O2.What could be its molecular formula?

    Solution

    Since empirical formula is multiplied by n to get molecular formula.
    CH2O2 will give only C2H4O4 as its molecular formula. (CH2O2)n where n = 1, 2, 3,... etc.

  • Question 4
    1 / -0

    A gas has molecular formula (CH)n. If vapour density of the gas is 39, what should be the formula of the compound?

    Solution

    Mol. wt. = 2 x V.D. = 2 x 39 = 78
    (CH)n = (13)n or 13 x n = 78
    n = 78/13 = 6
    ∴ Molecular formula = C6H6

  • Question 5
    1 / -0

    Choose the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively and its molecular mass is 160.

    Solution

    For element Fe, mole of atoms = 69.9/56 = 1.25
    For element O, Mole of atoms = 30.1/16 = 1.88
    Mole ratio of Fe = 1.25/1.25 = 1
    Mole ratio of O = 1.88/1.25 = 1.5
    Simplest whole number ratio of Fe and O = 2, 3
    Empirical formula of compoimd = Fe2O3
    Molecular mass of Fe2O3 = 160

    Molecular formula = Fe2O3

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