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21 December, 07:45

What is the mass (in grams) of 9.15 * 1024 molecules of methanol (ch3oh) ?

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Answers (2)
  1. 21 December, 09:52
    0
    Solution:

    first we have to calculate the mass of 1 mole of CH3OH?

    C = 12

    H3 = 3x1 = 3

    O = 16

    H = 1

    total mass 12+3+16+1=32g.

    Total 32g So we know 6.02x10^23 molecules weigh in32g

    therefore,

    1 molecule weighs 32/6.02x10^23 g

    now,

    9.95x10^24 molecules weighs 9.95x10^24 x (32/6.02x10^23) = 9.95x32x10/6.02=99.5x32/6.02

    =528.90g

    thus the mass in grams is 528.90g.
  2. 21 December, 10:12
    0
    486.68 g

    Solution:

    Data Given:

    Number of Molecules = 9.15 * 10²⁴

    M. Mass of Methanol = 32.04 g. mol⁻¹

    Mass of Methanol = ?

    Step 1: Calculate Moles of Methanol,

    Moles = Number of Molecules : 6.022 * 10²³ Molecules. mol⁻¹

    Putting value,

    Moles = 9.15 * 10²⁴ Molecules : 6.022 * 10²³ Molecules. mol⁻¹

    Moles = 15.19 mol

    Step 2: Calculate Mass of Methanol:

    Moles = Mass : M. Mass

    Solving for Mass,

    Mass = Moles * M. Mass

    Putting values,

    Mass = 15.19 mol * 32.04 g. mol⁻¹

    Mass = 486.68 g
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