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18 August, 06:11

A sample that weighs 103.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the sample?

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  1. 18 August, 06:33
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    1) we calculate the molar mass of He (helium) and Kr (Krypton).

    atomic mass (He) = 4 u

    atomic mas (Kr) = 83.8 u

    Therefore the molar mass will be:

    molar mass (He) = 4 g/mol

    molar mass (Kr) = 83.8 g/mol.

    1) We can find the next equation:

    mass=molar mass x number of moles.

    x=number of moles of helium

    y=number of moles of helium.

    (4 g/mol) x + (83.8 g/mol) y=103.75 g

    Therefore, we have the next equation:

    (1)

    4x+83.8y=103.75

    2) We can find other equation:

    We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.

    1 mol is always 6.022 * 10²³ atoms or molecules, (in this case atoms).

    Then:

    x=number of moles of helium

    y=number of moles of helium.

    (x+y) = number of moles of our sample.

    x=30% of (x+y)

    Therefore, we have this other equation:

    (2)

    x=0.3 (x+y)

    With the equations (1) and (2), we have the next system of equations:

    4x+83.8y=103.75

    x=0.3 (x+y) ⇒ x=0.3x+0.3y ⇒ x-0.3x=0.3y ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7

    ⇒x=3y/7

    We solve this system of equations by substitution method.

    x=3y/7

    4 (3y/7) + 83.8y=103.75

    lower common multiple) 7

    12y+586.6y=726.25

    598.6y=726.25

    y=1.21

    x=3y/7=3 (1.21) / 7=0.52

    We have 0.52 moles of helium and 1.21 moles of Krypton.

    1 mol=6.022 * 10²³ atoms

    Total number of particles = (6.022 * 10²³ atoms / 1 mol) (number of moles of He + number of moles of Kr).

    Total number of particles=6.022 * 10²³ (0.52+1.21) = 6.022 * 10²³ (1.73) =

    =1.044 * 10²⁴ atoms.

    Answer: The sample have 1.044 * 10²⁴ atoms.
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