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?

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.