Question

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?

Answer 1

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.

Answer 2

Answer: The total number of particles present in the sample are $53.815\times 10^{23}$ particles.

Explanation:

We are given a total mass of the sample 103.75 grams in which 30% are helium atoms and 70% are krypton atoms.

The mass of 30% of helium atoms in the given amount of sample is = $\frac{30}{100}\times 103.75=31.125g$

the mass of 7% of krypton atoms in the given amount of sample is = $\frac{70}{100}\times 103.75=72.625g$

According to mole concept:

1 mole of an element contains $6.022\times 10^{23}$ atoms/particles.

• For Helium:

1 mole of helium contains 4 grams.

4 grams of helium element contain $6.022\times 10^{23}$ atoms/particles.

So, 31.125 g of helium element contain $\frac{6.022\times 10^{23}}{4}\times 31.125=46.858\times 10^{23}$ atoms/particles.

• For Krypton:

1 mole of krypton contains 84 grams.

84 grams of krypton element contain $6.022\times 10^{23}$ atoms/particles.

So, 72.625 g of krypton element contain $\frac{6.022\times 10^{23}}{84}\times 72.625=5.026\times 10^{23}$ atoms/particles.

Total number of particles in the given sample = $(46.858+5.026)\times 10^{23}=51.884\times 10^{23}$ atoms/particles.