Question

1. how many particles are in 1.20 mol of sodium sulfide? n = ? n = 1.20 mol n = 7.23×10²³ particles nₐ = 6.02×10²³ part/mol 5. how many particles are in 1.25 mol of no₂? 6. a small pin contains 0.0178 mol of iron, fe. how many atoms of iron are in the pin? 7. how many formula units are in 0.21 mol of magnesium nitrate, mg(no₃)₂? 8. a litre of water contains 55.6 mol of water. how many molecules of water are in this sample? 9. ethyl acetate, c₄h₈o₂, is frequently used in nail polish remover. a typical bottle of nail polish remover contains about 2.5 mol of ethyl acetate. how many molecules are in the bottle of nail polish remover? 10. consider a 0.829 mol sample of sodium sulfate, na₂so₄. how many particles are in the sample? n = nnₐ n = 0.829 mol × 6.02×10²³ form units/mol n = 4.99×10²³ formula units

Explanation:

Step1: Recall Avogadro's number formula

The formula to find the number of particles $N$ is $N = n\times N_A$, where $n$ is the number of moles and $N_A=6.02\times 10^{23}\text{ particles/mol}$.

Step2: Solve problem 5

Given $n = 1.25\ mol$ of $NO_2$. Using the formula $N=n\times N_A$, we substitute $n = 1.25\ mol$ and $N_A = 6.02\times 10^{23}\text{ particles/mol}$. So $N=1.25\ mol\times6.02\times 10^{23}\text{ particles/mol}=7.525\times 10^{23}$ particles.

Step3: Solve problem 6

Given $n = 0.0178\ mol$ of $Fe$. Using $N=n\times N_A$, substitute $n = 0.0178\ mol$ and $N_A = 6.02\times 10^{23}\text{ atoms/mol}$. So $N=0.0178\ mol\times6.02\times 10^{23}\text{ atoms/mol}=1.07156\times 10^{22}$ atoms.

Step4: Solve problem 7

Given $n = 0.21\ mol$ of $Mg(NO_3)_2$. Using $N=n\times N_A$, substitute $n = 0.21\ mol$ and $N_A = 6.02\times 10^{23}\text{ formula - units/mol}$. So $N=0.21\ mol\times6.02\times 10^{23}\text{ formula - units/mol}=1.2642\times 10^{23}$ formula - units.

Step5: Solve problem 8

Given $n = 55.6\ mol$ of $H_2O$. Using $N=n\times N_A$, substitute $n = 55.6\ mol$ and $N_A = 6.02\times 10^{23}\text{ molecules/mol}$. So $N=55.6\ mol\times6.02\times 10^{23}\text{ molecules/mol}=3.34712\times 10^{25}$ molecules.

Step6: Solve problem 9

Given $n = 2.5\ mol$ of $C_4H_8O_2$. Using $N=n\times N_A$, substitute $n = 2.5\ mol$ and $N_A = 6.02\times 10^{23}\text{ molecules/mol}$. So $N=2.5\ mol\times6.02\times 10^{23}\text{ molecules/mol}=1.505\times 10^{24}$ molecules.

Step7: Solve problem 10

Given $n = 0.829\ mol$ of $Na_2SO_4$. Using $N=n\times N_A$, substitute $n = 0.829\ mol$ and $N_A = 6.02\times 10^{23}\text{ formula - units/mol}$. So $N=0.829\ mol\times6.02\times 10^{23}\text{ formula - units/mol}=4.99058\times 10^{23}$ formula - units.

Answer:

1. $7.525\times 10^{23}$ particles
2. $1.07156\times 10^{22}$ atoms
3. $1.2642\times 10^{23}$ formula - units
4. $3.34712\times 10^{25}$ molecules
5. $1.505\times 10^{24}$ molecules
6. $4.99058\times 10^{23}$ formula - units