Question

shown is a set of problems in a chemistry class.
(a) how many atoms are in 0.534 mol of nickel, ni?
a. 1.13×10²⁴ atoms b. 1.48×10²³ atoms c. 2.44×10²³ atoms
d. 3.22×10²³ atoms e. 6.98×10²³ atoms
(b) how many moles are in 8.73×10²³ atoms of boron, b?
a. 1.45 moles b. 3.84×10²³ c. 1.45 moles
d. 5.04×10⁻²³ moles e. 6.90×10⁻³ moles
(c) how many grams are present in 0.885 moles of manganese?
a. 62.1 g b. 48.6 g c. 21.5 g d. 27.5 g e. 0.016 g
after steven solved the above problems, which problem used the conversion - factor with 1 mole on top and 6.02×10²³ atoms on bottom?
o a question a
o b question b
o c question c
o d none of them

Explanation:

Step1: Recall Avogadro's number

Avogadro's number ($N_A$) is $6.022\times 10^{23}$ atoms/mol.

Step2: Solve part (a)

To find the number of atoms in 0.534 mol of nickel, use the formula $N = n\times N_A$, where $n = 0.534$ mol. So $N=0.534\times6.022\times 10^{23}=3.215\times 10^{23}\approx3.22\times 10^{23}$ atoms. So the answer for (a) is D.

Step3: Solve part (b)

To find the number of moles in $8.73\times 10^{22}$ atoms of boron, use the formula $n=\frac{N}{N_A}$. So $n=\frac{8.73\times 10^{22}}{6.022\times 10^{23}} = 0.145$ moles. So the answer for (b) is E.

Step4: Solve part (c)

The molar - mass of manganese (Mn) is approximately 54.94 g/mol. To find the mass in 0.885 moles of manganese, use the formula $m=n\times M$, where $n = 0.885$ mol and $M = 54.94$ g/mol. So $m=0.885\times54.94 = 48.6229\approx48.6$ g. So the answer for (c) is B.

Step5: Identify the question using the given conversion factor

The conversion factor $\frac{1\ mol}{6.02\times 10^{23}\text{ atoms}}$ is used when going from the number of atoms to the number of moles. In part (b), we went from the number of atoms of boron to moles of boron using this conversion factor. So the answer for the last question is b.

Answer:

(a) D. $3.22\times 10^{23}$ atoms
(b) E. $0.145$ moles
(c) B. $48.6$ g
Last question: b