Question

How many atoms of N are in 137.0 grams of N2O3?
a. 1.085x10^23
b. 1.802x10^23
c. 5.985x10^23
d. 2.171x10^24
e. 3.604x10^24

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Answer 1

Answer: Option (d) is the correct answer.

Explanation:

It is given that mass of $N_{2}O_{3}$ is 137.0 g.

Molar mass of  $N_{2}O_{3}$ is as follows.

    Molar mass of  $N_{2}O_{3}$ = 2 × 14 + 3 × 16

                                                                = 76 g

Thus, number of moles will be calculated as follows.

        No. of moles = $\frac{mass}{molar mass}$

                              = $\frac{137.0 g}{76 g/mol}$

                              = 1.803 mol

There are 2 nitrogen (N) atoms present in the formula of dinitrogen trioxide. Also Avogadro's number equals $6.022 \times 10^{23}$ atoms or molecules per mol.

Therefore, number of N molecules or atoms in $N_{2}O_{3}$ is as follows.

           Number of N atoms = $2 \times 1.803 mol \times 6.022 \times 10^{23} atoms/mol$

                                             = $2 \times 10.855 \times 10^{23}$ atoms

                                             = $2.171 \times 10^{24}$ atoms

Thus, we can conclude that there are $2.171 \times 10^{24}$ atoms of N in in 137.0 grams of $N_{2}O_{3}$.

Answer 2

Answer:

d 2.171x10^24

Explanation:

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