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How many atoms of N are in 137.0 grams of N2O3?


A.

1.085 × 1023

B.

1.802 × 1023

C.

5.985 × 1023

D.

2.171 × 1024

E.

3.604 × 1024

Respuesta :

Answer:

1.085 x 10²⁴

Explanation:

The answer is not in your choices, but it maybe due to a typo but to get the answer to this, you just need to convert the grams to moles, then moles to atoms.

First we get the mass of the molecule for every mole. Get the atomic mass of each element and multiply it by the number of atoms present then get their total.

N₂O₃

Element       number of atoms        Atomic mass       TOTAL

    N                          2                x          14.007            28.014

    O                          3                x          15.999           47.997

                                                                                      76.011 g/mole

So now we know for every 1 mole of N₂O₃ there are 76.011 g of N₂O₃.

Next we need to see how many moles of N₂O₃ are there in 137.0g of N₂O₃.

[tex]137.0g\times\dfrac{1mole}{76.011g}=1.802moles[/tex]

Now we know that we have 1.802moles of N₂O₃.

We use Avogadro's constant to find out how many atoms there are. Avogadro's constant states that for every mole of any substance, there are 6.022140857 × 10²³ atoms.

[tex]1.802moles\times\dfrac{6.022140857\times10^{23}atoms}{1 mole}=1.085\times10^{24}atoms[/tex]

Answer: [tex]2.17\times 10^{24}[/tex]

Explanation:-

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

For [tex]N_2O_3[/tex]

Mass  given = 137.0 g

Molar mass of  [tex]N_2O_3[/tex] = 76 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of}N_2O_3 =\frac{137g}{76g/mol}=1.80mol[/tex]

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles

1 mole of [tex]N_2O_3[/tex] contains =[tex]2\times 6.023\times 10^{23}=12.04\times 10^{23}[/tex] atoms of nitrogen

1.80 moles of [tex]N_2O_3[/tex] contains =[tex]\frac{12.04\times 10^{23}}{1}\times 1.80=2.17\times 10^{24}[/tex] atoms of nitrogen.