Question

Calculate the number of sodium ions present in 0.031 g of sodium oxide. ​

Answer 1

Answer:

Sodium sulfate is

N

a

2

S

O

4

First, calculate the moles of sodium sulfate you have in a 14.2g sample.

Find the molar mass of sodium sulfate

molar mass = atomic mass in grams

2(23) + 32 + 4(16) = 142 grams per mole

Divide the mass of the sample by the molar mass to obtain moles

14.2

g

142

g

m

o

l

−

1

= 0.1 mol

Now, for every 1 mol of

N

a

2

S

O

4

you have 2 moles of

N

a

+

ions. Multiple the moles of

N

a

2

S

O

4

that you found just then by 2 to obtain the moles of sodium ions.

Use Avogadro's number (6.022 x

10

23

) to find the number of ions present. (Recall that Avogrado's number is the number of particles per mole of a substance).

6.022 x

10

23

x 0.2 mol = 1.2044 x

10

23

ions

Answer 2

Hi dude

Your answer:

Molar mass of sodium oxide = 62 g/mol

In one mole of Sodium oxide, there will be 2 moles of sodium ions (since its formula is Na₂O)

So, in 0.031 g of  Na₂O, there will be: 0.031/62 = 0.0005 moles of Na₂O

Now, no. of Na⁺ ions =  2 x 5 x 10⁻⁴ x 6.022 x 10²³ = 6.022 x 10²⁰ is your answer

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