Question

How many sodium atoms are present in 0.2310 g of sodium?

Answer 1

The number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.

The given parameters;

• mass of the reacting sodium = 0.231 g
• 1 mole of an atom = 6.02 x 10²³ atoms
• the atomic mass of sodium is 23 g/mol

The number of moles of 0.231 g of sodium available is calculated as;

$no. \ of \ moles = \frac{Reacting \ mass}{Molar \ mass} \\\\no. \ of \ moles = \frac{0.231}{23} \\\\no. \ of \ moles = 0.01 \ mol.$

The number of atoms of 0.01 mole of sodium available is calculated as;

1 mole ---------- 6.02 x 10²³ atoms

0.01 ----------------- ?

= 0.01 x 6.02 x 10²³ atoms

= 6.02 x 10²¹ atoms.

Thus, the number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.

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Answer 2

Answer:

There are 6.048X10^21 atoms present in 0.2310 g of sodium.

Explanation:

We know,

23g Na contains 6.022X10^23 atoms

∴1g Na contains 6.022X10^23/23 atoms

So, 0.2310g Na contains (6.022X10^23/23)X0.2310 atoms

                                         = 6.048X10^21 atoms

There are 6.048X10^21 atoms present in 0.2310 g of sodium.