Question

calculate number of atoms of the constituent element in 53 gram of Na2Co3​

Answer 1

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your answer:-

Avagadro's constant states that 1 mole of any substance has 6.022 × 10²³ molecules/atoms.

Find the number of moles of 53 g of Na2CO3:

molar mass Na2CO3 = 23 × 2 + 12 + 16 × 3 = (46+12+48) = 106

moles = mass/molar mass

= 53/106

= 0.5 moles

1 mole = 6.022 ×10²³ atoms

Then 0.5 moles = 6.022 × 10²³ × 0.5 atoms

= 3.011 × 10²³ atoms of Na2CO3

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Answer 2

Na2CO3

Molecular Mass of Na2CO3:

(2×23)+12+(3×16)

=46+12+48

=106g

106g of Na2CO3 has 6.022×10²³ atoms

To find:

No.of atoms in 53g of Na2CO3

Now,

1g of Na2CO3 has

$\sf{ \frac{6.022 \times {10}^{ 23}}{106}atoms \: }$

Then,

53g of Na2CO3 has

$\sf{ \frac{6.022 \times 10 {}^{23} \times 53}{106} } \\ \\ = 3.011 \times 10 {}^{23}$

Therefore,53g of Na2CO3 has 3.01×10²³ atoms