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Calculate the no. of atoms of the constituent elements in 53g of Na2CO3

Posted by Nikhil Malik 8 years ago

CBSE > Class 11 > Chemistry

1 answers

Payal Singh 7 years, 11 months ago

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

Given mass of Na₂CO₃ = 53 g

106 g of Na₂CO₃Contain = 1 mol

53 g of Na₂CO₃ Contain = ${53\over 106} = {1\over 2}$ mol

No of atoms of Na = ${1\over 2}\times 2 \times 6.022 \times 10^{23}$ = $6.022 \times 10^{23}$

No of atoms of C = ${1\over 2}\times 6.022 \times 10^{23}$

= $3.011 \times 10^{23}$

No of atoms of O = ${1\over 2}\times 3 \times 6.022 \times 10^{23}$ = $9.033\times 10^{23}$

10Thank You

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