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Calculate the no. of atoms of …

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

  • 1 answers

Payal Singh 6 years, 10 months ago

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

Given mass of Na2CO3 = 53 g

106 g of Na2COContain = 1 mol

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

 

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

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

{tex}3.011 \times 10^{23}{/tex}

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

 

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