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Question

 Calculate the number o fatoms of each kind in 5.3 grams of sodium carbonate.
 


Solution

Moles of N a subscript 2 C O subscript 3=fraction numerator 5.3 over denominator 106 end fraction=0.05
so moles of Na=0.05×2=0.1
moles of C=0.05×1=0.05
moles of O=0.05×3=0.15
we can now find the no. of atoms of each type by multiplying the no. of moles by Avagadro's constant.
number of molecules of Na = 0.1 ×6.02×1023 = 6.02×1022
number of molecules of C = 0.05 ×6.02×1023 = 0.301×1023
number of molecules of O = 0.15 ×6.02×1023 = 0.903×1023

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