Calculate the number of Na2CO3 molecules present in 0.53g NA2CO3?
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mass of Na2CO3 = 23 + (2*12) + (16*3) = 95g
so 95g of Na2CO3 = 6.022*10^23 molecules
then, molecules present in 0.53g NA2CO3 = {(6.022*10^23) /95 } * 0.53 = 3.35964 * 10^23 molecules
so 95g of Na2CO3 = 6.022*10^23 molecules
then, molecules present in 0.53g NA2CO3 = {(6.022*10^23) /95 } * 0.53 = 3.35964 * 10^23 molecules
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