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| 1. |
Calculate the no. of atoms of the constituent elements in 53g of Na2CO3 |
| Answer» Molecular mass of Na2CO3\xa0= ( 2 × 23 ) + 12 + (3 × 16) = 46 + 12 + 48 = 106\xa0Given mass of Na2CO3\xa0= 53 g106 g of Na2CO3\xa0Contain = 1 mol53 g of Na2CO3 Contain = {tex}{53\\over 106} = {1\\over 2}{/tex}\xa0mol \xa0No of atoms of Na =\xa0{tex}{1\\over 2}\\times 2 \\times 6.022 \\times 10^{23}{/tex}=\xa0{tex} 6.022 \\times 10^{23}{/tex}No of atoms of C =\xa0{tex}{1\\over 2}\\times 6.022 \\times 10^{23}{/tex}=\xa0{tex}3.011 \\times 10^{23}{/tex}No of atoms of O =\xa0{tex}{1\\over 2}\\times 3 \\times 6.022 \\times 10^{23}{/tex}=\xa0{tex}9.033\\times 10^{23}{/tex}\xa0 | |