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3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution areA. `6.68xx10^(23)`B. `6.09xx10^(22)`C. `6.022xx10^(23)`D. `6.022xx10^(21)` |
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Answer» Correct Answer - A a no. of moles of sucrose in 3.442 g `=((3.42g))/((342 g "mol"^(-1)))` =0.01 mol No. of oxygen atoms in 1 mole of sucrose `(C_(12)H_(22)O_(11))=11xxN_(A)` No. of oxygen atoms in 0.01 moles of sucrose `=0.01 xx11 xxN_(A)=0.11 N_(A)` No. of moles of `H_(2)O` in 18 g =`((18g))/((18 g"mol"^(-1))) =1 ` mol No. of oxygen atoms in 1 mole of water =`N_(A)` Total no. of oxygen atoms =`(0.11+1) =1.11 N_(A)` `=1.11xx6.022xx10^(23)` `=6.68xx10^(23)` |
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