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Question
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
Options
`6.68 xx 10^23`
`6.09 xx 10^23`
`6.022 xx 10^23`
`6.022 xx 10^21`
Solution
Number of moles of sucrose
`3.42/342` = 0.01mol–
1 moll of sucrose (C12H22O11) = 11 × NA atoms of O
0.01 mol of C12H22O11 = 0.01 × 11 × NA
= 0.11 × NA atoms of O
Number of moles of water = `18/18`
= 1 mol
1 mol of water contains = 1 × NA atoms of O
Total number of oxygen atoms = 0.11 NA + 1.0 NA
= 1.11 NA
Number of oxygen atoms in solution = 1.11 NA
= 1.11 × 6..022 × 1023
= 6.68 × 1023
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