Question

3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
A. `6.68xx10^(23)`
B. `6.09xx10^(22)`
C. `6.022xx10^(23)`
D. `6.022xx10^(21)`

Answer 1

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)`