The mole fraction of sugar `(C_(12)H_(22)O_(11))` in an aqueous solution that is `5.30%` sugar by mass is

Options

A. 0.99706
B. 0.00294
C. 0.79906
D. 0.20094

Answer

2

Explanation

To calculate the mole fraction, we must find the number of moles of sugar and of water in some fixed quantity of solution.
From the given mass percent, we can say that there is `5.30 g` of sugar in every `100 f` of solution. Since the solution contains only sugar and water, `100 g` of solution must contain `100 g-5.30 g=94.70 g` of water.
A table of atomic masses shows that the molecular mass of the water is `18 u` and the molecular mass of sugar is `342 u`. Therefore, in the `100 g` sample of solution there is:
`n_(sugar)=("mass of sugar")/("molar mass of sugar")`
`=(5.30 g sugar)/(342 g sugar//"mol sugar")`
`=0.0155` mol sugar
`n_(water)=("mass of water")/("molar mass of water")`
`=(94.70" g water")/(18" g sugar"//"mol water")`
`=5.26 mol H_(2)O`
We can substitute thses date into Equations `(2.9 and 2.10)`, which give the mole fraction relationship for a two-component solution:
`chi_(C_(12)H_(22)O_(11))=n_(C_(12)H_(22)C_(11))/(n_(C_(12)H_(22)C_(11))+n_(H_(2)O))`
`=(0.0155 mol sugar)/(0.0155 mol sugar + 5.26 mol H_(2)O)`
`=0.00294`
Since `chi_(C_(12)H_(22)O_(11))+chi_(H_(2)O)=1`
`chi_(H_(2)O)=1-chi_(C_(12)H_(22)O_(11))`
`=1-0.00294`
`=0.99706`