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A sugar syrup of weight 183.42 g contain...

A sugar syrup of weight 183.42 g contains 3.42 g of sugar `(C_(12)H_(22)O_(11))`. Calulate the mole fraction of sugar.

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The correct Answer is:
0.00099
`"Mass of sugar" = 3.42 g: "Moles of sugar" =((3.42g))/((342 g mol^(-1)))=0.01 mol`
`"Mass of water"=183.42-3.42=180g : "Moles of water"=(180g)/((342 "g mol"^(-1)))=10 mol`
`"Mole fration of sugar" = ("No. of moles of sugar")/("No. of moles of sugar + No. of moles of water")`
`=((0.01mol))/((0.01mol)+(10 mol))=(0.01)/(10.01)=0.00099`.
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