What is the concentration of sugar `(C_(12)H_(22)O_(11))` in `mol L^(-1)` if its `20 g` are dissolved in enough water to make a final volume up to `2L`?

Explanation

The molar mass of sugar `(C_(12)H_(22)O_(11))`
`=(12xx12)+(1xx22)+(11xx16)`
`=144+22+176=342" g mol"^(-1)`
`therefore` Moles of sugar `=(20)/(342)=0.058` mol
Volume of solutoin = 2 L
`therefore` Molarity = `("Moles of solute")/("Volume of solutoin in litre")`
`=(0.058)/(2)=0.029 mol L^(-1)`