What is the molality of a solution made by dissolving 3.42 g of table sugar (sucrose, `C_(12H_(22)O_(11))` in 50.0 mL of water?
Strategy: Molality is the number of moles of solute dissolved per kilogram of solvent. Thus, we must find how many moles are present in 3.42 g of sucrose and how many kilograms are contained in 50.0 mL of water.

Explanation

`n_("sucrose")=(mass)_("sucrose")/("molar mass")`
`=(3.42 g)/(342 g mol^(-1))=0.01 mol`
Since the density of water is `1.00 g mL^(-1)`, we can find the mass of water as follows:
`d_(H_(2)O)=m_(H_(2)O)/V_(H_(2)O)`
or `m_(H_(2)O)=(d_(H_(2)O))(V_(H_(2)O))`
`=(1.00 g mL^(-1))(50.0 mL)`
`=50.0 g`
Thus, the molality of the solution is
`Molality=n_("sucrose")/g_(H_(2)O)xx(1000 g)/(kg)`
`=(0.01 mol)/(50.0 g)xx(1000 g)/(kg)`
`=0.2 m`