InterviewSolution
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InterviewSolution
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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. |
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Answer» `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` |
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