58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?

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58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?

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Answer 1 · 2014-05-15T23:13:52

The molality of the solution is 0.500 mol/kg.

The molality of a solution is the number of moles of solute per kilogram of solvent.

Molality = $\frac{moles of solute}{kilograms of solvent}$ $"moles of solute"/"kilograms of solvent"$

In this case, NaCl is the solute and water is the solvent.

Moles of NaCl = 58.44 g NaCl × $\frac{1 mol NaCl}{58.44 g NaCl}$ $(1" mol NaCl")/(58.44" g NaCl")$ = 1.0000 mol NaCl

Kilograms of water = 2.00 kg water

Molality = $\frac{moles of solute}{kilograms of solvent} \times \frac{1.0000 mol}{2.00 kg}$ $"moles of solute"/"kilograms of solvent" × (1.0000" mol")/(2.00" kg")$ = 0.500 mol/kg

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58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?

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