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Molality

QUESTION 2.26...

Question

If the density of some lake water is $1.25 g m L^{- 1}$ and contains 92 g of $N a^{+}$ ions per kg of water, calculate the molarity of $N a^{+}$ ions in the lake.

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Solution

Step 1:
Mass of sodium ions = 92 g
Molar mass of sodium ions = 23 g/mol
Number of moles of sodium ions = $\frac{92 g}{23 g {m o l}^{- 1}} = 4$
Mass of water = 1 kg
Volume = $\frac{m a s s}{d e n s i t y} = \frac{1000 g}{1.25 g {m L}^{- 1}} = 0.8 L$
Step 2:
As we know, molarity = $\frac{N u m b e r o f m o l e s}{v o l u m e o f s o l u t i o n}$
Molarity of $N a^{+}$ ions (m) $= \frac{N u m b e r o f m o l e s N a^{+} i o n s}{V o l u m e o f s o l u t i o n i n l i t r e s}$
$∴ M o l a r i t y = \frac{4 m o l}{0.8 L} = 5 m o l L^{- 1}$ or $5 M$

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If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.

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