If the density of water of a lake is 1.25 g/mL and 1 kg of water contains 92g of Na+ions , calculate the molarity of Na+ ions in this lake water.

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Solution

$D e n s i t y = \frac{1.25 g}{L}$
$W e i g h t i n g r a m s (s o d i u m) = 92 g r a m s$
$W e i g h t o f s o l v e n t = 1 K g$
$m o l a l i t y = \frac{N u m b e r o f m o l e s o f s o l u t e}{W e i g h t o f s o l v e n t i n K g}$
$N u m b e r o f m o l e s o f s o l u t e (s o d i u m) = \frac{W e i g h t i n g r a m s}{M o l e c u l a r w e i g h t} = \frac{92}{23} = 4 m o l e s$
$m o l a l i t y = \frac{4}{1} = 4$

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If the density of water of a lake is 1.25 g/mL and 1 kg of water contains 92g of Na+ions , calculate the molarity of Na+ ions in this lake water.

Density=1.25gLWeightingrams(sodium)=92gramsWeightofsolvent=1Kgmolality=NumberofmolesofsoluteWeightofsolventinKgNumberofmolesofsolute(sodium)=WeightingramsMolecularweight=9223=4molesmolality=41=4