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 molarity of `Na^+` ions in the lake.

Explanation

Number of moles in 92 g of `Na^+` ions ` = (92 g)/(23 g "mol"^(-1)) = 4` mole (as atomic mass of Na = 23)
Volume of 1000 g (1 kg) of solution = ` = (1000)/(1.25)= 800 mL^(-1)`
Number of moles of `Na^+` in 1000 `mL^(-1)` of solution = ` 4/800 xx 1000 = 5 `moles
Molarity = 5M