Question
Question: If the density of some lake water is \(1.25gm{L^{ - 1}}\) and contains \(92g\) of \(N{a^ + }\) ions ...
If the density of some lake water is 1.25gmL−1 and contains 92g of Na+ ions per kilogram of water, calculate molality of Na+ ions in the lake.
Solution
Density of a substance is the measure of the mass of that substance present per unit volume. Density is inversely proportional to the volume of a substance. Molality is the measure of the number of moles of a substance present per kilogram of solvent.
Formula used: number of moles=Mm
molality=Msn
Where m is given mass of substance, M is molecular mass of substance, n is number of moles,Ms is mass of solvent in kilogram
Complete step by step answer:
As we know the density of a substance is mass per unit volume of that substance and molality is the number of moles present per kilogram of solvent.
In this question we have given:
Density of lake water=1.25gmL−1
Mass of sodium ions (m)=92g
Mass of solvent (Ms)=1Kg
We know molecular mass of sodium (M) is 23g
Therefore number of moles of sodium ions (n) is Mm
n=2392 n=4
Now, we know n=4 and Ms=1 and we know formula to calculate molality (written above)
Molality of sodium ions=Msn
molality=14=4m
So, the molality of sodium ions is 4m.
Additional information: Sodium is a chemical element with symbol Na. Its atomic number is 11. It is highly reactive due to which it is stored in kerosene oil. It reacts violently with air and water. This reaction is highly exothermic due to which hydrogen evolved catches fire. Accidents may also happen because of these reasons. To avoid such accidents this metal is stored in kerosene oil. Free State of this metal does not occur in nature and must be prepared from compounds. Many salts of sodium are water-soluble.
Note:
The terms molality and molarity are different. Molality is the measure of a number of moles present per kilogram of solvent and molarity is the measure of a number of moles present per liter of solution. In molality mass of the solvent is used and in molarity volume of solution is used.