Question:

If the density of some lake water is \(1.25g mL^{–1}\) and contains 92 g of \(Na^+\) ions per kg of water, calculate the molality of \(Na^+\) ions in the lake

Updated On: Sep 28, 2023
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Solution and Explanation

Number of moles present in 92 g of \(Na ^+\) ions \(\frac{92g}{23g mol^{-1 }}\)
= 4 mol 
Therefore, molality of \(Na ^+\) ions in the lake \(= \frac{4 mol}{1kg}\)
=4 m
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Concepts Used:

Concentration of Solutions

It is the amount of solute present in one liter of solution.

Concentration in Parts Per Million - The parts of a component per million parts (106) of the solution.

Mass Percentage - When the concentration is expressed as the percent of one component in the solution by mass it is called mass percentage (w/w).

Volume Percentage - Sometimes we express the concentration as a percent of one component in the solution by volume, it is then called as volume percentage

Mass by Volume Percentage - It is defined as the mass of a solute dissolved per 100mL of the solution.

Molarity - One of the most commonly used methods for expressing the concentrations is molarity. It is the number of moles of solute dissolved in one litre of a solution.

Molality - Molality represents the concentration regarding moles of solute and the mass of solvent.

Normality - It is the number of gram equivalents of solute present in one liter of the solution and it is denoted by N.

Formality - It is the number of gram formula present in one litre of solution.