Molarity and Molality Calculations for 40% NaOH Solution

Interpretation of Compound Concentrations

In chemical calculations, concentration is often expressed as a percentage by weight, denoted as $(w/w)$. For a solution identified as $40\%$ $NaOH$ $(w/w)$, this value provides the specific ratio of the solute to the total solution mass. Specifically, it indicates that exactly $40\,g$ of pure sodium hydroxide ($NaOH$) is dissolved within $100\,g$ of the total solution. To perform further concentration calculations such as molarity or molality, one must also determine the mass of the solvent. In this case, the mass of the solvent (water) is calculated by subtracting the mass of the solute from the mass of the solution: $100\,g - 40\,g = 60\,g$.

Determining Solution Volume via Density

To calculate the molarity of a solution, the volume of the solution must be known. The volume is derived from the total mass of the solution and its given density. Based on the provided data, the density of the solution is $1.2\,g\,cm^{-3}$. Using the fundamental relationship between mass, volume, and density ($\text{Volume} = \frac{\text{Mass}}{\text{Density}}$), the volume of this solution is calculated by dividing the total solution mass ($100\,g$) by the density ($1.2\,g\,cm^{-3}$). This yields the following expression for volume:

$\text{Volume of solution} = \frac{100}{1.2}\,cm^3$

Calculation of Molarity ($M$)

Molarity ($M$), or molar concentration, is defined as the number of moles of solute per cubic decimeter (liter) of solution. The formula for molarity, incorporating the mass of the solute ($w$), the molar mass of the solute ($m$), and the volume of the solution in cubic centimeters ($V$), is expressed as:

$M = \frac{w}{m} \times \frac{1000}{V}$

For sodium hydroxide ($NaOH$), the molar mass ($m$) is calculated as the sum of the atomic masses: $Na=23$, $O=16$, and $H=1$, totaling $40\,g\,mol^{-1}$. Substituting the known variables ($w = 40\,g$, $m = 40\,g\,mol^{-1}$, and $V = \frac{100}{1.2}\,cm^3$) into the formula results in:

$M = \frac{40}{40} \times \frac{1000}{\frac{100}{1.2}}$

Through simplification, the expression becomes:

$M = 1 \times \frac{1000 \times 1.2}{100}$

$M = 12\,M$

Therefore, the molarity of the $40\%$ $NaOH$ $(w/w)$ solution is $12\,mol\,dm^{-3}$.

Calculation of Molality ($m$)

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is independent of the temperature and volume of the solution because it relies solely on mass. The formula for molality is given by:

$m = \frac{\text{moles of solute}}{\text{Mass of solvent in kg}}$

Alternatively, if the mass of the solvent is expressed in grams, the formula is:

$m = \frac{w}{m} \times \frac{1000}{\text{Mass of solvent (in g)}}$

As established, the mass of the solvent is the total solution mass minus the solute mass ($100\,g - 40\,g = 60\,g$). Given that the moles of solute ($\frac{w}{m}$) is $\frac{40}{40} = 1\,mol$, the calculation is as follows:

$m = \frac{40}{40} \times \frac{1000}{60}$

$m = 1 \times 16.666...$

$m = 16.66\,m$

Thus, the molality of the $40\%$ $NaOH$ $(w/w)$ solution is determined to be $16.66\,mol\,kg^{-1}$.