Understanding the Problem

We are given that the mole fraction of NaOH in an aqueous solution is 0.5. This means that half of the total moles in the solution are NaOH and the other half are water. We are also given the density of pure NaOH as 4 g/mL. We need to find the molarity of the NaOH solution.

Key Concepts

Molarity (M) is defined as the number of moles of solute per liter of solution. It can be calculated using the formula:

$M=\frac{n_{\mathrm{solute}}}{V\left(L\right)}$

Where $n_{\mathrm{solute}}$ is the moles of solute and V(L) is the volume of the solution in liters.

Mole fraction (χ) of a component is the ratio of the number of moles of that component to the total number of moles in the solution.

${\chi }_{\mathrm{NaOH}}=\frac{n_{\mathrm{NaOH}}}{n_{\mathrm{NaOH}}+n_{H_{2}O}}$

Given that ${\chi }_{\mathrm{NaOH}}=0.5$, we can deduce that $n_{\mathrm{NaOH}}=n_{H_{2}O}$.

Step-by-Step Solution

Step 1: Define Variables Based on Mole Fraction

Let the number of moles of NaOH be $n_{\mathrm{NaOH}}=n$.

Since ${\chi }_{\mathrm{NaOH}}=0.5$, the number of moles of water must also be $n_{H_{2}O}=n$.

Therefore, the total number of moles in the solution is $n_{\mathrm{total}}=n+n=2n$.

Step 2: Calculate the Mass of Each Component

Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = $40g/\mathrm{mol}$

Mass of NaOH = Moles × Molar Mass = $n\times 40=40ng$

Molar mass of H₂O = 2(1) + 16 = $18g/\mathrm{mol}$

Mass of water = $n\times 18=18ng$

Step 3: Calculate the Total Mass of the Solution

Total mass = Mass of NaOH + Mass of water = $40n+18n=58ng$

Step 4: Calculate the Volume of the Solution

This is the most critical step. The density given (4 g/mL) is for pure NaOH, not for the solution. We cannot use this density directly to find the volume of our aqueous solution. The volume of the solution is not simply the sum of the volumes of its pure components due to volume change on mixing.

However, the problem likely intends for us to assume that the volume of the solution is approximately equal to the volume of the solvent (water) because the amount of NaOH is significant but we have no other data. But this would be inaccurate. Let's re-examine the given data.

The density of pure NaOH is provided. Pure NaOH is a solid. The density of a solid is used to find its volume. In this context, the density is likely given so that we can find the volume occupied by the NaOH in its pure state, but this is not standard for calculating solution volume.

A more accurate approach is to realize that the density given might be a red herring, or perhaps it's for the solute. But in standard problems, when mole fraction is given and density of pure component is provided, it's often used to find the volume of the solution by assuming the volume is additive, even though it's not strictly true.

Assumption: We will assume that the volumes are additive. This is an approximation to solve the problem.

Volume of pure NaOH = Mass / Density = $\frac{40ng}{4g/\mathrm{mL}}=10n\mathrm{mL}$

Density of pure water is $1g/\mathrm{mL}$.

Volume of pure water = Mass / Density = $\frac{18ng}{1g/\mathrm{mL}}=18n\mathrm{mL}$

Therefore, total volume of solution (assuming additivity) = Volume of NaOH + Volume of water = $10n+18n=28n\mathrm{mL}=0.028nL$

Step 5: Calculate the Molarity

Molarity (M) = Moles of solute / Volume of solution in Liters

$M=\frac{n}{0.028n}=\frac{1}{0.028}\approx 35.714$

Rounding to two decimal places gives $35.71M$

Final Answer: The molarity of the aqueous NaOH solution is approximately $35.71M$. This corresponds to the option 35.71.

Related Topics & Formulae

Concentration Units:

• Molarity (M): $M=\frac{\text{moles of solute}}{\text{liters of solution}}$
• Molality (m): $m=\frac{\text{moles of solute}}{\text{kilograms of solvent}}$
• Mole Fraction (χ): ${\chi }_i=\frac{n_i}{n_{\text{total}}}$

Important Note on Density: The density of a solution is not always the weighted average of the densities of its components. Volume change can occur when solids dissolve in liquids or when liquids are mixed. The assumption of volume additivity is an approximation that may not hold for concentrated solutions.