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If 500 . g of water is added to 75 \(\mathrm{g}\) of a \(2.5-m\) NaCl solution, what is the mass percent of \(\mathrm{NaCl}\) in the diluted solution?

Short Answer

Expert verified
The mass percent of NaCl in the diluted solution can be found through these steps: 1. Find the initial amount of NaCl in the 75 g solution using the molality: - Calculate moles of NaCl: \(moles \, of \, NaCl = molality × mass \, of \, solvent (kg)\) - Convert moles of NaCl to mass: \(mass \, of \, NaCl = moles \, of \, NaCl × molar \, mass \, of \, NaCl\) 2. Calculate the total mass of the diluted solution: - \(mass \, of \, diluted \,solution = mass \, of \, NaCl + mass \, of \, added \,water + mass \, of \, initial \,water \) 3. Calculate the mass percent of NaCl in the diluted solution: - \(mass \,percent \, of \, NaCl = \frac{mass \, of \, NaCl}{mass \, of \, diluted \, solution} × 100\% \) Plug in the values from Steps 1 and 2 to find the answer.

Step by step solution

01

Find the initial amount of NaCl in the 75 g solution

To find the initial amount of NaCl in the 75 g NaCl solution, we can use the definition of molality. Molality is defined as the number of moles of solute per kilogram of solvent in the solution: \(molality = \frac{moles \, of \, solute}{mass \, of \, solvent (kg)}\) First, we need to convert the 75 g solution mass to the mass of the solvent (water): Since NaCl is completely dissociated in water, so the mass of solvent (water) can be found using the following formula: \(mass \, of \, solvent = mass\_total\_solution - mass \, of \, NaCl\) Next, we can rearrange the molality formula to find the moles of solute (NaCl): \(moles \, of \, NaCl = molality × mass \, of \, solvent (kg) \) Finally, we can convert the moles of NaCl into the initial mass of NaCl: \(mass \, of \, NaCl = moles \, of \, NaCl × molar \, mass \, of \, NaCl\) where the molar mass of NaCl is approximately 58.44 g/mol.
02

Calculate the total mass of the diluted solution

To find the mass of the diluted solution, we will add the mass of the solute (NaCl) and the mass of the solvent (water): \(mass \, of \, diluted \,solution = mass \, of \, NaCl + mass \, of \, added \,water + mass \, of \, initial \,water \)
03

Calculate the mass percent of NaCl in the diluted solution

Now that we have the mass of NaCl and the total mass of the diluted solution, we can calculate the mass percent of NaCl: \(mass \,percent \, of \, NaCl = \frac{mass \, of \, NaCl}{mass \, of \, diluted \, solution} × 100\% \) Now plug in the values obtained from Steps 1 and 2 to calculate the mass percent of NaCl.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Percent
Mass percent is a way to express the concentration of a component in a mixture. This concept is particularly useful when working with solutions. It answers the question: "What portion of the total solution's mass is made up by the solute?" In this context, we're interested in the mass percent of sodium chloride (NaCl) in the diluted solution.

To find the mass percent, use the formula:
  • Mass percent = \( \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100\% \)
This small equation tells you the mass of the solute, NaCl, as a percentage of the total mass of the solution after dilution. First, determine the mass of NaCl using initial solution data. Then, add this mass to the masses of initial and added water to get the total mass of the new diluted solution. Once you have these values, it's all about plugging them into the mass percent formula and solving for the percentage.

Understanding mass percent is valuable in chemistry and daily life. For instance, knowing the concentration of a salt solution is not only crucial in laboratory settings but also in culinary and industrial processes where specific concentrations are desired.
Molality
Molality is another way to represent the concentration of a solution. Unlike molarity, it doesn't depend on the total volume of the solution but instead focuses on the mass of the solvent. The importance of molality comes from its independence from temperature, making it a reliable measure under varying conditions.

Molality is defined by the formula:
  • Molality \( (m) = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \)
    • Using the given molality of the sodium chloride solution, you can determine the initial moles of NaCl present. Start by separating the mass of NaCl from the total solution to find the mass of just the solvent (water). With the known molality value, multiply it by this mass (converted into kilograms) to find the moles of NaCl in the solution.

      These steps not only teach you about concentration but also underline the difference between mass-based and volume-based concentrations, crucial for applications where precision under different conditions is essential.
Sodium Chloride Solution
A sodium chloride solution is an everyday example of a binary solution, comprising a solute (NaCl) and a solvent (water). Understanding its properties and concentration is very important in various fields from biological sciences to household applications.

Sodium chloride, commonly known as table salt, easily dissolves in water, forming a uniform solution. This homogeneous mixture is perfect for studying fundamental chemistry concepts like dilution, concentration, and solution stoichiometry.

When perfectly dissociated in the solution, the relationship between its concentration, given by measures like mass percent and molality, determines the properties and potential applications of the solution. Knowing how these concentrations shift with the addition of more solvent (like adding 500 g of water in this exercise) helps explain the physical and chemical behavior of such solutions under dilution.

Explore these concepts with practical exercises to encounter real-life scenarios where solution concentration modification is crucial, whether you're preparing a saline solution in medical settings or adjusting brine concentrations in food preservation.

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Most popular questions from this chapter

a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide. $$\begin{array}{lll}{\mathrm{CaCl}_{2}(s)} & {-2247 \mathrm{k} / \mathrm{mol}} & {-46 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Cal}_{2}(s)} & {-2059 \mathrm{k} / \mathrm{mol}} & {-104 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ b. Based on your answers to part a, which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-},\) is more strongly attracted to water?

Which ion in each of the following pairs would you expect to be more strongly hydrated? Why? a. \(\mathrm{Na}^{+}\) or \(\mathrm{Mg}^{2+}\) b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Be}^{2+}\) c. \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) d. \(\mathrm{F}^{-}\) or \(\mathrm{Br}^{-}\) e. \(\mathrm{Cl}^{-}\) or \(\mathrm{ClO}_{4}^{-}\) f. \(\mathrm{ClO}_{4}^{-}\) or \(\mathrm{SO}_{4}^{2-}\)

An aqueous solution containing 0.250 mole of \(\mathrm{Q},\) a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{g}\) water freezes at \(-2.79^{\circ} \mathrm{C} .\) What is the van't Hoff factor for \(\mathrm{Q} ?\) The molal freezing-point depression constant for water is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) . What is the formula of \(\mathrm{Q}\) if it is 38.68\(\%\) chlorine by mass and there are twice as many anions as cations in one formula unit of \(\mathrm{Q} ?\)

You and your friend are each drinking cola from separate 2-L bottles. Both colas are equally carbonated. You are able to drink 1 L of cola, but your friend can drink only about half a liter. You each close the bottles and place them in the refrigerator. The next day when you each go to get the colas, whose will be more carbonated and why?

n lab you need to prepare at least 100 mL of each of the following solutions. Explain how you would proceed using the given information. a. 2.0 \(\mathrm{mKCl}\) in water (density of \(\mathrm{H}_{2} \mathrm{O}=1.00 \mathrm{g} / \mathrm{cm}^{3} )\) b. 15\(\% \mathrm{NaOH}\) by mass in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\) c. 25\(\% \mathrm{NaOH}\) by mass in \(\mathrm{CH}_{3} \mathrm{OH}\left(d=0.79 \mathrm{g} / \mathrm{cm}^{3}\right)\) d. 0.10 mole fraction of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\)

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