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A \(60.0-\mathrm{mL} 0.513 \mathrm{M}\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) solution is mixed with \(120.0 \mathrm{~mL}\) of \(2.33 \mathrm{M}\) glucose solution. What is the concentration of the final solution? Assume the volumes are additive.

Short Answer

Expert verified
The final concentration of the solution is approximately 1.724 M.

Step by step solution

01

Identify the Given Information

We have two glucose solutions with the following volumes and concentrations: - Solution 1: Volume = 60.0 mL, Concentration = 0.513 M - Solution 2: Volume = 120.0 mL, Concentration = 2.33 M. We need to find the concentration of the final solution when these two are mixed.
02

Calculate Moles of Glucose in Solution 1

Use the concentration formula to find the moles in Solution 1: \[ \text{moles of glucose in Solution 1} = C_1 \times V_1 = 0.513 \, \text{M} \times 0.060 \, \text{L} = 0.03078 \, \text{mol} \]
03

Calculate Moles of Glucose in Solution 2

Similarly, calculate the moles in Solution 2: \[ \text{moles of glucose in Solution 2} = C_2 \times V_2 = 2.33 \, \text{M} \times 0.120 \, \text{L} = 0.2796 \, \text{mol} \]
04

Sum the Moles of Glucose

Add the moles from both solutions to get the total moles in the combined solution: \[ \text{Total moles of glucose} = 0.03078 \, \text{mol} + 0.2796 \, \text{mol} = 0.31038 \, \text{mol} \]
05

Calculate Total Volume of the Solution

Add the volumes of both solutions to get the total volume: \[ \text{Total Volume} = 60.0 \, \text{mL} + 120.0 \, \text{mL} = 180.0 \, \text{mL} = 0.180 \, \text{L} \]
06

Calculate Final Concentration

Using the formula for concentration, calculate the concentration of the final solution: \[ C_f = \frac{\text{Total moles of glucose}}{\text{Total Volume}} = \frac{0.31038}{0.180} \approx 1.724 \text{ M} \]

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Key Concepts

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

Mole Concept
The mole concept is a fundamental aspect of chemistry that helps in quantifying the amount of substance. The key idea is to use the "mole" as a "counting unit", just like a "dozen" is used to count eggs. One mole corresponds to Avogadro's number of particles, which is approximately \( 6.022 \times 10^{23} \) particles. It could be atoms, molecules, or ions — depending on the context.
  • Moles allow chemists to count particles by weighing them.
  • It is directly related to the molecular weight. For example, glucose \( (\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}) \) has a molar mass of about 180 grams per mole.
  • The mole enables the bridge between the atomic world and the macroscopic world, making it possible to calculate quantities like concentration and yield in chemical reactions.
In the context of the exercise, we calculated the moles of glucose using its concentration and the volume. For instance, the moles of glucose in Solution 1 were found using the formula \( \text{moles} = C \times V \), indicating how many moles of glucose are contained in the given volume. This knowledge is instrumental for combining solutions and understanding how the substances interact.
Molarity
Molarity is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution (\( ext{mol/L} \)), often represented with the symbol \( M \). Molarity provides a way to quantify the exact concentration of substances in a liquid.
  • A solution with a molarity of 1 M has 1 mole of solute in every liter of solution.
  • It is crucial for the preparation of solutions and for stoichiometric calculations in chemistry reactions.
In the problem we are looking at, we needed to calculate the final molarity of the combined glucose solutions. By summing up the individual moles from both solutions and dividing by the total volume, we find the molarity of the mixture. Understanding molarity aids in predicting how a substance will react, given the precise concentration of the reactants involved.
Volume Addition in Solutions
When dealing with solutions, particularly mixing them, understanding volume addition is significant. Typically, the volumes of solutions are considered additive unless there is significant interaction between solute and solvent, which would cause volume contraction or expansion.
  • When solutions are mixed, their volumes are summed up directly if no considerable interaction between molecules occurs.
  • This assumption enables easy calculation of final concentrations in solution mixtures.
  • It's important to convert volumes to consistent units for accuracy, often liters in chemical equations.
In the step we performed, we simply added the volumes of the two glucose solutions to determine the total solution volume. This straightforward arithmetic helps calculate the resultant concentration for a solution mixture, making it a critical part of finding the final molarity.

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Most popular questions from this chapter

Magnesium is a valuable, lightweight metal. It is used as a structural metal and in alloys, in batteries, and in chemical synthesis. Although magnesium is plentiful in Earth's crust, it is cheaper to "mine" the metal from seawater. Magnesium forms the second most abundant cation in the sea (after sodium); there are about \(1.3 \mathrm{~g}\) of magnesium in \(1 \mathrm{~kg}\) of seawater. The method of obtaining magnesium from seawater employs all three types of reactions discussed in this chapter: precipitation, acid-base, and redox reactions. In the first stage in the recovery of magnesium, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is heated at high temperatures to produce quicklime, or calcium oxide \((\mathrm{CaO})\) : $$ \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ When calcium oxide is treated with seawater, it forms calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right],\) which is slightly soluble and ionizes to give \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions: $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) $$ The surplus hydroxide ions cause the much less soluble magnesium hydroxide to precipitate: $$ \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s) $$ The solid magnesium hydroxide is filtered and reacted with hydrochloric acid to form magnesium chloride \(\left(\mathrm{MgCl}_{2}\right):\) $$ \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \underset{\mathrm{MgCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)}{\longrightarrow} $$ After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. In a process called electrolysis, an electric current is passed through the cell to reduce the \(\mathrm{Mg}^{2+}\) ions and oxidize the \(\mathrm{Cl}^{-}\) ions. The half-reactions are After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. In a process called electrolysis, an electric current is passed through the cell to reduce the \(\mathrm{Mg}^{2+}\) ions and oxidize the \(\mathrm{Cl}^{-}\) ions. The half-reactions are $$ \begin{aligned} \mathrm{Mg}^{2+}+2 e^{-} \longrightarrow & \mathrm{Mg} \\ 2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-} \end{aligned} $$ The overall reaction is $$ \mathrm{MgCl}_{2}(l) \longrightarrow \mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) $$ This is how magnesium metal is produced. The chlorine gas generated can be converted to hydrochloric acid and recycled through the process. (a) Identify the precipitation, acid-base, and redox processes. (b) Instead of calcium oxide, why don't we simply add sodium hydroxide to precipitate magnesium hydroxide? (c) Sometimes a mineral called dolomite (a combination of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) ) is substituted for limestone \(\left(\mathrm{CaCO}_{3}\right)\) to bring about the precipitation of magnesium hydroxide. What is the advantage of using dolomite? (d) What are the advantages of mining magnesium from the ocean rather than from Earth's crust?

Someone gave you a colorless liquid. Describe three chemical tests you would perform on the liquid to show that it is water.

A \(46.2-\mathrm{mL}, 0.568 M\) calcium nitrate \(\left[\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\right]\) solution is mixed with \(80.5 \mathrm{~mL}\) of \(1.396 \mathrm{M}\) calcium nitrate solution. Calculate the concentration of the final solution.

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