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Chapter 13: Problem 90

How many moles of solute particles are present in \(1 \mathrm{~L}\) of each of the following aqueous solutions? (a) \(0.3 M \mathrm{KBr}\) (b) \(0.065 \mathrm{M} \mathrm{HNO}_{3}\) (c) \(10^{-4} M \mathrm{KHSO}_{4}\) (d) \(0.06 M\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\)

Short Answer

Expert verified
(a) 0.6 moles (b) 0.13 moles (c) 2 x 10鈦烩伌 moles (d) 0.06 moles

Step by step solution

01

Understanding the concept

To find the number of moles of solute particles in a solution, we need to consider the dissociation in water if applicable. The molarity (M) of a solution indicates the number of moles of solute per liter of solution.
02

Calculation for (a) \(0.3 M \mathrm{KBr}\)

Potassium bromide (KBr) dissociates completely in water into K鈦� and Br鈦� ions.Total moles of ions produced = \(1 \mathrm{K}^{+} + 1 \mathrm{Br}^{-} = 2 \text{ions}\). Thus, \(0.3 M \mathrm{KBr}\) solution has \(0.3 \times 2 = 0.6 \text{moles of ions}\).
03

Calculation for (b) \(0.065 \mathrm{M} \mathrm{HNO}_{3}\)

Nitric acid (HNO鈧�) dissociates completely in water into H鈦� and \mathrm{NO}_{3}^{-}\ ions.Total moles of ions produced = \(1 \mathrm{H}^{+} + 1 \mathrm{NO}_{3}^{-} = 2 \text{ions}\). Thus, \(0.065 M \mathrm{HNO}_{3}\) solution has \(0.065 \times 2 = 0.13 \text{moles of ions}\).
04

Calculation for (c) \(10^{-4} \mathrm{M} \mathrm{KHSO}_{4}\)

Potassium bisulfate (KHSO鈧�) dissociates in water into \mathrm{K}^{+}\ and \mathrm{HSO}_{4}^{-}\ ions.Total moles of ions produced = \(1 \mathrm{K}^{+} + 1 \mathrm{HSO}_{4}^{-} = 2 \text{ions}\). Thus, \(10^{-4} M \mathrm{KHSO}_{4}\) solution has \(10^{-4} \times 2 = 2 \times 10^{-4}\ \text{moles of ions}\).
05

Calculation for (d) \(0.06 M \text{ethanol} \left(\text{C}_{2} \text{H}_{5} \text{OH}\right)\)

Ethanol (C鈧侶鈧匫H) does not dissociate in water. Therefore, the number of moles of solute particles remains the same as its molarity.Thus, \(0.06 M \text{ethanol}\) solution has \(0.06 \text{moles of solute particles}\).

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Key Concepts

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

Molarity
Molarity is a key concept in chemistry. It measures the concentration of a solute in a solution. The unit for molarity is moles per liter (M), which tells us how many moles of a solute are in one liter of solution. For example, a 0.3 M solution of KBr means there are 0.3 moles of KBr in every liter of the solution.
Calculating molarity helps determine how concentrated or diluted a solution is. Knowing this is essential for various applications in chemistry, such as preparing solutions, conducting reactions, and even understanding biological processes.
In simple terms, think of molarity as a way to express how 'packed' a solution is with its solute.
Dissociation in Water
When certain substances dissolve in water, they can split into ions. This process is called dissociation. For instance, KBr (potassium bromide) dissociates into K鈦� (potassium ions) and Br鈦� (bromide ions) when it dissolves in water.
Each molecule of KBr forms two ions. So, if you have a solution with 0.3 M KBr, it correlates to 0.6 moles of ions (0.3 moles of K鈦� and 0.3 moles of Br鈦�). This dissociation is crucial in calculating the total number of solute particles in a solution.
However, not all substances dissociate. Ethanol (C鈧侶鈧匫H) is an example where no dissociation occurs. For ethanol in a 0.06 M solution, it remains 0.06 moles of solute particles since it doesn't split into ions.
Solute Particles
Understanding solute particles is essential for solving chemistry problems involving solutions. Solute particles refer to the individual molecules or ions in a solution. The number of these particles can change depending on whether the solute dissociates in water.
In a 0.065 M HNO鈧� (nitric acid) solution, HNO鈧� dissociates into H鈦� and NO鈧冣伝 ions. Each HNO鈧� molecule becomes two ions, making the total concentration of particles 0.13 moles.
  • For substances that dissociate completely, you must consider the number of particles formed.
  • For substances that do not dissociate, the number of solute particles equals the molarity of the solution.

This understanding helps you determine the particle count in solutions, which is vital for predicting how reactions will proceed and for performing precise measurements in lab settings.

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Most popular questions from this chapter

\(\beta \)-Pinene \(\left(C_{10} H_{16}\right)\) and \(\alpha\) -terpineol \(\left(C_{10} H_{18} O\right)\) are used in cosmetics to provide a "fresh pine" scent. At \(367 \mathrm{~K},\) the pure substances have vapor pressures of 100.3 torr and 9.8 torr, respectively. What is the composition of the vapor (in terms of mole fractions) above a solution containing equal masses of these compounds at \(367 \mathrm{~K} ?\) (Assume ideal behavior.)

A chemist is studying small organic compounds to evaluate their potential for use as an antifreeze. When \(0.243 \mathrm{~g}\) of a compound is dissolved in \(25.0 \mathrm{~mL}\) of water, the freezing point of the solution is \(-0.201^{\circ} \mathrm{C}\). (a) Calculate the molar mass of the compound \((d\) of water \(=\) \(1.00 \mathrm{~g} / \mathrm{mL}\) ). (b) Analysis shows that the compound is 53.31 mass \(\% \mathrm{C}\) and 11.18 mass \(\% \mathrm{H},\) the remainder being \(\mathrm{O}\). Determine the empirical and molecular formulas of the compound. (c) Draw a Lewis structure for a compound with this formula that forms \(\mathrm{H}\) bonds and another for one that does not.

A biochemical engineer isolates a bacterial gene fragment and dissolves a 10.0 -mg sample in enough water to make \(30.0 \mathrm{~mL}\) of solution. The osmotic pressure of the solution is 0.340 torr at \(25^{\circ} \mathrm{C}\). (a) What is the molar mass of the gene fragment? (b) If the solution density is \(0.997 \mathrm{~g} / \mathrm{mL}\), how large is the freezing point depression for this solution \(\left(K_{\mathrm{f}}\right.\) of water \(\left.=1.86^{\circ} \mathrm{C} / \mathrm{m}\right) ?\)

Which ion in each pair has greater charge density? Explain. (a) \(\mathrm{Na}^{+}\) or \(\mathrm{Cs}^{+}\) (b) \(\mathrm{Sr}^{2+}\) or \(\mathrm{Rb}^{+}\) (c) \(\mathrm{Na}^{+}\) or \(\mathrm{Cl}^{-}\) (d) \(\mathrm{O}^{2-}\) or \(\mathrm{F}^{-}\) (e) \(\mathrm{OH}^{-}\) or \(\mathrm{SH}^{-}\) (f) \(\mathrm{Mg}^{2+}\) or \(\mathrm{Ba}^{2+}\) (g) \(\mathrm{Mg}^{2+}\) or \(\mathrm{Na}^{+}\) (h) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}\)

The solubility of \(\mathrm{N}_{2}\) in blood is a serious problem for divers breathing compressed air \(\left(78 \% \mathrm{~N}_{2}\right.\) by volume \()\) at depths greater than \(50 \mathrm{ft}\). (a) What is the molarity of \(\mathrm{N}_{2}\) in blood at \(1.00 \mathrm{~atm} ?\) (b) What is the molarity of \(\mathrm{N}_{2}\) in blood at a depth of \(50 . \mathrm{ft} ?\) (c) Find the volume (in \(\mathrm{mL}\) ) of \(\mathrm{N}_{2}\), measured at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\), released per liter of blood when a diver at a depth of \(50 . \mathrm{ft}\) rises to the surface \(\left(k_{\mathrm{H}}\right.\) for \(\mathrm{N}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is \(7.0 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{atm}\) and at \(37^{\circ} \mathrm{C}\) is \(6.2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) -atm; assume \(d\) of water is \(1.00 \mathrm{~g} / \mathrm{mL}\) ).
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