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The average adult human male has a total blood volume of 5.0 L. If the concentration of sodium ion in this average individual is \(0.135 \mathrm{M},\) what is the mass of sodium ion circulating in the blood?

Short Answer

Expert verified
The mass of sodium ions circulating in the blood of an average adult human male is \(15.47\,g\).

Step by step solution

01

Find the number of moles of sodium ion

Using the given concentration (\(0.135 M\)) and volume of blood (\(5.0 L\)), we can find the number of moles of sodium ions in the blood: Number of moles = Concentration × Volume
02

Calculate the number of moles of sodium ion

Now, let's calculate the number of moles using given values: Number of moles = \(0.135 M × 5.0 L\) Number of moles = \(0.675\, mol\, Na^+\)
03

Find the molar mass of sodium

Sodium (Na) has a molar mass of \(22.99\,g/mol\). This information is typically found on a periodic table.
04

Calculate the mass of sodium ions

To find the mass of sodium ions, we need to multiply the number of moles of sodium ions by the molar mass of sodium: Mass of sodium ions = Number of moles × Molar mass Mass of sodium ions = \((0.675\, mol\, Na^+) × (22.99\,g/mol)\) Mass of sodium ions = \(15.47\,g\) So, the mass of sodium ions circulating in the blood of an average adult human male is \(15.47\,g\).

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Key Concepts

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

Understanding Molarity
Molarity is a crucial concept in chemistry that helps us understand the concentration of a solution. It is defined as the number of moles of solute per liter of solution. This concept is particularly useful when we are trying to figure out how much of a particular substance is present in a given volume of solution.
For example, in our exercise above, we had a blood volume of 5.0 liters and a sodium ion concentration of 0.135 M. Here, "M" stands for molarity and means that there are 0.135 moles of sodium ions in every liter of blood.
To find the total moles of sodium ions in 5.0 liters of blood, we multiply the molarity (0.135 M) by the volume of blood (5.0 L):
  • Total moles of sodium ions = 0.135 moles/L × 5.0 L = 0.675 moles
By understanding molarity, you can easily calculate how concentrated a solution is and how much solute it contains, allowing you to perform various calculations in chemistry.
Exploring Molar Mass
Molar mass is another fundamental concept in chemistry that tells us how much one mole of a substance weighs. This is measured in grams per mole (g/mol) and is derived from the molecular weight of an element or compound.
Using the periodic table, we can find the molar mass of any element. For sodium (Na), the molar mass is 22.99 g/mol. This means that one mole of sodium weighs 22.99 grams.
In our exercise, after calculating the number of moles of sodium ions in the blood (0.675 moles), we multiply that by the molar mass of sodium to find the mass in grams:
  • Mass of sodium ions = 0.675 moles × 22.99 g/mol = 15.47 grams
Understanding molar mass allows you to convert between the number of moles and the mass of a substance, which is helpful for various chemical calculations.
Conversion of Units in Chemistry
Converting units is an essential skill required to solve chemistry problems effectively. It helps ensure that calculations are accurate and meaningful. Conversions can include changes in volume, mass, or concentration.
In the context of our exercise, we've dealt with volume in liters and concentration in molarity. It is vital to match these units for straightforward conversions.
  • For example, knowing that 1 liter equals 1,000 milliliters can help when volumes are given in different units.
  • Similarly, understanding conversions between moles and grams requires the use of molar mass (as discussed), which links the moles to the mass in grams.
Effective unit conversion ensures compatibility in calculations and allows you to navigate different scientific scenarios with ease. Practicing these conversions develops efficiency and precision in chemistry problem-solving.

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Most popular questions from this chapter

Magnesium carbonate, magnesium oxide, and magnesium hydroxide are all white solids that react with acidic solutions. (a) Write a balanced molecular equation and a netionic equation for the reaction that occurs when each substance reacts with a hydrochloric acid solution. (b) By observing the reactions in part (a), how could you distinguish any of the three magnesium substances from the other two?

The arsenic in a \(1.22-\) g sample of a pesticide was converted to \(\mathrm{AsO}_{4}^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of As in AsO \(_{4}^{3-2}(\mathbf{b})\) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took 25.0 \(\mathrm{mL}\) of 0.102 \(\mathrm{MAg}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

The commercial production of nitric acid involves the following chemical reactions: \begin{equation} \begin{array}{c}{4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)} \\ {3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)}\end{array} \end{equation} (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make 1000.0 L of a 0.150 \(\mathrm{M}\) aqueous solution of nitric acid? Assume all the reactions give 100\(\%\) yield.

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or neutralization reactions.\begin{equation} \begin{array}{c}{\text { (a) } \mathrm{P}_{4}(s)+10 \mathrm{HClO}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{\quad 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+10 \mathrm{HCl}(a q)}\end{array} \end{equation} \begin{equation} \begin{array}{l}\quad\quad\quad\quad\quad\quad{\text { (b) } \mathrm{Br}_{2}(l)+2 \mathrm{K}(s) \longrightarrow 2 \mathrm{KBr}(s)} \\\\\quad\quad\quad\quad\quad\quad {\text { (c) } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)}\end{array} \end{equation} \begin{equation} \begin{array}{r}\quad\quad\quad\quad{\text { (d) } \mathrm{ZnCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+} \\ {2 \mathrm{NaCl}(a q)}\end{array} \end{equation}

Using the activity series (Table 4.5), write balanced chemical equations for the following reactions. If no reaction occurs, write NR. (a) Iron metal is added to a solution of copper(II) nitrate, (b) zinc metal is added to a solution of magnesium sulfate, (c) hydrobromic acid is added to tin metal, (d) hydrogen gas is bubbled through an aqueous solution of nickel(II) chloride, (e) aluminum metal is added to a solution of cobalt(II) sulfate.

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