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Chapter 4: Problem 89

An average human being has about \(5.0 \mathrm{~L}\) of blood in his or her body. If an average person were to eat \(32.0 \mathrm{~g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{~g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

Short Answer

Expert verified
The change in molarity of blood sugar after an average person consumes 32.0 grams of sugar and it dissolves into their bloodstream would be approximately 0.0187 mol/L.

Step by step solution

01

Converting grams of sugar to moles

First, we need to convert the 32.0 grams of sugar to moles. To do this, we use the molecular weight of sucrose, which is 342.30 g/mol. Moles of sugar = (grams of sugar) / (molecular weight of sucrose) Moles of sugar = \( \frac{32.0~g}{342.30~g/mol} \)
02

Calculating the molarity of sugar in blood

Now, we'll use the volume of blood in the body and the moles of sugar to find the molarity of sugar in the bloodstream. Molarity = \( \frac{moles~of~sugar}{volume~of~blood~in~liters} \) We know the volume of blood in the body is 5.0 liters. Substituting the values we found in step 1: Molarity = \( \frac{\frac{32.0~g}{342.30~g/mol}}{5.0~L} \)
03

Simplifying to find the change in molarity of blood sugar

Now, we'll simplify the expression to find the change in molarity of blood sugar. Molarity = \( \frac{32.0~g}{342.30~g/mol \times 5.0~L} \) Molarity ≈ \( \frac{32.0~g}{1711.5~g \cdot L/mol} \) Molarity ≈ \( \frac{32.0}{1711.5} mol/L \) Molarity ≈ 0.0187 mol/L The change in molarity of blood sugar after the sugar is dissolved into the bloodstream would be approximately 0.0187 mol/L.

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Most popular questions from this chapter

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The VO \(^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q) $$ To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

Differentiate between the following terms. a. species reduced versus the reducing agent b. species oxidized versus the oxidizing agent c. oxidation state versus actual charge

Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal compound. It is composed solely of \(\mathrm{C}, \mathrm{F}\), and \(\mathrm{B}\); it is \(42.23 \% \mathrm{C}\) by mass and \(55.66 \% \mathrm{~F}\) by mass. a. What is the empirical formula of BARF? b. A \(2.251-g\) sample of BARF dissolved in \(347.0 \mathrm{~mL}\) of solution produces a \(0.01267 M\) solution. What is the molecular formula of BARF?

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Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 \mathrm{M} \mathrm{NaOH}\) from \(1.00 \mathrm{M} \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution
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