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Chapter 4: Problem 41

Which has the higher concentration of sucrose: a \(46 \%\) sucrose solution by mass \((d=1.21 \mathrm{g} / \mathrm{mL}),\) or \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ? Explain your reasoning.

Short Answer

Expert verified
The 1.50 M C12H22O11 solution has a higher sucrose concentration than the 46% sucrose solution by mass.

Step by step solution

01

Identify Given Information

In this exercise two different solutions are given. The first solution has 46% sucrose by mass and a density of 1.21 mg/mL. The second solution is 1.50 M sucrose (C12H22O11). The goal is to compare the concentrations of sucrose in these two solutions.
02

Conversion of Percent Solution into Molarity

Firstly, realize that 46% by mass implies that 100 ml solution contains 46 g of sucrose. This is because percentage by mass is defined as the mass of solute per 100 units of solution. So, we first need to calculate how many moles of sucrose (C12H22O11) are present in 46 g. From the molecular formula, the molar mass of sucrose is approximately 342.3 g/mol. Thus, the moles of sucrose can be calculated by dividing the mass by the molar mass: \( \frac{46 g}{342.3 g/mol} = 0.134 mol \). This gives the number of moles in 100 ml, but typically molarity is in moles per liter. Therefore, multiply the result by 10 to convert from 100 ml to 1 L, so the molarity of the first solution is \(0.134 mol/L x 10 = 1.34 M \)
03

Comparison of Molarities

Now compare the molarity of the sucrose in these two solutions. The first solution has a molarity of 1.34 M, while the second solution has a molarity of 1.50 M. Thus, the second solution, 1.50 M C12H22O11, has a higher concentration of sucrose.

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Most popular questions from this chapter

How much (a) glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in grams, must be dissolved in water to produce \(75.0 \mathrm{mL}\) of \(0.350 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?\) (b) methanol, \(\mathrm{CH}_{3} \mathrm{OH} \quad(d=0.792 \mathrm{g} / \mathrm{mL}),\) in milli- liters, must be dissolved in water to produce 2.25 L of \(0.485 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH} ?\)

An aqueous solution that is \(5.30 \%\) LiBr by mass has a density of \(1.040 \mathrm{g} / \mathrm{mL} .\) What is the molarity of this solution? (a) 0.563 M; (b) 0.635 M; (c) 0.0635 M; (d) \(0.0563 \mathrm{M} ;\) (e) \(12.0 \mathrm{M}\).

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

How ma475.15 grams \(\mathrm{HCl}\)ny grams of HCl are consumed in the reaction of \(425 \mathrm{g}\) of a mixture containing \(35.2 \% \mathrm{MgCO}_{3}\) and \(64.8 \% \mathrm{Mg}(\mathrm{OH})_{2},\) by mass? $$\begin{array}{c} \mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{MgCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}(\mathrm{g}) \end{array}$$

Write a chemical equation to represent the complete combustion of malonic acid, a compound with \(34.62 \% \mathrm{C}, 3.88 \% \mathrm{H},\) and \(61.50 \% \mathrm{O},\) by mass.
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