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Chapter 4: Problem 35

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 \(0.485 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH} ?\)

Short Answer

Expert verified
Thus, (a) 4.73 grams of glucose and (b) 44.1 milliliters of methanol are needed to prepare the solutions.

Step by step solution

01

Find amount of glucose in moles

To find the amount of glucose in moles, we can use the molarity equation: M = n / V. Where M is the molarity, n is the number of moles, and V is the volume. We are given M = 0.350 M and V = 75.0 mL or 0.075 L (converted from mL to L). Let's solve for n: n = M * V = 0.350 moles/L * 0.075 L = 0.02625 moles.
02

Convert moles to grams

To convert the number of moles to mass in grams, we can use the molecular weight of glucose which is 180.16 g/mol. The mass of glucose = number of moles * molecular weight. So the mass of glucose = 0.02625 mol * 180.16 g/mol = 4.73 g.
03

Find amount of methanol in moles

Similarly, for methanol, we use the molarity equation and given values: M = 0.485 M and V = 2.25 L. So, the amount of methanol in moles = M * V = 0.485 moles/L * 2.25 L = 1.09 moles.
04

Find volume of methanol

To find the volume of methanol, we use its density: d = m / V. We are given d = 0.792 g/mL. Rearranging the formula to solve for V: V = m / d. But first we need to calculate m: the mass of methanol = number of moles * molecular weight = 1.09 mol * 32.04 g/mol = 34.92 g. Now, we can find V: V = m / d = 34.92 g / 0.792 g/mL = 44.1 mL

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Most popular questions from this chapter

The manufacture of ethyl alcohol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) yields diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) as a by-product. The complete combustion of a \(1.005 \mathrm{g}\) sample of the product of this process yields \(1.963 \mathrm{g} \mathrm{CO}_{2} .\) What must be the mass percents of \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) and of \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) in this sample?

An essentially \(100 \%\) yield is necessary for a chemical reaction used to analyze a compound, but it is almost never expected for a reaction that is used to synthesize a compound. Explain this difference.

Ammonia can be generated by heating together the solids \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2} . \mathrm{CaCl}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are also formed. (a) If a mixture containing \(33.0 \mathrm{g}\) each of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) is heated, how many grams of \(\mathrm{NH}_{3}\) will form? (b) Which reactant remains in excess, and in what mass?

The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

Excess \(\mathrm{NaHCO}_{3}\) is added to \(525 \mathrm{mL}\) of \(0.220 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) These substances react as follows: \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow\) $$ \mathrm{CuCO}_{3}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) How many grams of the \(\mathrm{NaHCO}_{3}(\mathrm{s})\) will be consumed? (b) How many grams of \(\mathrm{CuCO}_{3}(\mathrm{s})\) will be produced?
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