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Chapter 4: Problem 46

A \(25.0 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) is diluted to a volume of 500.0 mL. If the concentration of the diluted solution is found to be \(0.085 \mathrm{M} \mathrm{HCl}\), what was the concentration of the original solution?

Short Answer

Expert verified
The concentration of the original solution was \(1.7 \mathrm{M} \mathrm{HCl}\).

Step by step solution

01

Understanding the Given Information

The initial volume of \( \mathrm{HCl} \) solution is \( 25.0 \mathrm{mL} \) or \( 0.0250 \mathrm{L} \). The diluted volume is \( 500.0 \mathrm{L} \). The molarity of the diluted solution (M2) is \( 0.085 \mathrm{M} \). We need to find the molarity of the initial solution (M1).
02

Use the Dilution Formula

The dilution formula is \( M1V1 = M2V2 \). Inserting the given values into this equation gives \( M1(0.0250L) = 0.085M(0.500L) \).
03

Solve for M1

We solve the equation for M1 by dividing both sides of the equation by \( 0.0250L \). That gives \( M1 = \frac{0.085M(0.500L)}{0.0250L} \).
04

Calculate the Result

Performing the calculation gives \( M1 = 1.7M \). So, the molarity of the original solution was \( 1.7M \) .

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Most popular questions from this chapter

A reaction mixture contains \(1.0 \mathrm{mol} \mathrm{CaCN}_{2}\) (calcium cyanamide) and \(1.0 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\). The maximum number of moles of \(\mathrm{NH}_{3}\) produced is (a) \(3.0 ;\) (b) 2.0 (c) between 1.0 and 2.0; (d) less than 1.0. $$\mathrm{CaCN}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{CaCO}_{3}+2 \mathrm{NH}_{3}(\mathrm{g})$$

Consider the chemical equation below. What is the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}, \mathrm{KI},\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) \(3.0 \mathrm{mol}\); (b) \(3.8 \mathrm{mol}\) (c) \(5.0 \mathrm{mol} ;\) (d) \(6.0 \mathrm{mol} ;\) (e) \(15 \mathrm{mol}\). $$\begin{array}{r} 2 \mathrm{KMnO}_{4}+10 \mathrm{KI}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{array}$$

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

Azobenzene, an intermediate in the manufacture of dyes, can be prepared from nitrobenzene by reaction with triethylene glycol in the presence of \(\mathrm{Zn}\) and KOH. In one reaction, 0.10 L of nitrobenzene \((d=1.20 \mathrm{g} / \mathrm{mL})\) and \(0.30 \mathrm{L}\) of triethylene glycol \((d=1.12 \mathrm{g} / \mathrm{mL})\) yields \(55 \mathrm{g}\) azobenzene. What are the (a) theoretical yield, (b) actual yield, and (c) percent yield of this reaction? $$ 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+4 \mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{4} \frac{\mathrm{Zn}}{\mathrm{KOH}} $$ nitrobenzene triethylene glycol $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}\right)_{2}+4 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{4}+4 \mathrm{H}_{2} \mathrm{O} $$ azobenzene
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