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Chapter 4: Problem 73

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

Short Answer

Expert verified
(a) To make a \(1000.0\,\mathrm{mL}\) solution of \(0.250\,\mathrm{M}\,\mathrm{NH}_{3}\), you will need approximately \(16.89\,\mathrm{mL}\) of the \(14.8\,\mathrm{M}\,\mathrm{NH}_{3}\) stock solution. (b) The concentration of the final solution after diluting \(10.0\,\mathrm{mL}\) of the stock solution to \(0.500\,\mathrm{L}\) is approximately \(0.296\,\mathrm{M}\).

Step by step solution

01

Identify given values and setup the formula

Given: \(c_{1} = 14.8\,\mathrm{M}\), \(c_{2} = 0.250\,\mathrm{M}\), and \(v_{2} = 1000.0\,\mathrm{mL}\). We need to find \(v_{1}\). Formula: \(c_{1}v_{1} = c_{2}v_{2}\).
02

Solve for \(v_{1}\)

Rearrange the formula to solve for \(v_{1}\): \(v_{1} = \frac{c_{2}v_{2}}{c_{1}}\) Now, plug in the values: \(v_{1} = \frac{(0.250\,\mathrm{M})(1000.0\,\mathrm{mL})}{14.8\,\mathrm{M}}\)
03

Calculate \(v_{1}\)

Perform the calculation: \(v_{1} \approx 16.89\,\mathrm{mL}\) So, approximately \(16.89\,\mathrm{mL}\) of the stock solution is needed to make \(1000.0\,\mathrm{mL}\) of \(0.250\,\mathrm{M}\,\mathrm{NH}_{3}\). #b) Calculate the concentration of the final solution after diluting the 10.0 mL stock solution#
04

Identify given values and setup the formula

Given: \(c_{1} = 14.8\,\mathrm{M}\), \(v_{1} = 10.0\,\mathrm{mL}\), and \(v_{2} = 0.500\,\mathrm{L} (= 500.0\,\mathrm{mL})\). We need to find \(c_{2}\). Formula: \(c_{1}v_{1} = c_{2}v_{2}\).
05

Solve for \(c_{2}\)

Rearrange the formula to solve for \(c_{2}\): \(c_{2} = \frac{c_{1}v_{1}}{v_{2}}\) Now, plug in the values: \(c_{2} = \frac{(14.8\,\mathrm{M})(10.0\,\mathrm{mL})}{500.0\,\mathrm{mL}}\)
06

Calculate \(c_{2}\)

Perform the calculation: \(c_{2} \approx 0.296\,\mathrm{M}\) So, the concentration of the final solution after diluting the \(10.0\,\mathrm{mL}\) stock solution to \(0.500\,\mathrm{L}\) is approximately \(0.296\,\mathrm{M}\).

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Most popular questions from this chapter

A medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is \(1.5 \times 10^{-9} \mathrm{M},\) and $1.00 \mathrm{~mL}\( of this solution will be delivered to a dish containing \)2.0 \times 10^{5}\( cancer cells in \)5.00 \mathrm{~mL}$ of aqueous fluid. What is the ratio of drug molecules to the number of cancer cells in the dish?

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and \(\mathrm{BaCl}_{2}\) Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: $\mathrm{Br}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-} ?$

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{FeCl}_{3}(a q) \longrightarrow\) (b) $\mathrm{ZnCl}_{2}(a q)+\mathrm{Cs}_{2} \mathrm{CO}_{3}(a q) \longrightarrow$ (c) $\mathrm{Na}_{2} \mathrm{~S}(a q)+\operatorname{CoSO}_{4}(a q) \longrightarrow$

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You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride, lead chloride, or zinc chloride. When you transfer the solid to a beaker and add water, the solid dissolves to give a clear solution. Next an $\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)$ solution is added and a white precipitate forms. What is the identity of the unknown white solid? [Section 4.2\(]\)
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