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Chapter 4: Problem 1

Household ammonia used for cleaning contains about \(10 \mathrm{~g}\) (two significant figures) of \(\mathrm{NH}_{3}\) in \(100 \mathrm{~mL}\) (two significant figures) of solution. What is the molarity of the \(\mathrm{NH}_{3}\) in solution?

Short Answer

Expert verified
Answer: The molarity of NH3 in the solution is approximately 5.87 M.

Step by step solution

01

Find moles of NH3

First, find the moles of NH3 using the mass and molar mass: Moles of NH3 = mass of NH3 / molar mass of NH3 The mass of NH3 is given as 10 g, and the molar mass of NH3 is 14.01 g/mol for nitrogen and 3 * 1.01 g/mol for three hydrogen atoms, which adds up to 17.03 g/mol. Moles of NH3 = (10 g) / (17.03 g/mol)
02

Calculate the volume of the solution in Liters

We are given that the volume of the solution is 100 mL. To convert this to Liters, we use the conversion factor: 1 L = 1000 mL. Volume in Liters = (100 mL) / (1000 mL/L) = 0.1 L
03

Calculate the molarity of NH3

Now that we have the moles of NH3 and the volume of the solution in Liters, we can find the molarity by dividing the moles by the volume. Molarity of NH3 = moles of NH3 / volume in Liters Molarity of NH3 = (10 g) / [(17.03 g/mol) * (0.1 L)]
04

Solve and report the molarity to the appropriate number of significant figures

When solving the equation above, make sure to consider significant figures. Since both the mass and the volume of the solution are given to two significant figures, the molarity of NH3 should also be reported to two significant figures. Molarity of NH3 ≈ 5.87 M (rounded to two significant figures)

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Most popular questions from this chapter

Write balanced net ionic equations for the following reactions in acid solution. (a) Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed. (b) Solid phosphorus \(\left(P_{4}\right)\) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right)\) ions. (c) Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese(II) ions are formed.

Consider the following balanced redox reaction in basic medium. \(3 \mathrm{Sn}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow\) \(3 \mathrm{Sn}^{4+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+8 \mathrm{OH}^{-}(a q)\) (a) What is the oxidizing agent? (b) What species has the element that increases its oxidation number? (c) What species contains the element with the highest oxidation number? (d) If the reaction were to take place in acidic medium, what species would not be included in the reaction?

Identify the type of aqueous reaction using the symbols PPT for precipitation, SA/SB for strong acid-strong base, SA/WB for strong acid-weak base, WA/SB for weak acid-strong base, and NR for no reaction. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{HCl}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{HF}\) (c) \(\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{Na}_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}+\mathrm{BaCl}_{2}\) (e) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NaCl}\)

Write balanced equations for the following reactions in acid solution. (a) \(\mathrm{P}_{4}(s)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (c) \(\mathrm{HBrO}_{3}(a q)+\mathrm{Bi}(s) \longrightarrow \mathrm{HBrO}_{2}(a q)+\mathrm{Bi}_{2} \mathrm{O}_{3}(s)\) (d) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\)

What volume of \(0.285 \mathrm{M}\) strontium hydroxide is required to neutralize \(25.00 \mathrm{~mL}\) of \(0.275 \mathrm{M}\) hydrofluoric acid (HF)?
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