9.7.2: Practice Solution Concentration
- Page ID
- 424718
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise \(\PageIndex{1}\)
How many grams of glucose (molar mass = 180.18 g/mol) are dissolved in 175 mL of a 2.00 M solution?
- Answer
-
63.1 g. (0.350 moles)
Exercise \(\PageIndex{1}\)
How many mL of 5.25 M HCl solution would be needed to react with 6.00 g of Zn in the reaction below?
Zn (s) + 2 HCl (aq) --> ZnCl2 (aq) + H2 (g)
- Answer
-
35.0 mL (0.0350 L, 0.184 moles)
Titration Problems
Exercise \(\PageIndex{1}\)
You perform the titration below on 12.00 mL of H2SO4 solution. You find that it takes 26.44 mL of 0.4483 M NaOH to react completely. What is the concentration of the H2SO4 solution?
H2SO4 (aq) + 2 NaOH (aq) \(\rightarrow\) Na2SO4 (aq) + 2 H2O (l)
- Answer
-
0.4939 M H2SO4
You perform the titration below on 20.00 mL of HCl solution. You find that it takes 18.72 mL of 0.3786 M NaOH to react completely. What is the concentration of the HCl solution?
HCl (aq) + NaOH (aq) \(\rightarrow\) NaCl (aq) + H2O (l)
- Answer
-
0.3544 M HCl
You perform the titration below on 15.00 mL of HCl solution. You find that it takes 26.44 mL of 0.1352 M Ba(OH)2 to react completely. What is the concentration of the HCl solution?
2 HCl (aq) + Ba(OH)2 (aq) \(\rightarrow\) BaCl2 (aq) + 2 H2O (l)
- Answer
-
0.4766 M HCl
Solution Stoichiometry
Exercise \(\PageIndex{1}\)
In the reaction below, what volume of 0.7644 M Pb(NO3)2 solution is needed in order to make 37.06 g PbCl2? (Molar mass of PbCl2 is 278.1 g/mol. NaCl is in excess.)
Pb(NO3)2 (aq) + 2 NaCl (aq) \(\rightarrow\) 2 NaNO3 (aq) + PbCl2 (s)
- Answer
-
174.3 mL (0.1743 L)
In the reaction below, what volume of 0.3805 M Pb(NO3)2 solution is needed to react with 455.0 mL of 0.6322 M NaCl solution?
Pb(NO3)2 (aq) + 2 NaCl (aq) \(\rightarrow\) 2 NaNO3 (aq) + PbCl2 (s)
- Answer
-
378.0 mL (0.3780 L)