Skip to main content
Chemistry LibreTexts

6.1.1: Practice Problems- Solution Concentration

  • Page ID
    217282
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    PROBLEM \(\PageIndex{1}\)

    Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.

    Answer

    The number of moles always stays the same in a dilution.

    The concentration and the volumes change in a dilution.

    PROBLEM \(\PageIndex{2}\)

    What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?

    Answer

    The two samples contain the same proportion of moles of salt to liters of solution, but have different numbers of actual moles.

    PROBLEM \(\PageIndex{3}\)

    Determine the molarity for each of the following solutions:

    1. 0.444 mol of CoCl2 in 0.654 L of solution
    2. 98.0 g of phosphoric acid, H3PO4, in 1.00 L of solution
    3. 0.2074 g of calcium hydroxide, Ca(OH)2, in 40.00 mL of solution
    4. 10.5 kg of Na2SO4·10H2O in 18.60 L of solution
    5. 7.0 × 10−3 mol of I2 in 100.0 mL of solution
    6. 1.8 × 104 mg of HCl in 0.075 L of solution
    Answer a

    0.679 M

    Answer b

    1.00 M

    Answer c

    0.06998 M

    Answer d

    1.75 M

    Answer e

    0.070 M

    Answer f

    6.6 M

    Click here to see a video of the solution

     

     

    PROBLEM \(\PageIndex{4}\)

    Determine the molarity of each of the following solutions:

    1. 1.457 mol KCl in 1.500 L of solution
    2. 0.515 g of H2SO4 in 1.00 L of solution
    3. 20.54 g of Al(NO3)3 in 1575 mL of solution
    4. 2.76 kg of CuSO4·5H2O in 1.45 L of solution
    5. 0.005653 mol of Br2 in 10.00 mL of solution
    6. 0.000889 g of glycine, C2H5NO2, in 1.05 mL of solution
    Answer a

    0.9713 M

    Answer b

    5.25 × 10-3 M

    Answer c

    6.122 × 10-2 M

    Answer d

    7.62 M

    Answer e

    0.5653 M

    Answer f

    1.13 × 10-2 M

    PROBLEM \(\PageIndex{5}\)

    Calculate the number of moles and the mass of the solute in each of the following solutions:

    (a) 2.00 L of 18.5 M H2SO4, concentrated sulfuric acid
    (b) 100.0 mL of 3.8 × 10−5 M NaCN, the minimum lethal concentration of sodium cyanide in blood serum
    (c) 5.50 L of 13.3 M H2CO, the formaldehyde used to “fix” tissue samples
    (d) 325 mL of 1.8 × 10−6 M FeSO4, the minimum concentration of iron sulfate detectable by taste in drinking water

    Answer a

    37.0 mol H2SO4

    3.63 × 103 g H2SO4

    Answer b

    3.8 × 10−6 mol NaCN

    1.9 × 10−4 g NaCN

    Answer c

    73.2 mol H2CO

    2.20 kg H2CO

    Answer d

    5.9 × 10−7 mol FeSO4

    8.9 × 10−5 g FeSO4

    Click here to see a video of the solution

     

     

    PROBLEM \(\PageIndex{6}\)

    Calculate the molarity of each of the following solutions:

    (a) 0.195 g of cholesterol, C27H46O, in 0.100 L of serum, the average concentration of cholesterol in human serum
    (b) 4.25 g of NH3 in 0.500 L of solution, the concentration of NH3 in household ammonia
    (c) 1.49 kg of isopropyl alcohol, C3H7OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol
    (d) 0.029 g of I2 in 0.100 L of solution, the solubility of I2 in water at 20 °C

    Answer a

    5.04 × 10−3 M

    Answer b

    0.499 M

    Answer c

    9.92 M

    Answer d

    1.1 × 10−3 M

    PROBLEM \(\PageIndex{7}\)

    There is about 1.0 g of calcium, as Ca2+, in 1.0 L of milk. What is the molarity of Ca2+ in milk?

    Answer

    0.025 M

    Click here to see a video of the solution

     

     

    PROBLEM \(\PageIndex{8}\)

    What volume of a 1.00-M Fe(NO3)3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M?

    Answer

    0.250 L

     

    PROBLEM \(\PageIndex{9}\)

    If 0.1718 L of a 0.3556-M C3H7OH solution is diluted to a concentration of 0.1222 M, what is the volume of the resulting solution?

    Answer

    0.5000 L

    Click here to see a video of the solution

     

     

    PROBLEM \(\PageIndex{10}\)

    What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?

    Answer

    1.9 mL

    PROBLEM \(\PageIndex{11}\)

    What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed to evaporate until the volume is reduced to 0.105 L?

    Answer

    0.794 M

    Click here to see a video of the solution

    PROBLEM \(\PageIndex{12}\)

    What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?

    1. 1.00 L of a 0.250-M solution of Fe(NO3)3 is diluted to a final volume of 2.00 L
    2. 0.5000 L of a 0.1222-M solution of C3H7OH is diluted to a final volume of 1.250 L
    3. 2.35 L of a 0.350-M solution of H3PO4 is diluted to a final volume of 4.00 L
    4. 22.50 mL of a 0.025-M solution of C12H22O11 is diluted to 100.0 mL
    Answer a

    0.125 M

    Answer b

    0.04888 M

    Answer c

    0.206 M

    Answer d

    0.0056 M

    PROBLEM \(\PageIndex{13}\)

    What is the final concentration of the solution produced when 225.5 mL of a 0.09988-M solution of Na2CO3 is allowed to evaporate until the solution volume is reduced to 45.00 mL?

    Answer

    0.5005 M

    Click here to see a video of the solution

     

     

    PROBLEM \(\PageIndex{14}\)

    A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?

    Answer

    11.9 M

    PROBLEM \(\PageIndex{15}\)

    An experiment in a general chemistry laboratory calls for a 2.00-M solution of HCl. How many mL of 11.9 M HCl would be required to make 250 mL of 2.00 M HCl?

    Answer

    42.0 mL

    Click here to see a video of the solution

    PROBLEM \(\PageIndex{16}\)

    What volume of a 0.20-M K2SO4 solution contains 57 g of K2SO4?

    Answer

    1.6 L

    PROBLEM \(\PageIndex{17}\)

    The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (K2Cr2O7), what is the maximum permissible molarity of that substance?

    Answer

    4.8 × 10−6 M

    Click here to see a video of the solution

     

    Contributors

     


    6.1.1: Practice Problems- Solution Concentration is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?