Molarity Notes

Molarity Practice Problems

Key Concepts

  • Molarity (): The concentration of a solution expressed as the number of moles of solute per liter of solution.

    • Formula: M=racextmolesofsoluteextlitersofsolutionM = rac{ ext{moles of solute}}{ ext{liters of solution}}

Problem 1: Molarity of NaCl in Sea Water

  • Given: 28.0 g of NaCl per liter.

  • Molar Mass of NaCl: 58.44 g/mol.

  • Calculation:

    1. Convert grams to moles:
      extmolesofNaCl=rac28.0extg58.44extg/mol=0.479extmolesext{moles of NaCl} = rac{28.0 ext{ g}}{58.44 ext{ g/mol}} = 0.479 ext{ moles}

    2. Molarity = rac0.4791.000=0.479extMrac{0.479}{1.000} = 0.479 ext{ M}

Problem 2: Molarity of H2SO4

  • Given: 245.0 g of H2SO4 in 1.000 L of solution.

  • Molar Mass of H2SO4: 98.079 g/mol.

  • Calculation:

    1. Convert grams to moles:
      extmolesofH2SO4=rac245.0extg98.079extg/mol=2.497extmolesext{moles of H2SO4} = rac{245.0 ext{ g}}{98.079 ext{ g/mol}} = 2.497 ext{ moles}

    2. Molarity = rac2.4971.000=2.497extMrac{2.497}{1.000} = 2.497 ext{ M}

Problem 3: Molarity of Na2CO3

  • Given: 5.30 g of Na2CO3 in 400.0 mL.

  • Molar Mass of Na2CO3: 105.99 g/mol.

  • Calculation:

    1. Convert grams to moles:
      extmolesofNa2CO3=rac5.30extg105.99extg/mol=0.0500extmolesext{moles of Na2CO3} = rac{5.30 ext{ g}}{105.99 ext{ g/mol}} = 0.0500 ext{ moles}

    2. Convert mL to L: 400.0 mL = 0.400 L

    3. Molarity = rac0.05000.400=0.125extMrac{0.0500}{0.400} = 0.125 ext{ M}

Problem 4: Molarity of NaOH

  • Given: 5.00 g of NaOH in 750.0 mL.

  • Molar Mass of NaOH: 40.00 g/mol.

  • Calculation:

    1. Convert grams to moles:
      extmolesofNaOH=rac5.00extg40.00extg/mol=0.125extmolesext{moles of NaOH} = rac{5.00 ext{ g}}{40.00 ext{ g/mol}} = 0.125 ext{ moles}

    2. Convert mL to L: 750.0 mL = 0.750 L

    3. Molarity = rac0.1250.750=0.167extMrac{0.125}{0.750} = 0.167 ext{ M}

Problem 5: Moles in 10.0 L of 2.00 M Na2CO3

  • Calculation:

    • extmolesofNa2CO3=2.00extMimes10.0extL=20.0extmolesext{moles of Na2CO3} = 2.00 ext{ M} imes 10.0 ext{ L} = 20.0 ext{ moles}

Problem 6: Moles in 10.0 mL of 2.00 M Na2CO3

  • Calculation:

    • extmolesofNa2CO3=2.00extMimesrac10.0extmL1000extmL=0.0200extmolesext{moles of Na2CO3} = 2.00 ext{ M} imes rac{10.0 ext{ mL}}{1000 ext{ mL}} = 0.0200 ext{ moles}

Problem 7: Moles of NaCl in 100.0 mL of 0.200 M solution

  • Calculation:

    • extmolesofNaCl=0.200extMimesrac100.0extmL1000extmL=0.0200extmolesext{moles of NaCl} = 0.200 ext{ M} imes rac{100.0 ext{ mL}}{1000 ext{ mL}} = 0.0200 ext{ moles}

Problem 8: Mass of NaCl in Problem #7

  • Calculation:

    • extmassofNaCl=0.0200extmolesimes58.44extg/mol=1.17extgNaClext{mass of NaCl} = 0.0200 ext{ moles} imes 58.44 ext{ g/mol} = 1.17 ext{ g NaCl}

Problem 9: Mass of H2SO4 for 750.0 mL of 2.00 M solution

  • Calculation:

    • extmass=2.00extmoles/Limes0.750extLimes98.079extg/mol=147.12extgH2SO4ext{mass} = 2.00 ext{ moles/L} imes 0.750 ext{ L} imes 98.079 ext{ g/mol} = 147.12 ext{ g H2SO4}

Problem 10: Volume of 18.0 M H2SO4 needed for 2.45 g H2SO4

  • Calculation:

    1. Convert grams to moles:
      extmolesofH2SO4=rac2.45extg98.079extg/mol=0.0250extmolesext{moles of H2SO4} = rac{2.45 ext{ g}}{98.079 ext{ g/mol}} = 0.0250 ext{ moles}

    2. Required Volume = rac0.0250extmoles18.0extM=0.00139extLrac{0.0250 ext{ moles}}{18.0 ext{ M}} = 0.00139 ext{ L}