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11 August, 09:04

Calculate the molarity of each solution.

a. 0.38 mol of lino3 in 6.14 l of solution

b. 72.8 g c2h6o in 2.34 l of solution

c. 12.87 mg ki in 112.4 ml of solution

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Answers (1)
  1. 11 August, 09:19
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    Q1)

    molarity is defined as the number of moles of solute in 1 L solution

    the number of moles of LiNO₃ - 0.38 mol

    volume of solution - 6.14 L

    since molarity is number of moles in 1 L

    number of moles in 6.14 L - 0.38 mol

    therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L

    molarity of solution is 0.0619 M

    Q2)

    the mass of C₂H₆O in the solution is 72.8 g

    molar mass of C₂H₆O is 46 g/mol

    number of moles = mass present / molar mass of compound

    the number of moles of C₂H₆O - 72.8 g / 46 g/mol

    number of C₂H₆O moles - 1.58 mol

    volume of solution - 2.34 L

    number of moles in 2.34 L - 1.58 mol

    therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M

    molarity of C₂H₆O is 0.675 M

    Q3)

    Mass of KI in solution - 12.87 x 10⁻³ g

    molar mass - 166 g/mol

    number of mole of KI = mass present / molar mass of KI

    number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol

    volume of solution - 112.4 mL

    number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol

    therefore number of moles in 1000 mL - 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL

    molarity of KI - 6.90 x 10⁻⁴ M
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