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

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