Question

Calculate the molarity of each of the following solutions:

A) 0.35 mol of LiNO3 in 6.34 L of solution

B) 74.9 g C2H6O in 2.36 L of solution

C) 11.56 mg KI in 113.4 mL of solution

Answer 1

Molarity , is equal to the moles of solute divided by the liters of solution.

A)

Moles of LiNO3 = 0.35 mol

Volume of the solution = 6.34 L

Molarity = moles of the solute / liters of solution

Molarity = 0.35 / 6.34 = 0.055 mol/L

B)

Weight of C2H6O = 74.9 g

Molecular weight of C2H6O = 46 g/mol

Number of moles = weight /mol.wt = 74.9/ 46 = 1.63 moles

Molarity = Moles /liters

Volume of the solution = 2.36 L

Therefore

Molarity = 1.63/2.36= 0.689 moles/L

C)

Weight of KI = 11.56 mg = 0.01156 g

Molecular weight of KI = 166 g/mol

Number of moles = 0.01156/ 166= 0.0000696 moles

Volume of the solution = 113.4 mL= 0.1134 L

There fore

Molarity = Moles / Liters = 0.0000696/ 0.1134 = 0.00061moles /Liter