A chemist prepares a solution by adding 422 mg of CO (NO3) 2 (MW = 182.94 g/mol) to a volumetric flask, and then adding water until the total volume of the contents of the flask reaches the calibration line that indicates 750 mL. Determine the molarity of the prepared solution.

0.003 M

Explanation:

Molarity of a substance, is the number of moles present in a liter of solution.

M = n / V

M = molarity (unit = mol / L or M)

V = volume of solution in liter (unit = L),

n = moles of solute (unit = mol),

Moles is denoted by given mass divided by the molecular mass,

Hence,

n = w / m

n = moles,

w = given mass,

m = molecular mass.

From the information of the question,

w = 422 mg

(Since, 1mg = 0.001 g)

w = 0.422 g

m = 182.94 g/mol

V = 750 mL

(since, 1 mL = 0.001 L)

V = 0.750 L

From the above formula,

n = w / m

n = 0.422 g / 182.94 g/mol = 0.002 mol

Molarity can be calculated by using the above formula,

M = n / V

M = 0.002 mol / 0.750 L

M = 0.003 M