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

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