A chemist prepares a solution of barium chlorate by measuring out of barium chlorate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium chlorate solution. Round your answer to significant digits.

Answer: The concentration C is given as;

C = n/V = 0.138/0.5 = 0.276mol/L

A chemist prepares a solution of barium chlorate BaClO32 by measuring out 42. g of barium chlorate into a 500. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in / molL of the chemist's barium chlorate solution.

Explanation:

To calculate the concentration C of this solution, in mol/L

C = n/V = mol/L

where, n is the number of moles

V is the volume

The data are:

mass m of barium chlorate = 42g

V=500mL equivalent to 0.5 L

The problem gives the Volume, there is to find the number of moles, n.

n from the stochiometry of molarity is equal to

n = m/M, where m is the mass in g, and M the molecular weight.

We have m, we must calculate M for Ba (ClO3) 2

Ba = 137.327 Cl = 35.453 O = 15.9994

Molar mass of Ba (ClO3) 2 = 137.327 + (35.453 + 15.9994*3) * 2 = 304.2294 g/mol

n = m/M = 42/304.2394 = 0.138moles

Finally the concentration will be given after converting the volume from mL to L.

The concentration C is given as;

C = n/V = 0.138/0.5 = 0.276mol/L