Achemist 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.

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 

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