A chemist adds 345.0 mL of a 0.520 mol/L of a barium chlorate solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.

54.54g

Explanation:

First, let us obtain the number of mole of Barium chlorate, Ba(ClO3)2 in the solution. This is illustrated below:

Volume = 345mL = 345/1000 = 0.345L

Molarity = 0.520mol/L

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.520 x 0.345

Mole = 0.1794mole

Now, we can easily find the mass of the Barium chlorate. This is illustrated below:

Molar Mass of Ba(ClO3)2 = 137 + 2[35.5 + (16x3)] = 137 +2[35.5 + 48] = 137 + 2[83.5] = 137 + 167 = 304g/mol

Number of mole of Ba(ClO3)2 = 0.1794mole

Mass =?

Mass = number of mole x Molar Mass

Mass of Ba(ClO3)2 = 0.1794 x 304

Mass of Ba(ClO3)2 = 54.54g

The mass of barium chlorate is 54.6 grams

Explanation:

Step 1: Data given

Barium chlorate = Ba(ClO3)2

Volume of the barium chlorate solution = 345.0 mL = 0.345 L

Concentration of the barium chlorate solution = 0.520 M

Molar mass of barium chlorate = 304.23 g/mol

Step 2: Calculate moles Ba(ClO3)2

Moles Ba(ClO3)2 = volume of Ba(ClO3)2 solution * concentration of Ba(ClO3)2

Moles Ba(ClO3)2 = 0.345 L * 0.520 M

Moles Ba(ClO3)2 = 0.1794 moles

Step 3: calculate mass of Ba(ClO3)2

Mass Ba(ClO3)2 = moles Ba(ClO3)2 * molar mass Ba(ClO3)2

Mass Ba(ClO3)2 = 0.1794 moles * 304.23 g/mol

Mass Ba(ClO3)2 = 54.6 grams

The mass of barium chlorate is 54.6 grams

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