Question

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. Be sure your answer has the correct number of significant digits.

Answer 1

Complete Question

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. Be sure your answer has the correct number of significant digits.

Answer:

The concentration is
`C = 0.28 \ mol/L`

Step-by-step explanation:

From the question we told that

The mass of
`Ba(ClO_(3))_2` is
`m_b = 42 \ g`

The volume of the solution
`V_s = 500 mL = 500*10^(-3) L`

Now the number f moles of
`Ba(ClO_(3))_2` in the solution is mathematically represented as

`n = (m_b)/(Z_b)`

Where
`Z_b` is the molar mass of
`Ba(ClO_(3))_2` which a constant with a value

`Z_b = 304.23 \ g/mol`

Thus

`n = (42)/(304.23)`

`n = 0.14 \ mol`

The concentration of
`Ba(ClO_(3))_2` in the solution is mathematically evaluated as

`C = (n)/(V_2)`

substituting values

`C = (0.14)/(500*10^(-3))`

`C = 0.28 \ mol/L`