Question

A chemist adds 25.0 mL of a 0.549 molL barium chloride solution to a reaction flask. Calculate the millimoles of barium chloride the chemist hasadded to the flask. Round your answer to 3 significant digits.

Answer 1

Given the following

`\begin{gathered} M=0.549\text{ mol/L} \\ V=25.0mL=25*10^(-3)L=0.025L \\ M=\text{Molarity} \\ V=\text{Volume} \end{gathered}`

To find the millimoles of Barium Chloride

To achieve, define molarity

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution. Molarity is also known as the molar concentration of a solution

So,

`M=(mole)/(Volume)`

Substitute the given into the formula above

`\begin{gathered} 0.549\text{ mol/L=}(mole)/(0.025L) \\ \text{mole}=0.549\text{ mol/L}*0.025L \\ \text{mole}=0.013725\text{mol} \\ \text{millimole}=(0.013725)/(1000) \\ \text{millimole}=0.000013725 \\ \text{millimole}=0.0000137(\text{nearest 3 significant digits)} \end{gathered}`

Hence, the millimoles of Barium Chloride correct to 3 significant digits is 0.0000137 millimoles