This is an incomplete question, here is a complete question.
Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.
Answer : The volume of iron(II) bromide solution is, [tex]2.13\times 10^2mL[/tex]
Explanation : Given,
Concentration of iron(II) bromide = 1.29 mo/L
Moles of iron(II) bromide = 275 mmol = 0.275 mol
conversion used : 1 mmol = 0.001 mol
Now we have to calculate the volume of iron(II) bromide.
[tex]\text{Volume of iron(II) bromide}=\frac{Moles of iron(II) bromide}}{\text{Concentration of iron(II) bromide}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Volume of iron(II) bromide}=\frac{0.275mol}{1.29mol/L}=0.213L=2.13\times 10^2mL[/tex]
Thus, the volume of iron(II) bromide solution is, [tex]2.13\times 10^2mL[/tex]
