Answer:
The correct answer is 0.1236 M.
Explanation:
What is the concentration of [tex]FeBr_{3}[/tex] in a prepared solution by dissolving 10.0 g of [tex]FeBr_{3}[/tex] insufficient water to obtain 275 mL of solution?
First, we must look for the molar mass of iron bromide (III).
Mm = 295.56 g/mol
Then, the amount of [tex]FeBr_{3}[/tex] moles in 10 g is equal to:
n= [tex]\frac{10g}{295.56 \frac{g}{mol} }[/tex]
n= 0.034 mol
Now to know the molar concentration we have to divide the number of moles of the compound by the volume of the solution in liters:
Volume= [tex]275mL * \frac{1L}{1000mL}[/tex]
M= [tex]\frac{0.034 mol}{0.275 L}[/tex]
M=0.1236
Have a nice day!
