Respuesta :
Answer: The molarity of bromine ions in the chemist's solution is [tex]4.06\times 10^{-3}M[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of vanadium(III) bromide = 0.12 g
Molar mass of vanadium(III) bromide = 295.65 g/mol
Volume of solution = 300 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{0.12\times 1000}{295.65g/mol\times 300}\\\\\text{Molarity of solution}=1.353\times 10^{-3}M[/tex]
The chemical formula of vanadium(III) bromide is [tex]VBr_3[/tex]
1 mole of vanadium(III) bromide produces 1 mole of [tex]V^{3+}[/tex] ions and 3 moles of [tex]Br^-[/tex] ions
Molarity of bromine ions = [tex](3\times 1.353\times 10^{-3})=4.06\times 10^{-3}M[/tex]
Hence, the molarity of bromine ions in the chemist's solution is [tex]4.06\times 10^{-3}M[/tex]
