Respuesta :
Answer: The molarity of bromide ions is 0.348 M.
Explanation:
To calculate the moles of cadmium nitrate, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of silver nitrate = 0.7170 M
Volume of silver nitrate = 24.52 mL = 0.02452 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.7170mol/L=\frac{\text{Moles of silver nitrate}}{0.02425L}\\\\\text{Moles of silver nitrate}=0.0174mol[/tex]
The chemical equation for the reaction of silver nitrate and bromide ions follows:
[tex]AgNO_3(aq.)+Br^-(aq.)\rightarrow AgBr(s)+NO_3^-(aq.)[/tex]
By Stoichiometry of the reaction:
1 mole of silver nitrate reacts with 1 mole of bromide ions.
So, 0.0174 moles of silver nitrate will react with = [tex]\frac{1}{1}\times 0.0174=0.0174mol[/tex] of bromide ions.
Now, calculating the molarity of bromide ions by using equation 1, we get:
Moles of bromide ions = 0.0174 moles
Volume of solution = 50 mL = 0.05 L
Putting values in equation 1, we get:
[tex]\text{Molarity of bromide ions}=\frac{0.0174mol}{0.05L}=0.348M[/tex]
Hence, the molarity of bromide ions is 0.348 M.
