Answer:
a) 24.85 grams of sodium sulfate is needed.
b) Mass of 0.00202 moles of chloride ions:
Mass percentage of chloride ion present in the sample is 0.7171%.
Explanation:
![M=\frac{n}{V(L)}](https://tex.z-dn.net/?f=M%3D%5Cfrac%7Bn%7D%7BV%28L%29%7D)
m = Molarity of the solution
n = moles of compound
V = volume of the solution in L.
a) Moles of sodium sulfate = n
Molarity of the solution , M= 0.500 M
Volume of the solution = V = 350.0 mL = 0.3500 L
![n=M\times V=0.500 M\times 0.3500 L=0.175 mol](https://tex.z-dn.net/?f=n%3DM%5Ctimes%20V%3D0.500%20M%5Ctimes%200.3500%20L%3D0.175%20mol)
Mass of 0.175 moles of sodium sulfate = 0.175 mol × 142 g/mol = 24.85 g
24.85 grams of sodium sulfate is needed.
b) Moles of silver ion = n
Molarity of the silver ions = M = 0.100 M
Volume of the solution = V = 20.2 mL = 0.0202 L
![n=M\times V= 0.100 M\times 0.0202 L=0.00202 mol](https://tex.z-dn.net/?f=n%3DM%5Ctimes%20V%3D%200.100%20M%5Ctimes%200.0202%20L%3D0.00202%20mol)
![Ag^++Cl^-\rightarrow AgCl](https://tex.z-dn.net/?f=Ag%5E%2B%2BCl%5E-%5Crightarrow%20AgCl)
According to reaction , 1 mole of silver ions reacts with 1 mole of chloride ions.
Then 0.00202 moles of silver ions will react with :
of chloride ions.
Mass of 0.00202 moles of chloride ions:
0.00202 mol × 35.5 g/mol = 0.07171 g
Mass percentage of chloride ions in 10.0 grams of water:
![=\frac{0.07171 g}{10.0 g}\times 100=0.7171\%](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.07171%20g%7D%7B10.0%20g%7D%5Ctimes%20100%3D0.7171%5C%25)
Mass percentage of chloride ion present in the sample is 0.7171%.