Answer
1.765% of $Cl^{-}$.
Work Step by Step
1 $Cl^{-}$ + 1 $Ag_{}N_{}O_{3}$
1. Find the amount of moles of Silver Nitrate.
nº of moles = Concentration (mol/L) * Volume(L)
nº of moles = 0.2997 * 0.04258
nº of moles = 0.012761 moles
2. Because the proportion is 1 to 1, the amount of moles of $Cl^{-}$ is 0.012761.
3. Now, find the mass of $Cl^{-}$.
mass = nº of moles * molar mass
mass = 0.012761 * 35.45
mass = 0.45238g
4. Calculate the mass of the solvent (water).
mass = density * volume
mass = 1.025 * 25
mass = 25.625g
5. Now, with the mass of the $Cl^{-}$ and the solvent, find the percentage.
Percentage = $\frac{0.45238}{25.625}$ $\times$ 100%
Percentage $\approx$ 1.765%
