4. A chemist carries out an experiment to determine the number of moles of $$ \mathrm{Cl}^{-} $$ions in a 1 L sample of water. The salt water reacts with excess $$ \mathrm{AgNO}_{3} $$ solution. The resulting precipitate is collected, dried and weighed. The mass of the AgCl precipitate is found to be 5.4 g . The chemical equation for the reaction is: $$ \mathrm{AgNO}_{3}(a q)+\mathrm{NaCl}(a q) \rightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q) $$ Calculate the moles of $$ \mathrm{Cl}^{-} $$ions in the sample, and describe how the number of moles was found. Re to the periodic table.

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To calculate the moles of $$ \mathrm{Cl}^{-} $$ ions in the sample, we need to use the chemical equation provided and the mass of the AgCl precipitate.

Given:
- Mass of AgCl precipitate = 5.4 g
- Chemical equation: $$ \mathrm{AgNO}_{3}(a q)+\mathrm{NaCl}(a q) ightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q) $$

First, we need to find the molar mass of AgCl. The molar mass of AgCl can be calculated by adding the atomic masses of Ag and Cl.

The atomic mass of Ag is 107.868 g/mol, and the atomic mass of Cl is 35.453 g/mol.

Molar mass of AgCl = 107.868 g/mol + 35.453 g/mol = 143.321 g/mol

Next, we can calculate the number of moles of AgCl precipitate using the mass of the precipitate and the molar mass of AgCl.

Number of moles of AgCl = Mass of AgCl precipitate / Molar mass of AgCl

Number of moles of AgCl = 5.4 g / 143.321 g/mol

Now, we need to find the number of moles of $$ \mathrm{Cl}^{-} $$ ions in the sample. Since the chemical equation shows that 1 mole of AgCl is formed from 1 mole of $$ \mathrm{Cl}^{-} $$ ions, the number of moles of $$ \mathrm{Cl}^{-} $$ ions is equal to the number of moles of AgCl.

Therefore, the number of moles of $$ \mathrm{Cl}^{-} $$ ions in the sample is 5.4 g / 143.321 g/mol.

Let's calculate this.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{5.4}{143.321}$$
- step1: Convert the expressions:
  $$\frac{\frac{27}{5}}{\frac{143321}{1000}}$$
- step2: Multiply by the reciprocal:
  $$\frac{27}{5}	imes \frac{1000}{143321}$$
- step3: Reduce the numbers:
  $$27	imes \frac{200}{143321}$$
- step4: Multiply:
  $$\frac{27	imes 200}{143321}$$
- step5: Multiply:
  $$\frac{5400}{143321}$$
The number of moles of $$ \mathrm{Cl}^{-} $$ ions in the sample is approximately 0.037678 moles.

To find the number of moles of $$ \mathrm{Cl}^{-} $$ ions, we first calculated the number of moles of AgCl precipitate using the mass of the precipitate and the molar mass of AgCl. Since the chemical equation shows that 1 mole of AgCl is formed from 1 mole of $$ \mathrm{Cl}^{-} $$ ions, the number of moles of $$ \mathrm{Cl}^{-} $$ ions is equal to the number of moles of AgCl. Therefore, the number of moles of $$ \mathrm{Cl}^{-} $$ ions in the sample is approximately 0.037678 moles.
The number of moles of $$ \mathrm{Cl}^{-} $$ ions in the sample is approximately 0.0377 moles.

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