Questions: Part C If 27.0 mL of an AgNO3 solution is needed to precipitate all the Cl- ions in a 800-mg sample of KCl (forming AgCl ), what is the molarity of the AgNO3 solution? Express the molarity in the moles of a solute per liters of a solution to three significant figures. Molarity = M Part D If 46.0 mL of 0.106 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution? Express the mass in grams to three significant figures. m = g

Solution Steps

First, calculate the moles of KCl in the 800 mg sample. The molar mass of KCl is approximately 74.55 g/mol.

\[ \text{Moles of KCl} = \frac{800 \, \text{mg}}{1000 \, \text{mg/g}} \times \frac{1}{74.55 \, \text{g/mol}} = 0.01073 \, \text{mol} \]

Since each KCl molecule contains one Cl\(^-\) ion, the moles of Cl\(^-\) is also 0.01073 mol.

The reaction between AgNO\(_3\) and Cl\(^-\) is a 1:1 stoichiometry:

\[ \text{AgNO}_3 + \text{Cl}^- \rightarrow \text{AgCl} + \text{NO}_3^- \]

Thus, the moles of AgNO\(_3\) needed is equal to the moles of Cl\(^-\), which is 0.01073 mol.

Molarity is defined as moles of solute per liter of solution. Given that 27.0 mL of AgNO\(_3\) solution is used, convert this volume to liters:

\[ 27.0 \, \text{mL} = 0.0270 \, \text{L} \]

Now, calculate the molarity:

\[ \text{Molarity of AgNO}_3 = \frac{0.01073 \, \text{mol}}{0.0270 \, \text{L}} = 0.3974 \, \text{M} \]

Final Answer