Question

What mass of Silver chloride, in grams, gets precipitated when 150 ml of 32% solution of Silver nitrate is reacted with 150 ml of 11% Sodium chloride solution?
Atomic mass in g/mol: Ag = 108, Na = 23, Cl = 35.5, N = 14, O = 16.

• A. 40.47
• B. 32.45
• C. 48.0
• D. 16.52

Hint:

When solving precipitation problems, always identify the limiting reactant first, as it dictates the amount of product formed.

Answer 1

The Correct Option is A

We are given the following information: - Volume of AgNO₃ solution = 150 ml - Concentration of AgNO₃ = 32% - Volume of NaCl solution = 150 ml - Concentration of NaCl = 11% - Molar mass of AgNO₃ = 108 g/mol (for Ag) - Molar mass of NaCl = 23 g/mol (for Na) + 35.5 g/mol (for Cl) = 58.5 g/mol First, we need to calculate the amount of Silver nitrate (AgNO₃) and Sodium chloride (NaCl) in moles: \[ \text{Mass of AgNO₃} = \frac{32}{100} \times 150 = 48 \, \text{g} \] \[ \text{Moles of AgNO₃} = \frac{48}{108} = 0.444 \, \text{moles} \] \[ \text{Mass of NaCl} = \frac{11}{100} \times 150 = 16.5 \, \text{g} \] \[ \text{Moles of NaCl} = \frac{16.5}{58.5} = 0.282 \, \text{moles} \] Since Silver chloride (AgCl) precipitates in a 1:1 molar ratio between Ag⁺ and Cl⁻, the limiting reactant will determine the amount of AgCl that forms. \[ \text{Moles of AgCl formed} = \min(0.444, 0.282) = 0.282 \, \text{moles} \] Now, calculate the mass of AgCl: \[ \text{Molar mass of AgCl} = 108 + 35.5 = 143.5 \, \text{g/mol} \] \[ \text{Mass of AgCl} = 0.282 \times 143.5 = 40.47 \, \text{g} \] Thus, the mass of Silver chloride precipitated is 40.47 g.