Question

What is the approximate mass of the precipitate formed when \(50\) mL of \(16.9\%\) solution of \(\text{AgNO}_3\) is mixed with \(50\) mL of \(7.45\%\) KCl solution? (Molar mass of \(\text{AgNO}_3 = 169\) g/mol, KCl \(= 74.5\) g/mol, AgCl \(= 143.3\) g/mol)

• A. \(3.5\) g
• B. \(7\) g
• C. \(14\) g
• D. \(28\) g

Hint:

Find the moles of each reactant from the mass percent in 50 mL, then use the limiting reagent.

Answer 1

The Correct Option is B

Step 1: Use weight by volume.
Take the percentages as grams per 100 mL. For 50 mL, masses are half of the percentage figures: 8.45 g and 3.725 g.

Step 2: Mole ratio.
$8.45/169 = 0.05$ and $3.725/74.5 = 0.05$ mol. These are equal, so the amounts match the 1:1 equation.

Step 3: Precipitate.
$0.05$ mol of AgCl $\times$ 143.3 g/mol $= 7.165$ g, close to 7 g.

Final Answer:
Option (B). \[ \boxed{\approx 7\text{ g}} \]