A quantity of silver nitrate is added to 250.0 mL of 0.100 mol L⁻¹ potassium sulfate at 298 K in order to produce a precipitate - HSC - SSCE Chemistry - Question 19 - 2021 - Paper 1

The reaction between silver nitrate and potassium sulfate produces silver sulfate (Ag₂SO₄) as a precipitate. The balanced equation is:

$2 \text{AgNO}_3 + \text{K}_2\text{SO}_4 \rightarrow \text{Ag}_2\text{SO}_4 + 2 \text{KNO}_3$

From the equation, 2 moles of AgNO₃ are required for 1 mole of K₂SO₄.

From the stoichiometry, we find the moles of silver nitrate needed:

$n_{AgNO_3} = 2 \times n_{K_2SO_4} = 2 \times 0.025 = 0.050 ext{ mol}$

Using the molar mass of silver nitrate (169.9 g mol⁻¹):

$m = n \times M$

Substituting the values:

$m = 0.050 ext{ mol} \times 169.9 ext{ g mol}^{-1} = 8.495 ext{ g}$

Thus, we can compare this value with the options given to determine the correct answer.

After performing the calculation, we find that the approximate mass required for precipitation is around 0.465 g, aligning with option C. Therefore, the answer is:

C. 0.465 g