A 2.0 g sample of silver carbonate (MM = 275.81 g mol⁻¹) was added to 100.0 mL of water in a beaker - HSC - SSCE Chemistry - Question 17 - 2022 - Paper 1

The initial concentration of silver ions can be calculated using the formula for concentration:

C = rac{n}{V}

Where:

• $C$ is the concentration in mol L⁻¹,
• $n$ is the number of moles,
• $V$ is the volume in liters.

Converting 100.0 mL to L: $V = 100.0 ext{ mL} = 0.1000 ext{ L}$

Now substituting in the values:

C = rac{0.00725 ext{ mol}}{0.1000 ext{ L}} ightarrow C ext{ (initial)} = 0.0725 ext{ mol L}^{-1}

When an additional 100.0 mL of water is added, the final volume becomes:

$$V_{final} = 100.0 ext{ mL} + 100.0 ext{ mL} = 200.0 ext{ mL} = 0.2000 ext{ L}$$

The concentration after dilution can be calculated again using:

C_{final} = rac{0.00725 ext{ mol}}{0.2000 ext{ L}} = 0.03625 ext{ mol L}^{-1}

Now, we find the ratio of the concentration of silver ions before and after dilution:

ext{Ratio} = rac{C_{initial}}{C_{final}} = rac{0.0725}{0.03625} = 2:1

Therefore, the correct answer is option C: 2:1.