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

In the initial 100.0 mL of water, the concentration of silver ions can be calculated as follows:

$C_{initial} = \frac{n}{V}$

where:

• n is the moles of silver carbonate = 0.00725 mol
• V is the volume in liters = 0.100 L

So,

$C_{initial} = \frac{0.00725 \, \text{mol}}{0.100 \, \text{L}} = 0.0725 \, \text{mol L}⁻¹$

After adding another 100.0 mL of water, the total volume becomes 200.0 mL or 0.200 L. The number of moles remains the same:

$C_{final} = \frac{n}{V_{total}}$

Calculating:

$C_{final} = \frac{0.00725 \, \text{mol}}{0.200 \, \text{L}} = 0.03625 \, \text{mol L}⁻¹$

Now, we find the ratio of the initial concentration to the final concentration:

$\text{Ratio} = \frac{C_{initial}}{C_{final}} = \frac{0.0725}{0.03625} = 2:1$

Thus, the ratio of the concentration of silver ions in solution before and after dilution is 2:1.