## Explanation

The conductivity of a salt solution is fundamentally related to the concentration of ions present in the solution. To decrease the conductivity, it is necessary to **reduce the number of freely moving ions**.

### Method to Decrease Conductivity

The most effective way to accomplish this is by **diluting the solution**. This involves adding more water to the existing solution, which increases the volume and spreads out the ions, thereby reducing their concentration.

### Mathematical Explanation

The relationship between conductivity ($\sigma$) and concentration ($c$) can be expressed as:

$\sigma = \lambda \cdot c$Where $\lambda$ is the molar conductivity. By adding more water, the concentration $c$ decreases, which in turn lowers the conductivity $\sigma$.

### Practical Approach

To achieve this:

**Measure the initial volume $V_i$ and concentration $c_i$**of the salt solution.**Determine the desired final concentration $c_f$**.**Calculate the final volume $V_f$**required to achieve this concentration: $V_f = \frac{c_i \times V_i}{c_f}$**Add the difference in volumes $V_{add}$**, calculated as: $V_{add} = V_f - V_i$ which is the amount of water to be added to decrease the conductivity.

#### Example Calculation

Suppose we have 1 L of salt solution with an initial concentration of 1 M. To dilute it to a 0.1 M solution:

$V_f = \frac{1 \text{ M} \times 1 \text{ L}}{0.1 \text{ M}} = 10 \text{ L}$ $V_{add} = 10 \text{ L} - 1 \text{ L} = 9 \text{ L}$**Adding 9 liters of water** will decrease the concentration, thereby reducing the conductivity.

### Conclusion

**Decreasing the conductivity** of a salt solution is effectively achieved by **adding more water, which dilutes the solution** and reduces the concentration of ions. This straightforward method is supported both practically and mathematically, ensuring a lower conductivity for the solution.