Calculate the concentration of sodium chloride solution if 24.1 cm³ of 0.1 mol dm⁻³ silver nitrate solution reacts completely with 25.0 cm³ of sodium chloride solution.

Step 1: Calculate the moles of silver nitrate in 24.1 cm³ of solution.
The volume of silver nitrate solution is 24.1 cm³, which is $0.0241 \text{ dm}^3$. The concentration is $0.1 \text{ mol dm}^{-3}$.

$$\text{Moles of AgNO}_3 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of AgNO}_3 = 0.1 \text{ mol dm}^{-3} \times 0.0241 \text{ dm}^3$$ $$\text{Moles of AgNO}_3 = 0.00241 \text{ mol}$$

The moles of silver nitrate in 24.1 cm³ of solution is $\boxed{\text{0.00241 mol}}$.

Step 2: Calculate the moles of sodium chloride in 25.0 cm³ of solution.
The reaction between silver nitrate and sodium chloride is:

$$\text{AgNO}_3\text{(aq)} + \text{NaCl(aq)} \rightarrow \text{AgCl(s)} + \text{NaNO}_3\text{(aq)}$$

From the balanced equation, the mole ratio of $\text{AgNO}_3$ to $\text{NaCl}$ is 1:1.
Therefore, the moles of $\text{NaCl}$ that reacted with $0.00241 \text{ mol}$ of $\text{AgNO}_3$ is also $0.00241 \text{ mol}$.

$$\text{Moles of NaCl in 25.0 cm}^3 = \text{Moles of AgNO}_3$$ $$\text{Moles of NaCl in 25.0 cm}^3 = 0.00241 \text{ mol}$$

The moles of sodium chloride in 25.0 cm³ of solution is $\boxed{\text{0.00241 mol}}$.

Step 3: Calculate the moles of sodium chloride in 250 cm³ of solution.
The 25.0 cm³ sample of sodium chloride solution was taken from a total volume of 250 cm³ of the dilute solution.
To find the total moles in 250 cm³, we scale up the moles from the 25.0 cm³ sample:

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