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Let's break down the information given in the problem:

1. **Mass of impure KOH**: \( 3 \, \text{g} \)
2. **Volume of solution prepared**: \( 250 \, \text{cm}^3 \)
3. **Volume of solution taken for titration**: \( 25.0 \, \text{cm}^3 \)
4. **Volume of HCl used in titration**: \( 25.80 \, \text{cm}^3 \) of \( 0.2 \, \text{mol·dm}^{-3} \) HCl
5. **Reaction equation**: \( \text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O} \)

### Step-by-step solution:

#### 1. Moles of HCl used in the titration:
- Molarity (concentration) of HCl: \( 0.2 \, \text{mol·dm}^{-3} \)
- Volume of HCl: \( 25.80 \, \text{cm}^3 = 25.80 \times 10^{-3} \, \text{dm}^3 \)

Moles of HCl used = \( \text{Molarity} \times \text{Volume} \)

\[
\text{Moles of HCl} = 0.2 \, \text{mol·dm}^{-3} \times 25.80 \times 10^{-3} \, \text{dm}^3 = 5.16 \times 10^{-3} \, \text{mol}
\]

#### 2. Moles of KOH in the titrated solution:
Since the reaction between KOH and HCl occurs in a 1:1 molar ratio (as per the balanced equation), the moles of KOH will be equal to the moles of HCl.

\[
\text{Moles of KOH in 25.0 cm}^3 \text{ of solution} = 5.16 \times 10^{-3} \, \text{mol}
\]

#### 3. Moles of KOH in the entire solution:
The titrated sample was 25.0 cm³, but the total solution was 250 cm³. To find the total moles of KOH in the original solution:

\[
\text{Moles of KOH in 250 cm}^3 = \frac{250}{25.0} \times 5.16 \times 10^{-3} \, \text{mol} = 10 \times 5.16 \times 10^{-3} = 5.16 \times 10^{-2} \, \text{mol}
\]

Thus, the moles of KOH in the original solution is **0.0516 mol**.

Would you like to explore more details about this calculation or ask any questions?

Here are 5 follow-up questions you may find useful:
1. How can we calculate the purity of KOH based on this result?
2. What role does the balanced equation play in solving this problem?
3. How can we convert mass of KOH into moles using its molar mass?
4. How would the answer change if the concentration of HCl were different?
5. What is the significance of titration in this type of analysis?

**Tip**: Always ensure units are consistent when calculating moles, especially converting between cm³ and dm³ for volume.

Dineo dissolves 3 g of impure potassium hydroxide in water and makes the solution up to 250 cm³. She then takes 25.0 cm³ of this solution and titrates it with dilute hydrochloric acid. She finds that she needs 25.80 cm³ of 0.2 mol·dm⁻³ hydrochloric acid to fully neutralize the potassium hydroxide. Calculate the moles of KOH in the original solution.

Learn how to calculate the moles of potassium hydroxide (KOH) using titration data and stoichiometry. This problem involves neutralizing KOH with hydrochloric acid and requires the use of mole concepts and stoichiometric ratios. Suitable for high school chemistry students in Grades 9-12.

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