A titration experiment is performed using solution W (0.100 mol/dm3) and solution L containing an unknown metal hydroxide (MOH). Assume the reaction has a 1:1 mole ratio. If 20.00 cm3 of solution W re

To solve this problem, crucial information regarding the concentration of solution W and the volume of solution W used in the titration is missing from the image. I will proceed by making reasonable assumptions for these values to demonstrate the calculation method.

Assumptions:

• Concentration of solution W ($C_W$) = $0.100 \text{ mol/dm}^3$
• Average volume of solution W used ($V_W$) = $20.00 \text{ cm}^3$
• The reaction between W and the metal hydroxide (MOH) is a 1:1 mole ratio (e.g., if W is a monoprotic acid).

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a) Number of moles of solution W that reacted with each $25 \text{cm}^3$ portion of solution L.

Step 1: Calculate the moles of solution W.

$$\text{Moles of W} = \text{Concentration of W} \times \text{Volume of W (in dm}^3)$$ $$\text{Moles of W} = 0.100 \text{ mol/dm}^3 \times \frac{20.00 \text{ cm}^3}{1000 \text{ cm}^3/\text{dm}^3}$$ $$\text{Moles of W} = 0.100 \times 0.02000 \text{ mol}$$ $$\text{Moles of W} = \boxed{\text{0.00200 mol}}$$

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b) Calculate the:

i) Number of moles of the metal hydroxide (MOH) in solution L that reacted with each portion of solution W.

Step 2: Determine the moles of MOH.
Based on the assumed 1:1 mole ratio between W and MOH, the moles of MOH that reacted are equal to the moles of W.

$$\text{Moles of MOH} = \text{Moles of W}$$ $$\text{Moles of MOH} = \boxed{\text{0.00200 mol}}$$

ii) Concentration of solution L.

Step 3: Calculate the concentration of solution L.
The volume of solution L used was $25 \text{ cm}^3$.

$$\text{Concentration of L} = \frac{\text{Moles of MOH}}{\text{Volume of L (in dm}^3)}$$ $$$$