A solution of sulfuric acid of concentration 0.100 mol/dm³ is reacted with 25.0 cm³ of a base XOH. 12.0 cm³ of the acid is required to neutralize the base. What is the concentration of XOH in mol/dm³?

Here's the solution to question 3:

3. (a) Concentration of A in g/dm³

Step 1: Calculate the molar mass of $\text{H}_2\text{SO}_4$.

$$\text{Molar mass of } \text{H}_2\text{SO}_4 = (2 \times 1.0) + 32.0 + (4 \times 16.0) = 2.0 + 32.0 + 64.0 = 98.0 \text{ g/mol}$$

Step 2: Convert the molar concentration of A to $\text{g/dm}^3$.

$$\text{Concentration of A in g/dm}^3 = \text{Molarity} \times \text{Molar mass}$$ $$\text{Concentration of A} = 0.100 \text{ mol/dm}^3 \times 98.0 \text{ g/mol} = 9.80 \text{ g/dm}^3$$

The concentration of A in $\text{g/dm}^3$ is $\boxed{\text{9.80 g/dm}^3}$.

3. (b) Concentration of B in mol/dm³

Step 1: Write the balanced chemical equation for the reaction.
Assuming $\text{XOH}$ is a monobasic hydroxide, the reaction with sulfuric acid is:

$$\text{H}_2\text{SO}_4(aq) + 2\text{XOH}(aq) \longrightarrow \text{X}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$$

From the equation, the mole ratio of $\text{H}_2\text{SO}_4 : \text{XOH}$ is $1:2$. So, $n_A = 1$ and $n_B = 2$.

Step 2: Use the titration formula to find the concentration of B.
Given:
$C_A = 0.100 \text{ mol/dm}^3$
$V_A = 12.0 \text{ cm}^3 = 0.012 \text{ dm}^3$
$V_B = 25.0 \text{ cm}^3 = 0.025 \text{ dm}^3$

$$\frac{C_A V_A}{n_A} = \frac{C_B V_B}{n_B}$$ $$C_B = \frac{C_A V_A n_B}{V_B n_A}$$ $$C_B = \frac{0.100 \text{ mol/dm}^3 \times 0.012 \text{ dm}^3 \times 2}{0.025 \text{ dm}^3 \times 1}$$ $$C_B = \frac{0.0024}{0.025} \text{ mol/dm}^3$$ $$C_B = 0.096 \text{ mol/dm}^3$$

The concentration of B in $\text{mol/dm}^3$ is $\boxed{\text{0.096 mol/dm}^3}$.

3. (c) Moles of B in 25.0 cm³

Step 1: Use the concentration of B and the volume to calculate the moles.

$$\text{Moles of B} = \text{Concentration of B} \times \text{Volume of B}$$ $$\text{Moles of B} = 0.096 \text{ mol/dm}^3 \times 0.025 \text{ dm}^3$$ $$\text{Moles of B} = 0.0024 \text{ mol}$$

The moles of B in $25.0 \text{ cm}^3$ is $\boxed{\text{0.0024 mol}}$.

3. (d) Relative atomic mass of X

Step 1: Calculate the moles of $\text{XOH}$ in $250 \text{ cm}^3$ of solution B.
The concentration of B is $0.096 \text{ mol/dm}^3$.
Volume of solution B = $250 \text{ cm}^3 = 0.250 \text{ dm}^3$.

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