From your results and information given, calculate: (i) the Concentration of A in mol/dm3 (ii) the molar mass of H2x (iii) the atomic mass of x (iv) the percentage of x in H2x [Na=23, H=1, O=6]

Here's a step-by-step solution to your chemistry problem.

First, let's clarify the given atomic masses. We are given $\text{Na} = 23$, $\text{H} = 1$, and $\text{O} = \text{b}$. Assuming 'b' is a handwritten '16', we will use $\text{O} = 16$.

Given information:

• Average volume of A ($\text{H}_2\text{X}$) used ($V_A$) = $26.50 \text{ cm}^3$
• Equation: $2\text{NaOH} + \text{H}_2\text{X} \rightarrow \text{Na}_2\text{X} + 2\text{H}_2\text{O}$
• Mass concentration of $\text{H}_2\text{X}$ (acid A) = $4.9 \text{ g/dm}^3$
• Solution B ($\text{NaOH}$) contains $2 \text{ g}$ of $\text{NaOH}$ per $500 \text{ cm}^3$
• Volume of B ($\text{NaOH}$) used ($V_B$) = $25 \text{ cm}^3$
• Atomic masses: $\text{Na} = 23$, $\text{H} = 1$, $\text{O} = 16$ (assumed)

Step 1: Calculate the molar concentration of $\text{NaOH}$ (solution B).
First, find the molar mass of $\text{NaOH}$:

$$M(\text{NaOH}) = \text{Na} + \text{O} + \text{H} = 23 + 16 + 1 = 40 \text{ g/mol}$$

Next, calculate the mass of $\text{NaOH}$ in $1 \text{ dm}^3$:

$$\text{Mass concentration of NaOH} = \frac{2 \text{ g}}{500 \text{ cm}^3} \times \frac{1000 \text{ cm}^3}{1 \text{ dm}^3} = 4 \text{ g/dm}^3$$

Now, calculate the molar concentration of $\text{NaOH}$ ($C_B$):

$$C_B = \frac{\text{Mass concentration}}{\text{Molar mass}} = \frac{4 \text{ g/dm}^3}{40 \text{ g/mol}} = 0.1 \text{ mol/dm}^3$$

Step 2: Calculate the molar concentration of $\text{H}_2\text{X}$ (solution A).
From the balanced chemical equation, the mole ratio of $\text{H}_2\text{X}$ to $\text{NaOH}$ is $1:2$. So, $n_A = 1$ and $n_B = 2$.
We use the titration formula:

$$\frac{C_A V_A}{n_A} = \frac{C_B V_B}{n_B}$$

Rearrange to solve for $C_A$:

$$C_A = \frac{C_B V_B n_A}{V_A n_B}$$

Substitute the known values:

$$C_A = \frac{(0.1 \text{ mol/dm}^3) \times (25 \text{ cm}^3) \times 1}{(26.50 \text{ cm}^3) \times 2}$$ $$C_A = \frac{2.5}{53} = \frac{5}{106} \text{ mol/dm}^3$$ $$C_A \approx 0.0471698 \text{ mol/dm}^3$$

Rounding to four significant figures: