Please translate and these to English

The atomic masses are: $\text{Na} = 22.99 \text{ g/mol}$, $\text{C} = 12.01 \text{ g/mol}$, $\text{O} = 16.00 \text{ g/mol}$.

$$M_{\text{Na}_{2}\text{CO}_{3}} = (2 \times 22.99) + 12.01 + (3 \times 16.00) = 45.98 + 12.01 + 48.00 = 105.99 \text{ g/mol}$$

According to Table 1, the mass of $\text{Na}_{2}\text{CO}_{3}$ is $2.55 \text{ g}$.

$$n_{\text{Na}_{2}\text{CO}_{3}} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.55 \text{ g}}{105.99 \text{ g/mol}} = 0.024059 \text{ mol}$$

According to the procedure, the solution was prepared in a volume of $250 \text{ cm}^{3}$, which is equivalent to $0.250 \text{ L}$.

$$\text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}} = \frac{0.024059 \text{ mol}}{0.250 \text{ L}} = 0.096236 \text{ M}$$

Rounding to three significant figures:

$$\text{Molarity} = \boxed{0.0962 \text{ M}}$$

According to Table 2, the two precise titres are $27.3275 \text{ cm}^{3}$ and $27.3 \text{ cm}^{3}$.

$$\text{Average titre} = \frac{27.3275 \text{ cm}^{3} + 27.3 \text{ cm}^{3}}{2} = \frac{54.6275 \text{ cm}^{3}}{2} = 27.31375 \text{ cm}^{3}$$

We will use this value for the calculations.
The volume of $\text{HCl}$ ($V_{acid}$) is $27.31375 \text{ cm}^{3}$.
The volume of $\text{Na}_{2}\text{CO}_{3}$ ($V_{base}$) used in the titration is $25 \text{ cm}^{3}$ (from step 3 of the procedure).

The reaction between sodium carbonate and hydrochloric acid is:

$$\text{Na}_{2}\text{CO}_{3}(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_{2}\text{O}(\text{l}) + \text{CO}_{2}(\text{g})$$

From the equation, 1 mole of $\text{Na}_{2}\text{CO}_{3}$ reacts with 2 moles of $\text{HCl}$.
The molar ratio ($\text{Na}_{2}\text{CO}_{3} : \text{HCl}$) is $1:2$. So, $n_{base} = 1$ and $n_{acid} = 2$.

We use the titration formula:

$$\frac{M_{acid} \times V_{acid}}{n_{acid}} = \frac{M_{base} \times V_{base}}{n_{base}}$$

Where:
$M_{acid}$ = Concentration of $\text{HCl}$ (unknown)
$V_{acid} = 27.31375 \text{ cm}^{3}$
$n_{acid} = 2$
$M_{base} = 0.096236 \text{ M}$ (calculated in part a)
$V_{base} = 25 \text{ cm}^{3}$
$n_{base} = 1$

Rearranging to solve for $M_{acid}$:

$$M_{acid} = \frac{n_{acid} \times M_{base} \times V_{base}}{n_{base} \times V_{acid}}$$ $$M_{acid} = \frac{2 \times 0.096236 \text{ M} \times 25 \text{ cm}^{3}}{1 \times 27.31375 \text{ cm}^{3}}$$ $$M_{acid} = \frac{4.8118}{27.31375} \text{ M}$$ $$M_{acid} = 0.17616 \text{ M}$$

Rounding to three significant figures:

$$M_{acid} = \boxed{0.176 \text{ M}}$$

3 de faites, encore 2 aujourd'hui. Tu progresses.