In a reaction 20cm3 of 0.1M sodium carbonate completely reacted with 13cm3 of dilute sulphuric acid.find the molarity of the sulphuric acid used

Step 1: Write the balanced chemical equation.
The reaction between sodium carbonate ($\text{Na}_2\text{CO}_3$) and sulphuric acid ($\text{H}_2\text{SO}_4$) is:

$$\text{Na}_2\text{CO}_3 (aq) + \text{H}_2\text{SO}_4 (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g)$$

From the balanced equation, the mole ratio of $\text{Na}_2\text{CO}_3$ to $\text{H}_2\text{SO}_4$ is $1:1$.

Step 2: Calculate the moles of sodium carbonate.
Given:
Volume of $\text{Na}_2\text{CO}_3 = 20 \text{ cm}^3 = 0.020 \text{ dm}^3$
Molarity of $\text{Na}_2\text{CO}_3 = 0.1 \text{ M}$
Moles of $\text{Na}_2\text{CO}_3 = \text{Molarity} \times \text{Volume}$

$$\text{Moles of } \text{Na}_2\text{CO}_3 = 0.1 \text{ mol/dm}^3 \times 0.020 \text{ dm}^3 = 0.002 \text{ mol}$$

Step 3: Determine the moles of sulphuric acid.
Since the mole ratio of $\text{Na}_2\text{CO}_3$ to $\text{H}_2\text{SO}_4$ is $1:1$, the moles of $\text{H}_2\text{SO}_4$ reacted are equal to the moles of $\text{Na}_2\text{CO}_3$.

$$\text{Moles of } \text{H}_2\text{SO}_4 = 0.002 \text{ mol}$$

Step 4: Calculate the molarity of sulphuric acid.
Given:
Volume of $\text{H}_2\text{SO}_4 = 13 \text{ cm}^3 = 0.013 \text{ dm}^3$
Molarity of $\text{H}_2\text{SO}_4 = \frac{\text{Moles of } \text{H}_2\text{SO}_4}{\text{Volume of } \text{H}_2\text{SO}_4 \text{ (in dm}^3)} $$

$$\text{Molarity of } \text{H}_2\text{SO}_4 = \frac{0.002 \text{ mol}}{0.013 \text{ dm}^3} \approx 0.1538 \text{ M}$$

Rounding to three significant figures:

$$\text{Molarity of } \text{H}_2\text{SO}_4 \approx 0.154 \text{ M}$$

The molarity of the sulphuric acid used is $\boxed{\text{0.154 M}}$.