0.4 g mixture of `NaOH, Na_2CO_3`, and some inert N impurities was first titrated with `N/(10)`HCl using phenolphthalein as an indicator, 17.5 ml. of HCI was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of `Na_2CO_3`, in the mixture is (Rounded-off to the nearest integer)

Answer

To solve the problem step-by-step, we will first analyze the titration reactions and then calculate the weight percentage of Na2CO3 in the mixture.

### Step 1: Understand the Reactions
1. **First Titration with Phenolphthalein:**
- The reaction involves NaOH and Na2CO3 reacting with HCl.
- The balanced equations are:
- For NaOH:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- For Na2CO3:
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
- Let \( x \) be the moles of NaOH and \( y \) be the moles of Na2CO3 in the mixture.

### Step 2: Set Up the First Equation
2. **Calculate Moles from First Titration:**
- The total volume of HCl used is 17.5 mL at a normality of \( \frac{1}{10} \) N.
- Moles of HCl used:
\[ \text{Moles of HCl} = \text{Normality} \times \text{Volume (L)} = \frac{1}{10} \times 0.0175 = 0.00175 \text{ moles} \]
- The moles of HCl reacting with NaOH and Na2CO3 can be expressed as:
\[ x + 2y = 0.00175 \quad \text{(Equation 1)} \]

### Step 3: Understand the Second Titration
3. **Second Titration with Methyl Orange:**
- After the first titration, Na2CO3 is converted to NaHCO3, which reacts with HCl:
\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
- Let \( y \) be the moles of Na2CO3, thus moles of NaHCO3 formed will also be \( y \).
- The volume of HCl used in this titration is 1.5 mL:
\[ \text{Moles of HCl} = \frac{1}{10} \times 0.0015 = 0.00015 \text{ moles} \]
- The moles of HCl reacting with NaHCO3 is:
\[ y = 0.00015 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
4. **Substituting Equation 2 into Equation 1:**
- From Equation 2, we have \( y = 0.00015 \).
- Substitute \( y \) into Equation 1:
\[ x + 2(0.00015) = 0.00175 \]
\[ x + 0.0003 = 0.00175 \]
\[ x = 0.00175 - 0.0003 = 0.00145 \]

### Step 5: Calculate the Mass of Na2CO3
5. **Calculate the mass of Na2CO3:**
- Moles of Na2CO3 \( y = 0.00015 \).
- Molar mass of Na2CO3 = 106 g/mol.
- Mass of Na2CO3:
\[ \text{Mass} = \text{moles} \times \text{molar mass} = 0.00015 \times 106 = 0.0159 \text{ g} \]

### Step 6: Calculate the Weight Percentage of Na2CO3
6. **Calculate the weight percentage:**
- Total mass of the mixture = 0.4 g.
- Weight percentage of Na2CO3:
\[ \text{Weight \%} = \left( \frac{\text{mass of Na2CO3}}{\text{total mass}} \right) \times 100 = \left( \frac{0.0159}{0.4} \right) \times 100 \approx 3.975 \% \]
- Rounding off to the nearest integer gives:
\[ \text{Weight \% of Na2CO3} \approx 4\% \]

### Final Answer
The weight percentage of Na2CO3 in the mixture is **4%**.