0.1g of a solution containing `Na_(2)CO_(3) and NaHCO_(3)` requires 10mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator, mass% of `Na_(2)CO_(3)` in solution is:

Options

A. 25
B. 32
C. 50
D. none of these

Answer

To find the mass percent of Na₂CO₃ in a solution containing Na₂CO₃ and NaHCO₃, we will follow these steps:

### Step 1: Calculate the moles of HCl used
Given:
- Volume of HCl = 10 mL = 0.01 L (since 1 L = 1000 mL)
- Normality of HCl = 0.01 N

Using the formula for moles:
\[ \text{Moles of HCl} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Moles of HCl} = 0.01 \, \text{N} \times 0.01 \, \text{L} = 0.0001 \, \text{moles} \]

### Step 2: Write the balanced chemical equation
The reaction between HCl and Na₂CO₃ is:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{CO}_3 \]
From the equation, we see that 1 mole of Na₂CO₃ reacts with 2 moles of HCl.

### Step 3: Calculate the moles of Na₂CO₃
From the balanced equation, we can find the moles of Na₂CO₃:
\[ \text{Moles of Na}_2\text{CO}_3 = \frac{1}{2} \times \text{Moles of HCl} \]
\[ \text{Moles of Na}_2\text{CO}_3 = \frac{1}{2} \times 0.0001 = 0.00005 \, \text{moles} \]

### Step 4: Calculate the weight of Na₂CO₃
The molar mass of Na₂CO₃ (sodium carbonate) is approximately 105 g/mol. We can calculate the weight using:
\[ \text{Weight of Na}_2\text{CO}_3 = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Weight of Na}_2\text{CO}_3 = 0.00005 \, \text{moles} \times 105 \, \text{g/mol} = 0.00525 \, \text{grams} \]

### Step 5: Calculate the mass percent of Na₂CO₃
The mass percent of Na₂CO₃ in the solution can be calculated using the formula:
\[ \text{Mass percent} = \left( \frac{\text{Weight of Na}_2\text{CO}_3}{\text{Total weight of solution}} \right) \times 100 \]
Given the total weight of the solution is 0.1 g:
\[ \text{Mass percent} = \left( \frac{0.00525}{0.1} \right) \times 100 = 5.25\% \]

### Final Answer:
The mass percent of Na₂CO₃ in the solution is **5.25%**.

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