50 gm of sample of sodium hydroxide required for complete neutralisation, 1 litre 1 N HCl. What is the percentage purity of `NaOH` is

Options

A. 50
B. 60
C. 70
D. 80

Explanation

To find the percentage purity of sodium hydroxide (NaOH) in the sample, we can follow these steps:

### Step 1: Determine the number of moles of HCl
Given that we have 1 liter of 1 N HCl, we can determine the number of moles of HCl.

- **Normality (N)** is defined as the number of equivalents of solute per liter of solution. For HCl, which is a strong acid, the number of equivalents is equal to the number of moles since it donates one proton (H⁺) per molecule.
- Therefore, 1 N HCl in 1 liter gives us:
\[ \text{Moles of HCl} = 1 \, \text{N} \times 1 \, \text{L} = 1 \, \text{mol} \]

### Step 2: Determine the number of moles of NaOH required for neutralization
In a neutralization reaction between NaOH and HCl, the reaction is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
From the balanced equation, we see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, for 1 mole of HCl, we need 1 mole of NaOH.

### Step 3: Calculate the mass of NaOH required
The molar mass of NaOH is approximately 40 g/mol. Since we need 1 mole of NaOH for complete neutralization:
\[ \text{Mass of NaOH required} = 1 \, \text{mol} \times 40 \, \text{g/mol} = 40 \, \text{g} \]

### Step 4: Calculate the percentage purity of NaOH
We have a sample of 50 g of NaOH, but only 40 g is pure NaOH. The percentage purity can be calculated using the formula:
\[ \text{Percentage purity} = \left( \frac{\text{mass of pure NaOH}}{\text{mass of sample}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage purity} = \left( \frac{40 \, \text{g}}{50 \, \text{g}} \right) \times 100 = 80\% \]

### Final Answer:
The percentage purity of NaOH in the sample is **80%**.

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