To find the normality of 10% (weight/volume) acetic acid, we can follow these steps:
### Step 1: Understand the given concentration
A 10% (weight/volume) solution means that there are 10 grams of acetic acid in 100 mL of solution.
### Step 2: Convert the volume to liters
Since normality is expressed in terms of liters, we need to convert 100 mL to liters:
\[
100 \, \text{mL} = 0.1 \, \text{L}
\]
### Step 3: Calculate the number of grams in 1 liter
To find out how many grams of acetic acid are present in 1 liter (1000 mL) of solution, we can set up a proportion:
\[
\text{Grams in 1000 mL} = \frac{10 \, \text{grams}}{100 \, \text{mL}} \times 1000 \, \text{mL} = 100 \, \text{grams}
\]
### Step 4: Calculate the molar mass of acetic acid
The molecular formula for acetic acid is \( \text{CH}_3\text{COOH} \). The molar mass can be calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
Adding these together:
\[
\text{Molar mass of acetic acid} = 24 + 4 + 32 = 60 \, \text{g/mol}
\]
### Step 5: Calculate the number of moles of acetic acid in 100 grams
Using the formula for moles:
\[
\text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
Substituting the values:
\[
\text{Number of moles} = \frac{100 \, \text{g}}{60 \, \text{g/mol}} = 1.67 \, \text{moles}
\]
### Step 6: Calculate the molarity
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[
\text{Molarity} = \frac{\text{Number of moles}}{\text{Volume of solution in liters}} = \frac{1.67 \, \text{moles}}{1 \, \text{L}} = 1.67 \, \text{M}
\]
### Step 7: Determine the normality
For acetic acid, which is a weak acid, the n-factor (number of H+ ions it can donate) is 1. Therefore, normality (N) is equal to molarity (M):
\[
\text{Normality} = \text{Molarity} \times \text{n-factor} = 1.67 \, \text{M} \times 1 = 1.67 \, \text{N}
\]
### Final Answer
The normality of 10% (weight/volume) acetic acid is approximately **1.67 N**.
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