Equivalent weight of a metal chloride is 75.5. How many moles of NaOH is required to completely precipitate one mole of metal hydroxide. Atomic weight of the metal is 120.

Answer

To solve the problem, we need to determine how many moles of NaOH are required to completely precipitate one mole of metal hydroxide from the metal chloride, given the equivalent weight and atomic weight of the metal.

### Step-by-Step Solution:

1. **Understanding Equivalent Weight**:
The equivalent weight of a compound is defined as the mass of the compound that combines with or displaces 1 mole of hydrogen or 1/2 mole of oxygen or 1/3 mole of nitrogen. In this case, we know the equivalent weight of the metal chloride is 75.5 g.

2. **Setting Up the Formula**:
Let's denote the metal as "M" and its chloride as MCl_n, where "n" is the number of chlorine atoms. Since chlorine is monovalent, the valency of the metal (M) will be +n.

3. **Finding Molar Mass**:
The molar mass of the metal chloride can be expressed as:
\[ \text{Molar Mass} = \text{Atomic Weight of M} + n \times \text{Atomic Weight of Cl} \]
Given that the atomic weight of the metal (M) is 120 g and the atomic weight of chlorine (Cl) is approximately 35.5 g, we can write:
\[ \text{Molar Mass} = 120 + 35.5n \]

4. **Using the Equivalent Weight Formula**:
The equivalent weight is also given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
Substituting the known values:
\[ 75.5 = \frac{120 + 35.5n}{n} \]

5. **Solving for n**:
Rearranging the equation:
\[ 75.5n = 120 + 35.5n \]
Simplifying gives:
\[ 75.5n - 35.5n = 120 \]
\[ 40n = 120 \]
\[ n = 3 \]

6. **Identifying the Metal Chloride Formula**:
Now that we have found n = 3, the formula for the metal chloride is MCl₃.

7. **Determining the Reaction with NaOH**:
When MCl₃ reacts with NaOH, it will produce the metal hydroxide (M(OH)₃) and sodium chloride (NaCl). The balanced reaction can be written as:
\[ MCl_3 + 3NaOH \rightarrow M(OH)_3 + 3NaCl \]
From this reaction, we can see that 1 mole of MCl₃ reacts with 3 moles of NaOH to produce 1 mole of M(OH)₃.

8. **Conclusion**:
Therefore, to completely precipitate one mole of metal hydroxide, 3 moles of NaOH are required.

### Final Answer:
**3 moles of NaOH are required to completely precipitate one mole of metal hydroxide.**