A solution of acid having 62% by mass of acid has specific gravity of 1.8 g. If the solution is to be diluted to attain a specific gravity of 1.2 then what volume of water should be added to 100 ml of this solution?

Options

A. 200 ml
B. 1000 ml
C. 500 ml
D. 300 ml

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the mass of the acid solution.
Given that the specific gravity of the acid solution is 1.8, we can find the density of the solution:
- Density of solution = Specific gravity × Density of water
- Density of water = 1 g/mL
- Density of solution = 1.8 g/mL

Now, we can calculate the mass of the solution in 100 mL:
- Mass of solution = Density × Volume
- Mass of solution = 1.8 g/mL × 100 mL = 180 g

### Step 2: Calculate the mass of the acid in the solution.
The solution is 62% acid by mass:
- Mass of acid = 62% of mass of solution
- Mass of acid = (62/100) × 180 g = 111.6 g

### Step 3: Determine the mass of the water added.
Let \( x \) be the mass of water added (in grams). The new mass of the solution after adding water will be:
- Mass of new solution = Mass of original solution + Mass of water added
- Mass of new solution = 180 g + \( x \)

### Step 4: Calculate the new volume of the solution.
Since the density of water is 1 g/mL, the volume of water added will also be \( x \) mL. Therefore, the new volume of the solution will be:
- Volume of new solution = Original volume + Volume of water added
- Volume of new solution = 100 mL + \( x \) mL

### Step 5: Set up the equation for the new specific gravity.
We want the new specific gravity to be 1.2:
- Density of new solution = Specific gravity × Density of water
- Density of new solution = 1.2 g/mL

Using the formula for density (mass/volume):
- Density = (Mass of new solution) / (Volume of new solution)
- 1.2 = (180 + \( x \)) / (100 + \( x \))

### Step 6: Solve the equation.
Cross-multiplying gives:
- 1.2(100 + \( x \)) = 180 + \( x \)
- 120 + 1.2\( x \) = 180 + \( x \)

Rearranging the equation:
- 1.2\( x \) - \( x \) = 180 - 120
- 0.2\( x \) = 60
- \( x \) = 60 / 0.2
- \( x \) = 300

### Conclusion:
The volume of water that should be added to 100 mL of the acid solution is **300 mL**.

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