60 ml of a ''x'' % w/w alcohol by weight (d = 0.6 `g//cm^(3)`) must be used to prepare `200 cm^(3)` of 12% alcohol by weight `(d = 0.90 g//cm^(3))`. Calculate the value of ''x'' ?

Answer

To solve the problem, we need to find the value of "x" in the context of preparing a solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
We have two solutions:
1. **Solution 1**: 60 ml of x% w/w alcohol with a density of 0.6 g/cm³.
2. **Solution 2**: 200 cm³ of 12% w/w alcohol with a density of 0.9 g/cm³.

### Step 2: Calculate the Mass of the First Solution
Using the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the mass of the first solution:
\[ \text{Mass of Solution 1} = \text{Density} \times \text{Volume} = 0.6 \, \text{g/cm}^3 \times 60 \, \text{ml} = 36 \, \text{g} \]

### Step 3: Calculate the Mass of Solute in Solution 1
Since the solution is x% w/w, the mass of the solute (alcohol) in Solution 1 is:
\[ \text{Mass of Solute in Solution 1} = \frac{x}{100} \times \text{Mass of Solution 1} = \frac{x}{100} \times 36 = 0.36x \, \text{g} \]

### Step 4: Calculate the Mass of the Second Solution
Now, for the second solution (200 cm³ of 12% w/w alcohol):
\[ \text{Mass of Solution 2} = \text{Density} \times \text{Volume} = 0.9 \, \text{g/cm}^3 \times 200 \, \text{ml} = 180 \, \text{g} \]

### Step 5: Calculate the Mass of Solute in Solution 2
The mass of the solute (alcohol) in Solution 2 is:
\[ \text{Mass of Solute in Solution 2} = \frac{12}{100} \times \text{Mass of Solution 2} = \frac{12}{100} \times 180 = 21.6 \, \text{g} \]

### Step 6: Set Up the Equation
Since the amount of solute in both solutions must be equal, we can set up the equation:
\[ 0.36x = 21.6 \]

### Step 7: Solve for x
To find x, divide both sides of the equation by 0.36:
\[ x = \frac{21.6}{0.36} = 60 \]

### Conclusion
Thus, the value of x is **60%**.

---