What volume of 75% alcohol by weight `(d-0.80g//cm^(3))` must be used to prepare 150 `cm^(3)` of 30 % alcohal by mass `(d=0.90g//cm^(3))` ?

Options

A. (a) 67.5 mL
B. (b) 56.25 mL
C. (c) 44.44 mL
D. (d) None of these

Answer

To solve the problem, we need to determine the volume of 75% alcohol by weight required to prepare 150 cm³ of 30% alcohol by mass. We will use the relationship between mass, volume, and density.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Density of 75% alcohol (d₁) = 0.80 g/cm³
- Density of 30% alcohol (d₂) = 0.90 g/cm³
- Volume of 30% alcohol (V₂) = 150 cm³
- Percentage of alcohol by weight in the first case (C₁) = 75%
- Percentage of alcohol by weight in the second case (C₂) = 30%

2. **Calculate the Mass of 30% Alcohol:**
\[ \text{Mass of 30% alcohol} = \text{Density} \times \text{Volume} = d₂ \times V₂ \]
\[ \text{Mass of 30% alcohol} = 0.90 \, \text{g/cm³} \times 150 \, \text{cm³} = 135 \, \text{g} \]

3. **Calculate the Mass of Alcohol in 30% Solution:**
\[ \text{Mass of alcohol in 30% solution} = C₂ \times \text{Mass of 30% alcohol} \]
\[ \text{Mass of alcohol in 30% solution} = 0.30 \times 135 \, \text{g} = 40.5 \, \text{g} \]

4. **Set Up Equation for 75% Alcohol:**
Let \( V_1 \) be the volume of 75% alcohol required. The mass of 75% alcohol can be calculated as:
\[ \text{Mass of 75% alcohol} = C₁ \times (\text{Density} \times V_1) = 0.75 \times (0.80 \, \text{g/cm³} \times V_1) \]
\[ \text{Mass of 75% alcohol} = 0.75 \times 0.80 \times V_1 = 0.6 \times V_1 \]

5. **Equate the Masses:**
Since the mass of alcohol from both solutions must be equal:
\[ 0.6 \times V_1 = 40.5 \, \text{g} \]

6. **Solve for \( V_1 \):**
\[ V_1 = \frac{40.5 \, \text{g}}{0.6} = 67.5 \, \text{cm³} \]

### Final Answer:
The volume of 75% alcohol required is **67.5 cm³**.