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

Options

A. 67.5 mL
B. 56.25 mL
C. 44.44 mL
D. None of these

Answer

To solve the problem of how much volume of 75% alcohol by weight is needed to prepare 150 cm³ of 30% alcohol by mass, we can follow these steps:

### Step 1: Understand the Given Information
- We have a solution of 75% alcohol by weight with a density of 0.80 g/cm³.
- We want to prepare 150 cm³ of a solution that is 30% alcohol by mass, with a density of 0.90 g/cm³.

### Step 2: Calculate the Mass of the Final Solution
To find the mass of the 30% alcohol solution, we can use the formula:
\[ \text{Mass} = \text{Volume} \times \text{Density} \]
\[ \text{Mass of 30% solution} = 150 \, \text{cm}^3 \times 0.90 \, \text{g/cm}^3 = 135 \, \text{g} \]

### Step 3: Calculate the Mass of Alcohol in the Final Solution
Since the solution is 30% alcohol by mass, we can find the mass of the alcohol:
\[ \text{Mass of alcohol} = 30\% \times \text{Mass of solution} \]
\[ \text{Mass of alcohol} = 0.30 \times 135 \, \text{g} = 40.5 \, \text{g} \]

### Step 4: Calculate the Mass of the 75% Alcohol Solution Needed
In the 75% alcohol solution, 75 g of alcohol is present in 100 g of the solution. We can set up a proportion to find the mass of the 75% solution needed to get 40.5 g of alcohol:
Let \( x \) be the mass of the 75% alcohol solution needed:
\[ 0.75x = 40.5 \]
\[ x = \frac{40.5}{0.75} = 54 \, \text{g} \]

### Step 5: Calculate the Volume of the 75% Alcohol Solution
Now we need to find the volume of the 75% alcohol solution using its density:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
\[ \text{Volume of 75% solution} = \frac{54 \, \text{g}}{0.80 \, \text{g/cm}^3} = 67.5 \, \text{cm}^3 \]

### Final Answer
The volume of 75% alcohol by weight required to prepare 150 cm³ of 30% alcohol by mass is **67.5 cm³**.

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