Calculating Required Volume of 75% Alcohol by Mass to Prepare a Specific Alcohol Solution

Understanding the principles of density, concentration, and mass balance is crucial in chemistry and pharmaceutical science, particularly when preparing alcohol solutions of specific concentrations. This article will walk you through the steps required to determine the volume of 75% alcohol by mass (65% w/v) needed to prepare a solution of a specific concentration.

Step 1: Understanding the Given Data and Required Calculation

Letrsquo;s consider the scenario where we need to prepare 150 cm3 of a 0.30% (0.3%) alcohol solution, given:

Final solution volume: 150 cm3 Final alcohol solution density: 0.9980 g/cm3 (corrected from 0.9 g/cm3) Initial alcohol solution density: 0.85564 g/cm3 at 20°C (corrected from 0.8 g/cm3) Desired concentration of the final solution: 0.30% (0.3%) by mass

Step 2: Calculating the Mass of the Final Solution

The first step is to determine the mass of the final solution. This can be calculated using the final solution volume and its density.

m_{text{final}}  V_{text{final}} times rho_{text{final}}  150 , text{cm}^3 times 0.9980 , text{g/cm}^3  149.7 , text{g}

Step 3: Determining the Mass of Alcohol in the Final Solution

Next, we calculate the mass of alcohol in the final solution using the desired concentration. The formula for this is:

text{mass of alcohol}  frac{text{Concentration} times text{mass of solution}}{100}

After substituting the given values:

text{mass of alcohol}  frac{0.30 times 149.7}{100}  0.4491 , text{g}

Step 4: Calculating the Mass of 75% Alcohol Needed

Now, we need to find the mass of the 75% alcohol-by-mass solution (65% w/v) required. Since it is 75% alcohol by mass, the mass of alcohol in this solution can be determined using:

0.75 times m_{text{75}}  0.4491 , text{g}

Solving for m_{text{75}}:

m_{text{75}}  frac{0.4491}{0.75}  0.5988 , text{g}

Step 5: Finding the Volume of the 75% Alcohol Solution

To find the volume of 75% alcohol solution, we use the density of the 75% alcohol solution:

V_{text{75}}  frac{m_{text{75}}}{rho_{text{75}}}  frac{0.5988 , text{g}}{0.85564 , text{g/cm}^3}  0.701 , text{cm}^3

Conclusion

The volume of 75% alcohol by mass required to prepare 150 cm3 of a 0.30% alcoholic solution is approximately 0.701 cm3.

Understanding these principles is crucial for ensuring accuracy in chemical and pharmaceutical processes, and this process can be applied to a wide range of similar problems.