6 liters of a chemical mixture containing 24% solvent is combined with 10 liters of another mixture containing 42% solvent. What will be the concentration percentage of solvent in the final mixture?

The correct option is 35.25%.

Step-by-Step Explanation:

Step 1: Calculate the amount of solvent in the first mixture.

The first mixture has a volume of 6 liters and contains 24% solvent.

$\text{Amount of solvent in 1st mixture}=6\times \frac{24}{100}=1.44\text{liters}$

Step 2: Calculate the amount of solvent in the second mixture.

The second mixture has a volume of 10 liters and contains 42% solvent.

$\text{Amount of solvent in 2nd mixture}=10\times \frac{42}{100}=4.2\text{liters}$

Step 3: Calculate the total volume of the combined mixture and total amount of solvent.

$\text{Total Volume}=6+10=16\text{liters}$

$\text{Total Solvent}=1.44+4.2=5.64\text{liters}$

Step 4: Calculate the concentration percentage of solvent in the final mixture.

$\text{Concentration Percentage}=\left(\frac{\text{Total Solvent}}{\text{Total Volume}}\right)\times 100$

$\text{Concentration Percentage}=\frac{5.64}{16}\times 100=0.3525\times 100=35.25\%$

Thus, the concentration percentage of solvent in the final mixture is 35.25%.