Question

Find both the volume percent of a solution that has 10.0 g of ethanol (D = 0.7893 g/mL) and 90.0 g of water (D = 0.9987 g/mL). Assume volumes are additive.

Answer 1

Answer:

Volume percent of a solution that has 10.0 g of ethanol= 12.37%

Volume percent of a solution that has 90.0 g of water = 87.69%

Explanation:

Volume Percent:

The volume percent helps us  to indicate the concentration of the solution when the volume of the solute and volume of the solution is given by

$Volume Percent=\frac{\text{Volume of the Solute}}{\text{Volume of Solution}}\times100%$

With respect to density 1mL of ethanol

=> 0.7893g or 10g of methanol contains

=> $\frac{1}{0.7893}\times10$

=> 12.67mL

And for water = $\frac{1}{0.9987} \times 90.0$

=> 90.12mL

Now total volume = 90.12+ 12.67 = 102.79mL

%(v/v) for water = $\frac{90.12}{102.79}\times 100$

 = 87.69%

%(v/v) for ethanol = $\frac{12.67}{102.79}\times 100$

 = 12.37%