Question

250ml of isopropyl alcohol is dissolved in 350  ml of water. Calculate the percentage of alcohol in the solution. Is its strength enough to be an effective hand sanitiser for protection against COVID-19? How much more alcohol(ml) is needed to make the solution an effective protective agent.​

Answer 1

alchol upto 60percent is enough one

Answer 2

Answer:

Explanation:

Volume of solute ( isopropyl alcohol) = 250mL

Volume of solution = 250 + 350 = 600mL

Percentage of alcohol in solution $=\frac{volume of alcohol}{volume of solution} \times 100$

                                                        $= \frac{250}{600}\times 100$

                                                        $= 41.66%$ %

For protection against COVID-19 , strength of alcohol must be atleast 60%

Therefore to increase the strength of this solution , $\frac{x}{x+350} = \frac{60}{100}$ , where $x$ is the new volume of alcohol that should be added.

$40x = 60\times 350 \\ x = 15\times 35 = 525mL$

Therefore volume of alcohol that needs to be added to the original volume;

$=525-250 = 275mL$

275mL of isopropyl alcohol should be added to the solution to make it an effective protective agent.