# Question 91d1f

Sep 11, 2016

Let the density of water be $1 \frac{g}{m L}$

So the volume of $56 g$ of water =$56 m L = 0.056 L$

Volume of alcohol in the mixture $0.17 L$

So total Volume of the mixture $0.17 L + 0.056 L = 0.226 L$

So the concentration of water in percent by volume =0.056/0.226xx100=24.78%

And the concentration of alcohol in percent by volume (100-24.78)%=75.22%#