A solution is prepared by dissolving 20.2 mL of methanol
(CH3OH) ...

Question

A solution is prepared by dissolving 20.2 mL of methanol
(CH3OH) in 100.0 mL of water at 25 °C. The final volume of the
solution is 118 mL. The densities of methanol and water at this
temperature are 0.782 g>mL and 1.00 g>mL, respectively. For
this solution, calculate the concentration in each unit.
e. mole percent

Accepted Answer

We were asked the mole fraction of a solution Prepared by the 15.6 million of Ethanol and 250 millions of water at room temperature. We're going to assume that the final volume is the sum of each component and the densities of ethanol and water are given below. We call that the equation for more fraction. It's gonna be more fraction equal to the malls of the salute. What about the total moles in the solution? They need to first calculate animals of ethanol and water. 15.6 male leaders. We have the density of Ethanol which is 0.789 g. Are male leader And in one ball at the mall have them all mass. And this is 2 12.011 g Plus six was 1.008g Plus 15.999 g. We get 46.07 g. You get 0.267 moles ethanol An hour. 250 male leaders. We have the density of water which is 1.00 g from the leader. And in one wall of water, one of the Molar Mass. And this is two and 1.008 g Plus 15.999 g. Gonna get 18.02 g. We get 13.9 balls of water. And now we can calculate the mole fraction. It's gonna be a more fraction of ethanol. You can turn malls up at the mall. What about most ethanol? Most of water. We're going to get 0.267 balls five x 0.267 balls plus 13.9 balls. The more fraction of ethanol, I'm gonna get 0.0189. Thanks for watching my video. And I hope it was helpful.

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