For this problem, they are asking for the vapor pressure of the kool aid. Can so

Question

For this problem, they are asking for the vapor pressure of the kool aid. Can someone please explain for the mole fraction part, why does then moles of water (110.9 moles) go on the numerator instead of the moles of kool aid (1.5 moles)?
Where C= 0.00 105M, PN-1 atm, thus giving u(-0.00105 Mla https://chem.libretexts.org/Core/Physical and Theoretical Chemistry/Physical_Propertie s_of Matter/Solutions and_Mixtures/Ideal Solutions/Dissolving Gases In_Liquids9620 Henry's Lavw solule Problem #4:1.5 moles of cherry Kool-Aid are added to a pit hercontaining water on a nice day at 25° C. The vapor pressure of water alone is 23.8 mm Hg at 25 What is the new vapor pressure of Kool-Aid? soi vent 2 liters of C Solution: PH20 23.8 mm Hg To solve for the mole fraction, you must first convert the 2 L of water into moles: 1L=1000 mL = 1000 g Knowing this, you can convert the mass of water (2000 g) into moles: 2000 g /18.02 g (molar mass of water) f 110.9 moles H2O dots tg olve for the mole fraction, xH20 Ive for the mole fraction, XH20 xH20- moles H,0/total moles mol Bract.on Finally, apply Raoult's Law PKool-Aid-xH20PH20-(0.979)(23.8 mm Hg)23.3 mm Hg 1 10.9 moles / 1 10.9 + 1.5 moles ( 0.979

Explanation / Answer

By raoult's law partial pressure , partial vapor pressure of each component of liquid is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture at equilibrium.

Pa= xa P*a , Here only liquid is water so the pressure will be due to water only that is supressed by non volatile solute (cherry).

so mole fraction of water =moles of water/ moles of water + moles of cherry=110.9/110.9 + 1.5

general formula for mole fraction of a = moles of a divided by total number of moles present in the solution.

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