Ethanol (C2H5OH, M = 46) and methanol (CH3OH, M = 32) form an ideal solution when mixed. What is the vapor pressure of a solution prepared by mixing equal masses of ethanol and methanol? (The vapor pressures of ethanol and methanol are 44.5 mm Hg and 88.7 mm Hg, respectively.) (A) 133 mm Hg (C) 66.6 mm Hg (B) 70.6 mm Hg (D) 44.5 mm Hg help pleas

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Ethanol (C2H5OH, M = 46) and methanol (CH3OH, M = 32) form an ideal solution when mixed. What is the vapor pressure of a solution prepared by mixing equal masses of ethanol and methanol? (The vapor pressures of ethanol and methanol are 44.5 mm Hg and 88.7 mm Hg, respectively.) (A) 133 mm Hg (C) 66.6 mm Hg (B) 70.6 mm Hg (D) 44.5 mm Hg help pleas

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When you mix two volatile compounds as you have here, the vapor pressure of the solution will be somewhere in between the vapor pressures of the two pure compounds. This means already you can eliminate options A and D. To calculate the final answer, use Raoult's Law, which states that the vapor pressure over the solution is equal to the vapor pressure of each pure compound multiplied by its mole fraction. We have the vapor pressures given in the question, but need to find the mole fraction. Luckily we're told the two compounds are mixed in equal masses. You can choose any mass you want, but I'll just use pretend we have 50 grams of each. moles of ethanol = 50/46 = 1.087 moles of methanol = 50/32 = 1.562 mole FRACTION of ethanol = 1.087/(1.087+1.562) = 0.41 mole FRACTION of methanol = 1.562/(1.087+1.562) = 0.59 Now we can use Raoult's Law: P = 0.41(44.5) + 0.59(88.7) = 70.6 mmHg Answer is B. Let me know if you have any questions!
Thank you so much, yes it is the right answere but I wanted to understand more about it * * It is clear now^u^

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