**Problem:**Calculate the vapor pressure caused by the addition of 100.0 g of nonvolatile solid naphthalene (C₁₀H₈, 128.17 g/mol) into 500.0 g of benzene (C₆H₆, 78.11 g/mol). The vapor pressure of pure benzene at this condition is 42.8 torr.**Explanation:**To find the vapor pressure of the solution, we need to use Raoult's Law, which states that the vapor pressure of a solvent above a solution (\(P_{\text{solution}}\)) is equal to the vapor pressure of the pure solvent (\(P_{\text{pure}}\)) multiplied by its mole fraction (\(X_{\text{solvent}}\)) in the solution:\[ P_{\text{solution}} = X_{\text{solvent}} \times P_{\text{pure}} \]**Steps to Solve:**1. Calculate the moles of naphthalene and benzene: - Moles of naphthalene = \(\frac{100.0 \, \text{g}}{128.17 \, \text{g/mol}}\) - Moles of benzene = \(\frac{500.0 \, \text{g}}{78.11 \, \text{g/mol}}\)2. Determine the mole fraction of benzene: \[ X_{\text{benzene}} = \frac{\text{moles of benzene}}{\text{moles of benzene} + \text{moles of naphthalene}} \]3. Calculate the vapor pressure of the solution using Raoult’s Law.This calculation provides the theoretical vapor pressure of benzene when naphthalene is added, utilizing the principle that adding a nonvolatile solute decreases the total vapor pressure of the solvent.

Given that: Mass of solvent (benzene) = 500.0 gMass of nonvolatile solute (naphthalene) = 100.0 g Molar mass of solvent (benzene) = 78.11g/molMolar mass of nonvolatile solute (naphthalene) = 128.17 g/mol Vapor pressure of pure solvent (pure benzene) = 42.8 torr ? Vapor pressure of the solution can be calculated using Raoult’s law for non-volatile solute. p0−psp0=n1n1+n2Here,p0=Vapour pressure of pure solventps=Vapour pressure of the solutionn1=Moles of soluten2=Moles of solventCalculate the number of moles of solute (naphthalene) and solvent (benzene). Moles of solute, n1=Mass of soluteMolar mass of soluteMoles of solute, n1=100.0 g128.17 g/molMoles of solute, n1=0.780 mol Moles of solvent, n2=Mass of solventMolar mass of solventMoles of solvent, n2=500.0 g78.11 g/molMoles of solvent, n2=6.401 molSubstitute n1 = 0.780 mol, n2 = 6.401 mol, p0 = 42.8 torr in the above formula. ⇒p0−psp0=n1n1+n2⇒42.8 torr−ps42.8 torr=0.780 mol0.780 mol+6.401 mol⇒42.8 torr−ps=42.8 torr0.780 mol7.181 mol⇒ps=42.8 torr−42.8 torr0.780 mol7.181 mol⇒ps=38.15 torr Thus, vapor pressure of the solution is 38.15 torr

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2) Calculate the vapor pressure caused by the addition of 100.0 g of nonvolatile solid naphthalene (C10H8, 128.17g/mol) into 500.0 g of benzene (C,H6, 78.11g/mol). The vapor pressure of pure benzene at this condition is 42.8 torr.

2) Calculate the vapor pressure caused by the addition of 100.0 g of nonvolatile solid naphthalene (C10H8, 128.17g/mol) into 500.0 g of benzene (C,H6, 78.11g/mol). The vapor pressure of pure benzene at this condition is 42.8 torr.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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